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pretraining/mathematica/linear_algebra/null_space_w_steps/1013.txt

@@ -0,0 +1,270 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 4 & 10 & -8 \\
+ 10 & -2 & 10 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 4 & 10 & -8 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{5}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 0 & \frac{54}{5} & -12 \\
+ 4 & 10 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{5}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 0 & \frac{54}{5} & -12 \\
+ 0 & \frac{54}{5} & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }2 \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 0 & \frac{54}{5} & -12 \\
+ 0 & 0 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }3 \text{by }10: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 0 & \frac{54}{5} & -12 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }12\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 10 \\
+ 0 & \frac{54}{5} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 0 \\
+ 0 & \frac{54}{5} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{5}{54}: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -2 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }2\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 10 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }10: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1045.txt

@@ -0,0 +1,414 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 6 & 0 & 6 \\
+ 10 & 4 & 2 & -10 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 9 & 6 & 0 & 6 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{9}{10}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & \frac{12}{5} & -\frac{9}{5} & 15 \\
+ 10 & 1 & -3 & 8 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }1 \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & \frac{12}{5} & -\frac{9}{5} & 15 \\
+ 0 & -3 & -5 & 18 \\
+ -2 & 2 & -3 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{5}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & \frac{12}{5} & -\frac{9}{5} & 15 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & \frac{14}{5} & -\frac{13}{5} & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & \frac{12}{5} & -\frac{9}{5} & 15 \\
+ 0 & \frac{14}{5} & -\frac{13}{5} & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{5}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{29}{5} & \frac{147}{5} \\
+ 0 & \frac{14}{5} & -\frac{13}{5} & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{14}{15}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{29}{5} & \frac{147}{5} \\
+ 0 & 0 & -\frac{109}{15} & \frac{119}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{109}{15} & \frac{119}{5} \\
+ 0 & 0 & -\frac{29}{5} & \frac{147}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{87}{109}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{109}{15} & \frac{119}{5} \\
+ 0 & 0 & 0 & \frac{1134}{109} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{109}{1134}: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{109}{15} & \frac{119}{5} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{119}{5}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 18 \\
+ 0 & 0 & -\frac{109}{15} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }18\, \times \, \text{(row }4) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & -10 \\
+ 0 & -3 & -5 & 0 \\
+ 0 & 0 & -\frac{109}{15} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }4) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & 0 \\
+ 0 & -3 & -5 & 0 \\
+ 0 & 0 & -\frac{109}{15} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{15}{109}: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & 0 \\
+ 0 & -3 & -5 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 2 & 0 \\
+ 0 & -3 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 0 & 0 \\
+ 0 & -3 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-3: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 4 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }10: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1097.txt

@@ -0,0 +1,354 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 4 & -5 & 5 & -6 \\
+ 8 & -8 & -5 & 0 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 4 & -5 & 5 & -6 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{1}{2}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 0 & -1 & \frac{15}{2} & -6 \\
+ -5 & -9 & -7 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{8}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 0 & -1 & \frac{15}{2} & -6 \\
+ 0 & -14 & -\frac{81}{8} & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }14\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 0 & -1 & \frac{15}{2} & -6 \\
+ 0 & 0 & -\frac{921}{8} & 84 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{8}{921}: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 0 & -1 & \frac{15}{2} & -6 \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{15}{2}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & -5 & 0 \\
+ 0 & -1 & 0 & -\frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & 0 & -\frac{1120}{307} \\
+ 0 & -1 & 0 & -\frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -8 & 0 & -\frac{1120}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }8\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 0 & 0 & \frac{176}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }8: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{22}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{22}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }4 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{22}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{22}{307} \\
+ 0 & 1 & 0 & \frac{162}{307} \\
+ 0 & 0 & 1 & -\frac{224}{307} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{22 x_4}{307} \\
+ x_2+\frac{162 x_4}{307} \\
+ x_3-\frac{224 x_4}{307} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{22 x_4}{307}=0 \\
+ x_2+\frac{162 x_4}{307}=0 \\
+ x_3-\frac{224 x_4}{307}=0 \\
+\end{array}
+  \text{for }x_1,x_2 \text{and }x_3: \\
+ \{
+\begin{array}{l}
+ x_1=-\frac{22 x_4}{307} \\
+ x_2=-\frac{162 x_4}{307} \\
+ x_3=\frac{224 x_4}{307} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_4, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{22 x_4}{307} \\
+ -\frac{162 x_4}{307} \\
+ \frac{224 x_4}{307} \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{22 x}{307} \\
+ -\frac{162 x}{307} \\
+ \frac{224 x}{307} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }307 x: \\
+ \left(
+\begin{array}{c}
+ -\frac{22 x}{307} \\
+ -\frac{162 x}{307} \\
+ \frac{224 x}{307} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ -\frac{22}{307} (307 x) \\
+ -\frac{162}{307} (307 x) \\
+ \frac{224 (307 x)}{307} \\
+ 307 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -22 x \\
+ -162 x \\
+ 224 x \\
+ 307 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -22 x \\
+ -162 x \\
+ 224 x \\
+ 307 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-22 x,-162 x,224 x,307 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1215.txt

@@ -0,0 +1,177 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 5 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{6}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 0 & -\frac{85}{6} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{6}{85}: \\
+ \left(
+\begin{array}{cc}
+ 6 & 5 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ 6 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }6: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1250.txt

@@ -0,0 +1,342 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ -7 & 0 & -9 \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{7}{9}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & \frac{28}{9} & -\frac{11}{9} \\
+ -9 & 4 & -4 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & \frac{28}{9} & -\frac{11}{9} \\
+ 0 & 8 & 6 \\
+ 2 & 4 & -9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{9}\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & \frac{28}{9} & -\frac{11}{9} \\
+ 0 & 8 & 6 \\
+ 0 & \frac{28}{9} & -\frac{101}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & \frac{28}{9} & -\frac{11}{9} \\
+ 0 & \frac{28}{9} & -\frac{101}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{18}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & 0 & -\frac{32}{9} \\
+ 0 & \frac{28}{9} & -\frac{101}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{18}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & 0 & -\frac{32}{9} \\
+ 0 & 0 & -\frac{122}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & 0 & -\frac{122}{9} \\
+ 0 & 0 & -\frac{32}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{16}{61}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & 0 & -\frac{122}{9} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{9}{122}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 6 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 10 \\
+ 0 & 8 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 0 \\
+ 0 & 8 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }8: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 4 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1264.txt

@@ -0,0 +1,218 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ -7 & 3 \\
+ -4 & -5 \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{4}{7}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -7 & 3 \\
+ 0 & -\frac{47}{7} \\
+ 0 & 0 \\
+ 0 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{42}{47}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -7 & 3 \\
+ 0 & -\frac{47}{7} \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{7}{47}: \\
+ \left(
+\begin{array}{cc}
+ -7 & 3 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }3\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -7 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-7: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/141.txt

@@ -0,0 +1,275 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 3 & 4 & 2 & 3 \\
+ -7 & 5 & -9 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 5 & -9 & 5 \\
+ 3 & 4 & 2 & 3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{7}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 5 & -9 & 5 \\
+ 0 & \frac{43}{7} & -\frac{13}{7} & \frac{36}{7} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{7}{43}: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 5 & -9 & 5 \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 0 & -\frac{322}{43} & \frac{35}{43} \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-7: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{46}{43} & -\frac{5}{43} \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{46}{43} & -\frac{5}{43} \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{46}{43} & -\frac{5}{43} \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{46}{43} & -\frac{5}{43} \\
+ 0 & 1 & -\frac{13}{43} & \frac{36}{43} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{46 x_3}{43}-\frac{5 x_4}{43} \\
+ x_2-\frac{13 x_3}{43}+\frac{36 x_4}{43} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{46 x_3}{43}-\frac{5 x_4}{43}=0 \\
+ x_2-\frac{13 x_3}{43}+\frac{36 x_4}{43}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{5 x_4}{43}-\frac{46 x_3}{43} \\
+ x_2=\frac{13 x_3}{43}-\frac{36 x_4}{43} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{5 x_4}{43}-\frac{46 x_3}{43} \\
+ \frac{13 x_3}{43}-\frac{36 x_4}{43} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{5 y}{43}-\frac{46 x}{43} \\
+ -\frac{36 y}{43}+\frac{13 x}{43} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }43 x \text{and }43 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{5 y}{43}-\frac{46 x}{43} \\
+ -\frac{36 y}{43}+\frac{13 x}{43} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{5 (43 y)}{43}-\frac{46 (43 x)}{43} \\
+ -\frac{36}{43} (43 y)+\frac{13 (43 x)}{43} \\
+ 43 x \\
+ 43 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 5 y-46 x \\
+ -36 y+13 x \\
+ 43 x \\
+ 43 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 5 y-46 x \\
+ -36 y+13 x \\
+ 43 x \\
+ 43 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (5 y-46 x,-36 y+13 x,43 x,43 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1434.txt

@@ -0,0 +1,329 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ -7 & -6 & -9 \\
+ 6 & 8 & -4 \\
+ -10 & 2 & 2 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 6 & 8 & -4 \\
+ -7 & -6 & -9 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{5}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ -7 & -6 & -9 \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{10}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & -\frac{37}{5} & -\frac{52}{5} \\
+ 8 & -1 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{5}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & -\frac{37}{5} & -\frac{52}{5} \\
+ 0 & \frac{3}{5} & -\frac{2}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{37}{46}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & 0 & -\frac{291}{23} \\
+ 0 & \frac{3}{5} & -\frac{2}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{3}{46}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & 0 & -\frac{291}{23} \\
+ 0 & 0 & -\frac{5}{23} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{291}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & 0 & -\frac{291}{23} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{23}{291}: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & -\frac{14}{5} \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{14}{5}\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 2 \\
+ 0 & \frac{46}{5} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 0 \\
+ 0 & \frac{46}{5} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{5}{46}: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 2 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-10: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1507.txt

@@ -0,0 +1,258 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 6 & 2 & 7 \\
+ -7 & 1 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{6}{7}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 0 & \frac{74}{7} & \frac{67}{7} \\
+ -7 & 1 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }1 \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 0 & \frac{74}{7} & \frac{67}{7} \\
+ 0 & -9 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{63}{74}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 0 & \frac{74}{7} & \frac{67}{7} \\
+ 0 & 0 & \frac{1121}{74} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{74}{1121}: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 0 & \frac{74}{7} & \frac{67}{7} \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{67}{7}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 3 \\
+ 0 & \frac{74}{7} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }3\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 0 \\
+ 0 & \frac{74}{7} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{7}{74}: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 10 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -7 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-7: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1627.txt

@@ -0,0 +1,354 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 4 & 4 & -5 & 0 \\
+ 9 & -6 & -4 & -1 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 4 & 4 & -5 & 0 \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{4}{9}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 0 & \frac{20}{3} & -\frac{29}{9} & \frac{4}{9} \\
+ -5 & 8 & -10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{9}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 0 & \frac{20}{3} & -\frac{29}{9} & \frac{4}{9} \\
+ 0 & \frac{14}{3} & -\frac{110}{9} & -\frac{95}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{10}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 0 & \frac{20}{3} & -\frac{29}{9} & \frac{4}{9} \\
+ 0 & 0 & -\frac{299}{30} & -\frac{163}{15} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{30}{299}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 0 & \frac{20}{3} & -\frac{29}{9} & \frac{4}{9} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{29}{9}\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & -4 & -1 \\
+ 0 & \frac{20}{3} & 0 & \frac{3550}{897} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & 0 & \frac{1005}{299} \\
+ 0 & \frac{20}{3} & 0 & \frac{3550}{897} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{3}{20}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -6 & 0 & \frac{1005}{299} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 0 & 0 & \frac{90}{13} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{10}{13} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{10}{13} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }4 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{10}{13} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{10}{13} \\
+ 0 & 1 & 0 & \frac{355}{598} \\
+ 0 & 0 & 1 & \frac{326}{299} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{10 x_4}{13} \\
+ x_2+\frac{355 x_4}{598} \\
+ x_3+\frac{326 x_4}{299} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{10 x_4}{13}=0 \\
+ x_2+\frac{355 x_4}{598}=0 \\
+ x_3+\frac{326 x_4}{299}=0 \\
+\end{array}
+  \text{for }x_1,x_2 \text{and }x_3: \\
+ \{
+\begin{array}{l}
+ x_1=-\frac{10 x_4}{13} \\
+ x_2=-\frac{355 x_4}{598} \\
+ x_3=-\frac{326 x_4}{299} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_4, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{10 x_4}{13} \\
+ -\frac{355 x_4}{598} \\
+ -\frac{326 x_4}{299} \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{10 x}{13} \\
+ -\frac{355 x}{598} \\
+ -\frac{326 x}{299} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }598 x: \\
+ \left(
+\begin{array}{c}
+ -\frac{10 x}{13} \\
+ -\frac{355 x}{598} \\
+ -\frac{326 x}{299} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ -\frac{10}{13} (598 x) \\
+ -\frac{355}{598} (598 x) \\
+ -\frac{326}{299} (598 x) \\
+ 598 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -460 x \\
+ -355 x \\
+ -652 x \\
+ 598 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -460 x \\
+ -355 x \\
+ -652 x \\
+ 598 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-460 x,-355 x,-652 x,598 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1720.txt

@@ -0,0 +1,283 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 7 & 8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ -1 & -8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 4 & -4 \\
+ -5 & -1 \\
+ 7 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -36 \\
+ -5 & -1 \\
+ 7 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -36 \\
+ 0 & 39 \\
+ 7 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -36 \\
+ 0 & 39 \\
+ 0 & -48 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -48 \\
+ 0 & 39 \\
+ 0 & -36 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{13}{16}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -48 \\
+ 0 & 0 \\
+ 0 & -36 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{3}{4}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & -48 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-48: \\
+ \left(
+\begin{array}{cc}
+ -1 & -8 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }8\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1869.txt

@@ -0,0 +1,258 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ -5 & -9 & 2 \\
+ 1 & -3 & 6 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ -5 & -9 & 2 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ 0 & -24 & 32 \\
+ -6 & -3 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ 0 & -24 & 32 \\
+ 0 & -21 & 40 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{8}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ 0 & -24 & 32 \\
+ 0 & 0 & 12 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }3 \text{by }12: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ 0 & -24 & 32 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }32\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 6 \\
+ 0 & -24 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 0 \\
+ 0 & -24 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-24: \\
+ \left(
+\begin{array}{ccc}
+ 1 & -3 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1910.txt

@@ -0,0 +1,227 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 5 & 4 \\
+ -9 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 0 & 29 \\
+ -9 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 0 & 29 \\
+ 0 & -38 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 0 & -38 \\
+ 0 & 29 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{29}{38}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 0 & -38 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-38: \\
+ \left(
+\begin{array}{cc}
+ -1 & 5 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/1941.txt

@@ -0,0 +1,275 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ -3 & -3 & -10 & 7 \\
+ -9 & -10 & -10 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -9 & -10 & -10 & -6 \\
+ -3 & -3 & -10 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{1}{3}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -9 & -10 & -10 & -6 \\
+ 0 & \frac{1}{3} & -\frac{20}{3} & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }3: \\
+ \left(
+\begin{array}{cccc}
+ -9 & -10 & -10 & -6 \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -9 & 0 & -210 & 264 \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-9: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{70}{3} & -\frac{88}{3} \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{70}{3} & -\frac{88}{3} \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{70}{3} & -\frac{88}{3} \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{70}{3} & -\frac{88}{3} \\
+ 0 & 1 & -20 & 27 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{70 x_3}{3}-\frac{88 x_4}{3} \\
+ x_2-20 x_3+27 x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{70 x_3}{3}-\frac{88 x_4}{3}=0 \\
+ x_2-20 x_3+27 x_4=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{88 x_4}{3}-\frac{70 x_3}{3} \\
+ x_2=20 x_3-27 x_4 \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{88 x_4}{3}-\frac{70 x_3}{3} \\
+ 20 x_3-27 x_4 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{88 y}{3}-\frac{70 x}{3} \\
+ -27 y+20 x \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }3 x \text{and }3 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{88 y}{3}-\frac{70 x}{3} \\
+ -27 y+20 x \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{88 (3 y)}{3}-\frac{70 (3 x)}{3} \\
+ -27 (3 y)+20 (3 x) \\
+ 3 x \\
+ 3 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 88 y-70 x \\
+ -81 y+60 x \\
+ 3 x \\
+ 3 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 88 y-70 x \\
+ -81 y+60 x \\
+ 3 x \\
+ 3 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (88 y-70 x,-81 y+60 x,3 x,3 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2067.txt

@@ -0,0 +1,177 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 8 & 4 \\
+ -5 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{8}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 8 & 4 \\
+ 0 & \frac{7}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{2}{7}: \\
+ \left(
+\begin{array}{cc}
+ 8 & 4 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ 8 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }8: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2168.txt

@@ -0,0 +1,414 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 5 & 1 & -8 & 9 \\
+ 9 & -5 & -3 & 10 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 5 & 1 & -8 & 9 \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{9}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & \frac{34}{9} & -\frac{19}{3} & \frac{31}{9} \\
+ -3 & 5 & 7 & 4 \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{3}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & \frac{34}{9} & -\frac{19}{3} & \frac{31}{9} \\
+ 0 & \frac{10}{3} & 6 & \frac{22}{3} \\
+ -4 & -7 & -3 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{9}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & \frac{34}{9} & -\frac{19}{3} & \frac{31}{9} \\
+ 0 & \frac{10}{3} & 6 & \frac{22}{3} \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & \frac{10}{3} & 6 & \frac{22}{3} \\
+ 0 & \frac{34}{9} & -\frac{19}{3} & \frac{31}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{30}{83}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & \frac{368}{83} & \frac{922}{83} \\
+ 0 & \frac{34}{9} & -\frac{19}{3} & \frac{31}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{34}{83}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & \frac{368}{83} & \frac{922}{83} \\
+ 0 & 0 & -\frac{673}{83} & \frac{641}{83} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & -\frac{673}{83} & \frac{641}{83} \\
+ 0 & 0 & \frac{368}{83} & \frac{922}{83} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{368}{673}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & -\frac{673}{83} & \frac{641}{83} \\
+ 0 & 0 & 0 & \frac{10318}{673} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{673}{10318}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & -\frac{673}{83} & \frac{641}{83} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{641}{83}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & \frac{94}{9} \\
+ 0 & 0 & -\frac{673}{83} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{94}{9}\, \times \, \text{(row }4) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 10 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & 0 \\
+ 0 & 0 & -\frac{673}{83} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }4) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 0 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & 0 \\
+ 0 & 0 & -\frac{673}{83} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{83}{673}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 0 \\
+ 0 & -\frac{83}{9} & -\frac{13}{3} & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{13}{3}\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & -3 & 0 \\
+ 0 & -\frac{83}{9} & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & 0 & 0 \\
+ 0 & -\frac{83}{9} & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{9}{83}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & -5 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2191.txt

@@ -0,0 +1,239 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 3 & 0 \\
+ -8 & 5 \\
+ -7 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 3 & 0 \\
+ -7 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{8}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 0 & \frac{15}{8} \\
+ -7 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{8}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 0 & \frac{15}{8} \\
+ 0 & -\frac{83}{8} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 0 & -\frac{83}{8} \\
+ 0 & \frac{15}{8} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{15}{83}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 0 & -\frac{83}{8} \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{8}{83}: \\
+ \left(
+\begin{array}{cc}
+ -8 & 5 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -8 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-8: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2194.txt

@@ -0,0 +1,330 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ -9 & 9 & -2 & -2 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ 0 & 11 & 0 & -5 \\
+ -8 & -4 & -7 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{8}{9}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ 0 & 11 & 0 & -5 \\
+ 0 & -\frac{20}{9} & -\frac{47}{9} & -\frac{23}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{20}{99}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ 0 & 11 & 0 & -5 \\
+ 0 & 0 & -\frac{47}{9} & -\frac{859}{99} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{9}{47}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 2 & -3 \\
+ 0 & 11 & 0 & -5 \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 0 & -\frac{3269}{517} \\
+ 0 & 11 & 0 & -5 \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }11: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 2 & 0 & -\frac{3269}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 0 & 0 & -\frac{2799}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{311}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{311}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }4 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{311}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{311}{517} \\
+ 0 & 1 & 0 & -\frac{5}{11} \\
+ 0 & 0 & 1 & \frac{859}{517} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{311 x_4}{517} \\
+ x_2-\frac{5 x_4}{11} \\
+ x_3+\frac{859 x_4}{517} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{311 x_4}{517}=0 \\
+ x_2-\frac{5 x_4}{11}=0 \\
+ x_3+\frac{859 x_4}{517}=0 \\
+\end{array}
+  \text{for }x_1,x_2 \text{and }x_3: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{311 x_4}{517} \\
+ x_2=\frac{5 x_4}{11} \\
+ x_3=-\frac{859 x_4}{517} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_4, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{311 x_4}{517} \\
+ \frac{5 x_4}{11} \\
+ -\frac{859 x_4}{517} \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{311 x}{517} \\
+ \frac{5 x}{11} \\
+ -\frac{859 x}{517} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }517 x: \\
+ \left(
+\begin{array}{c}
+ \frac{311 x}{517} \\
+ \frac{5 x}{11} \\
+ -\frac{859 x}{517} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{311 (517 x)}{517} \\
+ \frac{5 (517 x)}{11} \\
+ -\frac{859}{517} (517 x) \\
+ 517 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 311 x \\
+ 235 x \\
+ -859 x \\
+ 517 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 311 x \\
+ 235 x \\
+ -859 x \\
+ 517 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (311 x,235 x,-859 x,517 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2214.txt

@@ -0,0 +1,342 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 7 & 8 & 6 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ -10 & -9 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ -8 & -8 & 10 \\
+ 3 & 6 & -3 \\
+ 7 & 8 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{4}{5}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & -\frac{4}{5} & 18 \\
+ 3 & 6 & -3 \\
+ 7 & 8 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{10}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & -\frac{4}{5} & 18 \\
+ 0 & \frac{33}{10} & -6 \\
+ 7 & 8 & 6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{7}{10}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & -\frac{4}{5} & 18 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & \frac{17}{10} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & -\frac{4}{5} & 18 \\
+ 0 & \frac{17}{10} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{8}{33}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & 0 & \frac{182}{11} \\
+ 0 & \frac{17}{10} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{17}{33}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & 0 & \frac{182}{11} \\
+ 0 & 0 & \frac{23}{11} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{23}{182}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & 0 & \frac{182}{11} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{11}{182}: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & -6 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & -10 \\
+ 0 & \frac{33}{10} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & 0 \\
+ 0 & \frac{33}{10} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{10}{33}: \\
+ \left(
+\begin{array}{ccc}
+ -10 & -9 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }9\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-10: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/240.txt

@@ -0,0 +1,282 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 3 & -10 & 3 \\
+ -10 & 6 & -3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 3 & -10 & 3 \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{10}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -\frac{41}{5} & \frac{21}{10} \\
+ -10 & -10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }1 \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -\frac{41}{5} & \frac{21}{10} \\
+ 0 & -16 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -16 & 1 \\
+ 0 & -\frac{41}{5} & \frac{21}{10} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{41}{80}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -16 & 1 \\
+ 0 & 0 & \frac{127}{80} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{80}{127}: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -16 & 1 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }3 \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & -3 \\
+ 0 & -16 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & 0 \\
+ 0 & -16 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-16: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 6 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -10 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-10: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2470.txt

@@ -0,0 +1,188 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 6 & 2 \\
+ 10 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 10 & 7 \\
+ 6 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{3}{5}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 10 & 7 \\
+ 0 & -\frac{11}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{5}{11}: \\
+ \left(
+\begin{array}{cc}
+ 10 & 7 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ 10 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }10: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2493.txt

@@ -0,0 +1,275 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 8 & -6 & -1 & -6 \\
+ -1 & -8 & 6 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -8 & 6 & -7 \\
+ 8 & -6 & -1 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }8\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -8 & 6 & -7 \\
+ 0 & -70 & 47 & -62 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-70: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -8 & 6 & -7 \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }8\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & \frac{22}{35} & \frac{3}{35} \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{22}{35} & -\frac{3}{35} \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{22}{35} & -\frac{3}{35} \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{22}{35} & -\frac{3}{35} \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{22}{35} & -\frac{3}{35} \\
+ 0 & 1 & -\frac{47}{70} & \frac{31}{35} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{22 x_3}{35}-\frac{3 x_4}{35} \\
+ x_2-\frac{47 x_3}{70}+\frac{31 x_4}{35} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{22 x_3}{35}-\frac{3 x_4}{35}=0 \\
+ x_2-\frac{47 x_3}{70}+\frac{31 x_4}{35}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{22 x_3}{35}+\frac{3 x_4}{35} \\
+ x_2=\frac{47 x_3}{70}-\frac{31 x_4}{35} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{22 x_3}{35}+\frac{3 x_4}{35} \\
+ \frac{47 x_3}{70}-\frac{31 x_4}{35} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{3 y}{35}+\frac{22 x}{35} \\
+ -\frac{31 y}{35}+\frac{47 x}{70} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }70 x \text{and }35 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{3 y}{35}+\frac{22 x}{35} \\
+ -\frac{31 y}{35}+\frac{47 x}{70} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{3 (35 y)}{35}+\frac{22 (70 x)}{35} \\
+ -\frac{31}{35} (35 y)+\frac{47 (70 x)}{70} \\
+ 70 x \\
+ 35 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 3 y+44 x \\
+ -31 y+47 x \\
+ 70 x \\
+ 35 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 3 y+44 x \\
+ -31 y+47 x \\
+ 70 x \\
+ 35 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (3 y+44 x,-31 y+47 x,70 x,35 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2523.txt

@@ -0,0 +1,275 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 5 & 9 & -5 & 8 \\
+ -9 & 9 & -1 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -9 & 9 & -1 & 2 \\
+ 5 & 9 & -5 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{9}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -9 & 9 & -1 & 2 \\
+ 0 & 14 & -\frac{50}{9} & \frac{82}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }14: \\
+ \left(
+\begin{array}{cccc}
+ -9 & 9 & -1 & 2 \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -9 & 0 & \frac{18}{7} & -\frac{27}{7} \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-9: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{2}{7} & \frac{3}{7} \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{2}{7} & \frac{3}{7} \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{2}{7} & \frac{3}{7} \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{2}{7} & \frac{3}{7} \\
+ 0 & 1 & -\frac{25}{63} & \frac{41}{63} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{2 x_3}{7}+\frac{3 x_4}{7} \\
+ x_2-\frac{25 x_3}{63}+\frac{41 x_4}{63} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{2 x_3}{7}+\frac{3 x_4}{7}=0 \\
+ x_2-\frac{25 x_3}{63}+\frac{41 x_4}{63}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{2 x_3}{7}-\frac{3 x_4}{7} \\
+ x_2=\frac{25 x_3}{63}-\frac{41 x_4}{63} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{2 x_3}{7}-\frac{3 x_4}{7} \\
+ \frac{25 x_3}{63}-\frac{41 x_4}{63} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{3 y}{7}+\frac{2 x}{7} \\
+ -\frac{41 y}{63}+\frac{25 x}{63} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }63 x \text{and }63 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ -\frac{3 y}{7}+\frac{2 x}{7} \\
+ -\frac{41 y}{63}+\frac{25 x}{63} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ -\frac{3}{7} (63 y)+\frac{2 (63 x)}{7} \\
+ -\frac{41}{63} (63 y)+\frac{25 (63 x)}{63} \\
+ 63 x \\
+ 63 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -27 y+18 x \\
+ -41 y+25 x \\
+ 63 x \\
+ 63 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -27 y+18 x \\
+ -41 y+25 x \\
+ 63 x \\
+ 63 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-27 y+18 x,-41 y+25 x,63 x,63 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2617.txt

@@ -0,0 +1,355 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -5 & -1 \\
+ -7 & -4 & -3 \\
+ 9 & -6 & -4 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ -7 & -4 & -3 \\
+ -8 & -5 & -1 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{7}{9}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{26}{3} & -\frac{55}{9} \\
+ -8 & -5 & -1 \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{8}{9}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{26}{3} & -\frac{55}{9} \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ -2 & -4 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{2}{9}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{26}{3} & -\frac{55}{9} \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & -\frac{16}{3} & \frac{64}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & -\frac{26}{3} & -\frac{55}{9} \\
+ 0 & -\frac{16}{3} & \frac{64}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{26}{31}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & 0 & -\frac{71}{31} \\
+ 0 & -\frac{16}{3} & \frac{64}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{16}{31}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & 0 & -\frac{71}{31} \\
+ 0 & 0 & \frac{880}{93} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & 0 & \frac{880}{93} \\
+ 0 & 0 & -\frac{71}{31} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{213}{880}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & 0 & \frac{880}{93} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{93}{880}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & -\frac{41}{9} \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{41}{9}\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & -4 \\
+ 0 & -\frac{31}{3} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & 0 \\
+ 0 & -\frac{31}{3} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{3}{31}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & -6 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2777.txt

@@ -0,0 +1,177 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 2 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{5}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 0 & \frac{22}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{5}{22}: \\
+ \left(
+\begin{array}{cc}
+ 5 & 9 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ 5 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }5: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2935.txt

@@ -0,0 +1,362 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ -3 & 0 & 4 & 3 \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{5}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & -3 & \frac{32}{5} & \frac{9}{5} \\
+ -5 & -1 & 4 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & -3 & \frac{32}{5} & \frac{9}{5} \\
+ 0 & -6 & 8 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{3}{7}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & 0 & \frac{149}{35} & \frac{108}{35} \\
+ 0 & -6 & 8 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{6}{7}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & 0 & \frac{149}{35} & \frac{108}{35} \\
+ 0 & 0 & \frac{26}{7} & -\frac{3}{7} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{130}{149}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & 0 & \frac{149}{35} & \frac{108}{35} \\
+ 0 & 0 & 0 & -\frac{465}{149} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }-\frac{149}{465}: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & 0 & \frac{149}{35} & \frac{108}{35} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{108}{35}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & -3 \\
+ 0 & 0 & \frac{149}{35} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }4) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & -2 \\
+ 0 & -7 & 5 & 0 \\
+ 0 & 0 & \frac{149}{35} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }2\, \times \, \text{(row }4) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & 0 \\
+ 0 & -7 & 5 & 0 \\
+ 0 & 0 & \frac{149}{35} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{35}{149}: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & 0 \\
+ 0 & -7 & 5 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 4 & 0 \\
+ 0 & -7 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 0 & 0 \\
+ 0 & -7 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-7: \\
+ \left(
+\begin{array}{cccc}
+ 5 & -5 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 5 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }5: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2970.txt

@@ -0,0 +1,177 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 0 & -60 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-60: \\
+ \left(
+\begin{array}{cc}
+ -1 & -5 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }5\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2974.txt

@@ -0,0 +1,264 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 5 & 0 & -6 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{6}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 0 & \frac{10}{3} & -\frac{31}{6} & \frac{16}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{3}{10}: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -4 & -1 & -10 \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 0 & -\frac{36}{5} & -\frac{18}{5} \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }6: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{6}{5} & -\frac{3}{5} \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{6}{5} & -\frac{3}{5} \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{6}{5} & -\frac{3}{5} \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{6}{5} & -\frac{3}{5} \\
+ 0 & 1 & -\frac{31}{20} & \frac{8}{5} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{6 x_3}{5}-\frac{3 x_4}{5} \\
+ x_2-\frac{31 x_3}{20}+\frac{8 x_4}{5} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{6 x_3}{5}-\frac{3 x_4}{5}=0 \\
+ x_2-\frac{31 x_3}{20}+\frac{8 x_4}{5}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{6 x_3}{5}+\frac{3 x_4}{5} \\
+ x_2=\frac{31 x_3}{20}-\frac{8 x_4}{5} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{6 x_3}{5}+\frac{3 x_4}{5} \\
+ \frac{31 x_3}{20}-\frac{8 x_4}{5} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{3 y}{5}+\frac{6 x}{5} \\
+ -\frac{8 y}{5}+\frac{31 x}{20} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }20 x \text{and }5 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{3 y}{5}+\frac{6 x}{5} \\
+ -\frac{8 y}{5}+\frac{31 x}{20} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{3 (5 y)}{5}+\frac{6 (20 x)}{5} \\
+ -\frac{8}{5} (5 y)+\frac{31 (20 x)}{20} \\
+ 20 x \\
+ 5 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 3 y+24 x \\
+ -8 y+31 x \\
+ 20 x \\
+ 5 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 3 y+24 x \\
+ -8 y+31 x \\
+ 20 x \\
+ 5 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (3 y+24 x,-8 y+31 x,20 x,5 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/2994.txt

@@ -0,0 +1,248 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 4 & -5 & 10 & 9 \\
+ 6 & -8 & 0 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -8 & 0 & -6 \\
+ 4 & -5 & 10 & 9 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{3}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -8 & 0 & -6 \\
+ 0 & \frac{1}{3} & 10 & 13 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }3: \\
+ \left(
+\begin{array}{cccc}
+ 6 & -8 & 0 & -6 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }8\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 0 & 240 & 306 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }6: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 40 & 51 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 40 & 51 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 40 & 51 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 40 & 51 \\
+ 0 & 1 & 30 & 39 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+40 x_3+51 x_4 \\
+ x_2+30 x_3+39 x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+40 x_3+51 x_4=0 \\
+ x_2+30 x_3+39 x_4=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=-40 x_3-51 x_4 \\
+ x_2=-30 x_3-39 x_4 \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -40 x_3-51 x_4 \\
+ -30 x_3-39 x_4 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -51 y-40 x \\
+ -39 y-30 x \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -51 y-40 x \\
+ -39 y-30 x \\
+ x \\
+ y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-51 y-40 x,-39 y-30 x,x,y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3027.txt

@@ -0,0 +1,264 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 6 & 6 & -7 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{5}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 0 & \frac{12}{5} & -\frac{41}{5} & -\frac{12}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{5}{12}: \\
+ \left(
+\begin{array}{cccc}
+ -10 & -6 & -2 & 1 \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -10 & 0 & -\frac{45}{2} & -5 \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-10: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{9}{4} & \frac{1}{2} \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{9}{4} & \frac{1}{2} \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{9}{4} & \frac{1}{2} \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{9}{4} & \frac{1}{2} \\
+ 0 & 1 & -\frac{41}{12} & -1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{9 x_3}{4}+\frac{x_4}{2} \\
+ x_2-\frac{41 x_3}{12}-x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{9 x_3}{4}+\frac{x_4}{2}=0 \\
+ x_2-\frac{41 x_3}{12}-x_4=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=-\frac{9 x_3}{4}-\frac{x_4}{2} \\
+ x_2=\frac{41 x_3}{12}+x_4 \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{9 x_3}{4}-\frac{x_4}{2} \\
+ \frac{41 x_3}{12}+x_4 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{y}{2}-\frac{9 x}{4} \\
+ y+\frac{41 x}{12} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }12 x \text{and }2 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ -\frac{y}{2}-\frac{9 x}{4} \\
+ y+\frac{41 x}{12} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ -\frac{1}{2} (2 y)-\frac{9 (12 x)}{4} \\
+ 2 y+\frac{41 (12 x)}{12} \\
+ 12 x \\
+ 2 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -y-27 x \\
+ 2 y+41 x \\
+ 12 x \\
+ 2 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -y-27 x \\
+ 2 y+41 x \\
+ 12 x \\
+ 2 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-y-27 x,2 y+41 x,12 x,2 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3079.txt

@@ -0,0 +1,257 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 4 & -1 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 27 \\
+ -1 & 9 \\
+ 9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 27 \\
+ 0 & 2 \\
+ 9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 27 \\
+ 0 & 2 \\
+ 0 & 73 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 73 \\
+ 0 & 2 \\
+ 0 & 27 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{73}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 73 \\
+ 0 & 0 \\
+ 0 & 27 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{27}{73}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 73 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }73: \\
+ \left(
+\begin{array}{cc}
+ 1 & -7 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3119.txt

@@ -0,0 +1,282 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 4 & -1 & -6 \\
+ 10 & -3 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 4 & -1 & -6 \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{5}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & \frac{1}{5} & -\frac{18}{5} \\
+ 4 & -5 & 3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{5}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & \frac{1}{5} & -\frac{18}{5} \\
+ 0 & -\frac{19}{5} & \frac{27}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & -\frac{19}{5} & \frac{27}{5} \\
+ 0 & \frac{1}{5} & -\frac{18}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{19}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & -\frac{19}{5} & \frac{27}{5} \\
+ 0 & 0 & -\frac{63}{19} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{19}{63}: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & -\frac{19}{5} & \frac{27}{5} \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{27}{5}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & -6 \\
+ 0 & -\frac{19}{5} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & 0 \\
+ 0 & -\frac{19}{5} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{5}{19}: \\
+ \left(
+\begin{array}{ccc}
+ 10 & -3 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 10 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }10: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3166.txt

@@ -0,0 +1,342 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 3 & -9 & 2 \\
+ -8 & -4 & -6 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 3 & -9 & 2 \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{8}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 6 & 9 & 8 \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{4}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 6 & \frac{7}{2} \\
+ 8 & -5 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 6 & \frac{7}{2} \\
+ 0 & -9 & -4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{7}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 0 & \frac{47}{14} \\
+ 0 & -9 & -4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{6}{7}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 0 & \frac{47}{14} \\
+ 0 & 0 & -\frac{53}{14} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 0 & -\frac{53}{14} \\
+ 0 & 0 & \frac{47}{14} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{47}{53}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 0 & -\frac{53}{14} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{14}{53}: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & -\frac{1}{4} \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{4}\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & -6 \\
+ 0 & -\frac{21}{2} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 0 \\
+ 0 & -\frac{21}{2} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{2}{21}: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -8 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-8: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3230.txt

@@ -0,0 +1,275 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 3 & -10 & -6 \\
+ -1 & 4 & 8 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 4 & 8 & 4 \\
+ 6 & 3 & -10 & -6 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }6\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 4 & 8 & 4 \\
+ 0 & 27 & 38 & 18 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }27: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 4 & 8 & 4 \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & \frac{64}{27} & \frac{4}{3} \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{64}{27} & -\frac{4}{3} \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{64}{27} & -\frac{4}{3} \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{64}{27} & -\frac{4}{3} \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{64}{27} & -\frac{4}{3} \\
+ 0 & 1 & \frac{38}{27} & \frac{2}{3} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{64 x_3}{27}-\frac{4 x_4}{3} \\
+ x_2+\frac{38 x_3}{27}+\frac{2 x_4}{3} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{64 x_3}{27}-\frac{4 x_4}{3}=0 \\
+ x_2+\frac{38 x_3}{27}+\frac{2 x_4}{3}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{64 x_3}{27}+\frac{4 x_4}{3} \\
+ x_2=-\frac{38 x_3}{27}-\frac{2 x_4}{3} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{64 x_3}{27}+\frac{4 x_4}{3} \\
+ -\frac{38 x_3}{27}-\frac{2 x_4}{3} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{4 y}{3}+\frac{64 x}{27} \\
+ -\frac{2 y}{3}-\frac{38 x}{27} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }27 x \text{and }3 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{4 y}{3}+\frac{64 x}{27} \\
+ -\frac{2 y}{3}-\frac{38 x}{27} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{4 (3 y)}{3}+\frac{64 (27 x)}{27} \\
+ -\frac{2}{3} (3 y)-\frac{38 (27 x)}{27} \\
+ 27 x \\
+ 3 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 4 y+64 x \\
+ -2 y-38 x \\
+ 27 x \\
+ 3 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 4 y+64 x \\
+ -2 y-38 x \\
+ 27 x \\
+ 3 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (4 y+64 x,-2 y-38 x,27 x,3 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/327.txt

@@ -0,0 +1,354 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 10 & 2 & -4 & -3 \\
+ -1 & -4 & -4 & -1 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 10 & 2 & -4 & -3 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 0 & -38 & -44 & -13 \\
+ 7 & -5 & 0 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 0 & -38 & -44 & -13 \\
+ 0 & -33 & -28 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{33}{38}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 0 & -38 & -44 & -13 \\
+ 0 & 0 & \frac{194}{19} & \frac{353}{38} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{19}{194}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 0 & -38 & -44 & -13 \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }44\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & -4 & -1 \\
+ 0 & -38 & 0 & \frac{2622}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & 0 & \frac{256}{97} \\
+ 0 & -38 & 0 & \frac{2622}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-38: \\
+ \left(
+\begin{array}{cccc}
+ -1 & -4 & 0 & \frac{256}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & 0 & -\frac{20}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{20}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{20}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }4 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{20}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & \frac{20}{97} \\
+ 0 & 1 & 0 & -\frac{69}{97} \\
+ 0 & 0 & 1 & \frac{353}{388} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{20 x_4}{97} \\
+ x_2-\frac{69 x_4}{97} \\
+ x_3+\frac{353 x_4}{388} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{20 x_4}{97}=0 \\
+ x_2-\frac{69 x_4}{97}=0 \\
+ x_3+\frac{353 x_4}{388}=0 \\
+\end{array}
+  \text{for }x_1,x_2 \text{and }x_3: \\
+ \{
+\begin{array}{l}
+ x_1=-\frac{20 x_4}{97} \\
+ x_2=\frac{69 x_4}{97} \\
+ x_3=-\frac{353 x_4}{388} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_4, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{20 x_4}{97} \\
+ \frac{69 x_4}{97} \\
+ -\frac{353 x_4}{388} \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -\frac{20 x}{97} \\
+ \frac{69 x}{97} \\
+ -\frac{353 x}{388} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }388 x: \\
+ \left(
+\begin{array}{c}
+ -\frac{20 x}{97} \\
+ \frac{69 x}{97} \\
+ -\frac{353 x}{388} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ -\frac{20}{97} (388 x) \\
+ \frac{69 (388 x)}{97} \\
+ -\frac{353}{388} (388 x) \\
+ 388 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ -80 x \\
+ 276 x \\
+ -353 x \\
+ 388 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ -80 x \\
+ 276 x \\
+ -353 x \\
+ 388 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (-80 x,276 x,-353 x,388 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3287.txt

@@ -0,0 +1,401 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ -5 & 8 & 3 & 10 \\
+ -7 & -1 & -1 & 10 \\
+ -1 & 10 & 3 & 1 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ -7 & -1 & -1 & 10 \\
+ -5 & 8 & 3 & 10 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ -5 & 8 & 3 & 10 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & -42 & -12 & 5 \\
+ 1 & 1 & 10 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\text{row }1 \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & -42 & -12 & 5 \\
+ 0 & 11 & 13 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{42}{71}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{72}{71} & \frac{229}{71} \\
+ 0 & 11 & 13 & -1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{11}{71}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{72}{71} & \frac{229}{71} \\
+ 0 & 0 & \frac{681}{71} & -\frac{38}{71} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{681}{71} & -\frac{38}{71} \\
+ 0 & 0 & \frac{72}{71} & \frac{229}{71} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{24}{227}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{681}{71} & -\frac{38}{71} \\
+ 0 & 0 & 0 & \frac{745}{227} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{227}{745}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{681}{71} & -\frac{38}{71} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{38}{71}\, \times \, \text{(row }4) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 3 \\
+ 0 & 0 & \frac{681}{71} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }3\, \times \, \text{(row }4) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 1 \\
+ 0 & -71 & -22 & 0 \\
+ 0 & 0 & \frac{681}{71} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }4 \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 0 \\
+ 0 & -71 & -22 & 0 \\
+ 0 & 0 & \frac{681}{71} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{71}{681}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 0 \\
+ 0 & -71 & -22 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }22\, \times \, \text{(row }3) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 3 & 0 \\
+ 0 & -71 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }3\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 0 & 0 \\
+ 0 & -71 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-71: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3297.txt

@@ -0,0 +1,375 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ -2 & -2 & 8 & -3 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ 8 & 6 & -3 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 7 & -1 & 3 & 5 \\
+ 0 & 5 & 4 & 2 \\
+ -2 & -2 & 8 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{7}{8}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 5 & 4 & 2 \\
+ -2 & -2 & 8 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{4}\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 5 & 4 & 2 \\
+ 0 & -\frac{1}{2} & \frac{29}{4} & -\frac{7}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{5}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 0 & \frac{17}{2} & \frac{5}{2} \\
+ 0 & -\frac{1}{2} & \frac{29}{4} & -\frac{7}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{25}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 0 & \frac{17}{2} & \frac{5}{2} \\
+ 0 & 0 & \frac{34}{5} & -\frac{9}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{4}{5}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 0 & \frac{17}{2} & \frac{5}{2} \\
+ 0 & 0 & 0 & -\frac{19}{5} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }-\frac{5}{19}: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 0 & \frac{17}{2} & \frac{5}{2} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{2}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & \frac{5}{8} \\
+ 0 & 0 & \frac{17}{2} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{8}\, \times \, \text{(row }4) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 5 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & 0 \\
+ 0 & 0 & \frac{17}{2} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }4) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 0 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & 0 \\
+ 0 & 0 & \frac{17}{2} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{2}{17}: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 0 \\
+ 0 & -\frac{25}{4} & \frac{45}{8} & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{45}{8}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & -3 & 0 \\
+ 0 & -\frac{25}{4} & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }3\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & 0 & 0 \\
+ 0 & -\frac{25}{4} & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{4}{25}: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 6 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 8 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }8: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3305.txt

@@ -0,0 +1,414 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ -7 & 5 & 10 & -6 \\
+ -1 & 8 & -7 & 9 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ -7 & 5 & 10 & -6 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -51 & 59 & -69 \\
+ -7 & 3 & 4 & 2 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -51 & 59 & -69 \\
+ 0 & -53 & 53 & -61 \\
+ -4 & -8 & 2 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }4\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -51 & 59 & -69 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & -40 & 30 & -31 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & -51 & 59 & -69 \\
+ 0 & -40 & 30 & -31 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{51}{53}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & 8 & -\frac{546}{53} \\
+ 0 & -40 & 30 & -31 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{40}{53}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & 8 & -\frac{546}{53} \\
+ 0 & 0 & -10 & \frac{797}{53} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & -10 & \frac{797}{53} \\
+ 0 & 0 & 8 & -\frac{546}{53} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{4}{5}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & -10 & \frac{797}{53} \\
+ 0 & 0 & 0 & \frac{458}{265} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{265}{458}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & -10 & \frac{797}{53} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{797}{53}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & -61 \\
+ 0 & 0 & -10 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }61\, \times \, \text{(row }4) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 9 \\
+ 0 & -53 & 53 & 0 \\
+ 0 & 0 & -10 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }4) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 0 \\
+ 0 & -53 & 53 & 0 \\
+ 0 & 0 & -10 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }3 \text{by }-10: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 0 \\
+ 0 & -53 & 53 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }53\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & -7 & 0 \\
+ 0 & -53 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & 0 & 0 \\
+ 0 & -53 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-53: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 8 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }8\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3370.txt

@@ -0,0 +1,264 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 5 & 5 & -3 & 2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{6}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 0 & \frac{25}{6} & \frac{1}{3} & \frac{19}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }\frac{6}{25}: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 1 & -4 & -9 \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }2 \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 6 & 0 & -\frac{102}{25} & -\frac{282}{25} \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }6: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{17}{25} & -\frac{47}{25} \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{17}{25} & -\frac{47}{25} \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{17}{25} & -\frac{47}{25} \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & -\frac{17}{25} & -\frac{47}{25} \\
+ 0 & 1 & \frac{2}{25} & \frac{57}{25} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{17 x_3}{25}-\frac{47 x_4}{25} \\
+ x_2+\frac{2 x_3}{25}+\frac{57 x_4}{25} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{17 x_3}{25}-\frac{47 x_4}{25}=0 \\
+ x_2+\frac{2 x_3}{25}+\frac{57 x_4}{25}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{17 x_3}{25}+\frac{47 x_4}{25} \\
+ x_2=-\frac{2 x_3}{25}-\frac{57 x_4}{25} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{17 x_3}{25}+\frac{47 x_4}{25} \\
+ -\frac{2 x_3}{25}-\frac{57 x_4}{25} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{47 y}{25}+\frac{17 x}{25} \\
+ -\frac{57 y}{25}-\frac{2 x}{25} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }25 x \text{and }25 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{47 y}{25}+\frac{17 x}{25} \\
+ -\frac{57 y}{25}-\frac{2 x}{25} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{47 (25 y)}{25}+\frac{17 (25 x)}{25} \\
+ -\frac{57}{25} (25 y)-\frac{2 (25 x)}{25} \\
+ 25 x \\
+ 25 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 47 y+17 x \\
+ -57 y-2 x \\
+ 25 x \\
+ 25 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 47 y+17 x \\
+ -57 y-2 x \\
+ 25 x \\
+ 25 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (47 y+17 x,-57 y-2 x,25 x,25 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3410.txt

@@ -0,0 +1,329 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 4 & -8 & 1 \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{4}{9}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{92}{9} & -\frac{11}{9} \\
+ -8 & -2 & 7 \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{8}{9}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{92}{9} & -\frac{11}{9} \\
+ 0 & \frac{22}{9} & \frac{103}{9} \\
+ 6 & -10 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{3}\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{92}{9} & -\frac{11}{9} \\
+ 0 & \frac{22}{9} & \frac{103}{9} \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+ 0 & \frac{22}{9} & \frac{103}{9} \\
+ 0 & -\frac{92}{9} & -\frac{11}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{11}{60}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+ 0 & 0 & \frac{38}{3} \\
+ 0 & -\frac{92}{9} & -\frac{11}{9} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{23}{30}\, \times \, \text{(row }2) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+ 0 & 0 & \frac{38}{3} \\
+ 0 & 0 & -\frac{19}{3} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{2}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+ 0 & 0 & \frac{38}{3} \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }\frac{3}{38}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & \frac{20}{3} \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{20}{3}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 5 \\
+ 0 & -\frac{40}{3} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 0 \\
+ 0 & -\frac{40}{3} & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{3}{40}: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 5 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }5\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 9 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }9: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+ 0 & 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3419.txt

@@ -0,0 +1,270 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 2 & -5 \\
+ -6 & -3 \\
+ -9 & 0 \\
+ -3 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ -6 & -3 \\
+ 2 & -5 \\
+ -3 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{3}\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & -3 \\
+ 2 & -5 \\
+ -3 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{2}{9}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & -3 \\
+ 0 & -5 \\
+ -3 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{1}{3}\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & -3 \\
+ 0 & -5 \\
+ 0 & 7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & 7 \\
+ 0 & -5 \\
+ 0 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{7}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & 7 \\
+ 0 & 0 \\
+ 0 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{7}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & 7 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }7: \\
+ \left(
+\begin{array}{cc}
+ -9 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-9: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3459.txt

@@ -0,0 +1,263 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ -3 & 7 & -3 \\
+ 6 & -2 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 6 & -2 & -3 \\
+ -3 & 7 & -3 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{1}{2}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ 6 & -2 & -3 \\
+ 0 & 6 & -\frac{9}{2} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }6: \\
+ \left(
+\begin{array}{ccc}
+ 6 & -2 & -3 \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }2\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ 6 & 0 & -\frac{9}{2} \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }6: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & -\frac{3}{4} \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & -\frac{3}{4} \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }3 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{ccc}
+ 1 & 0 & -\frac{3}{4} \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & -\frac{3}{4} \\
+ 0 & 1 & -\frac{3}{4} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{3 x_3}{4} \\
+ x_2-\frac{3 x_3}{4} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{3 x_3}{4}=0 \\
+ x_2-\frac{3 x_3}{4}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{3 x_3}{4} \\
+ x_2=\frac{3 x_3}{4} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_3, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{3 x_3}{4} \\
+ \frac{3 x_3}{4} \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{3 x}{4} \\
+ \frac{3 x}{4} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }4 x: \\
+ \left(
+\begin{array}{c}
+ \frac{3 x}{4} \\
+ \frac{3 x}{4} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{3 (4 x)}{4} \\
+ \frac{3 (4 x)}{4} \\
+ 4 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 3 x \\
+ 3 x \\
+ 4 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 3 x \\
+ 3 x \\
+ 4 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (3 x,3 x,4 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3476.txt

@@ -0,0 +1,401 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 7 & 8 & 8 & -7 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ -1 & 6 & 10 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ -2 & -7 & -4 & -2 \\
+ -6 & -1 & -7 & -8 \\
+ 7 & 8 & 8 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & -19 & -24 & 18 \\
+ -6 & -1 & -7 & -8 \\
+ 7 & 8 & 8 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & -19 & -24 & 18 \\
+ 0 & -37 & -67 & 52 \\
+ 7 & 8 & 8 & -7 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & -19 & -24 & 18 \\
+ 0 & -37 & -67 & 52 \\
+ 0 & 50 & 78 & -77 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & -37 & -67 & 52 \\
+ 0 & -19 & -24 & 18 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{37}{50}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & 0 & -\frac{232}{25} & -\frac{249}{50} \\
+ 0 & -19 & -24 & 18 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{19}{50}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & 0 & -\frac{232}{25} & -\frac{249}{50} \\
+ 0 & 0 & \frac{141}{25} & -\frac{563}{50} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{141}{232}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & 0 & -\frac{232}{25} & -\frac{249}{50} \\
+ 0 & 0 & 0 & -\frac{6629}{464} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }-\frac{464}{6629}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & 0 & -\frac{232}{25} & -\frac{249}{50} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{249}{50}\, \times \, \text{(row }4) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & -77 \\
+ 0 & 0 & -\frac{232}{25} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }77\, \times \, \text{(row }4) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & -10 \\
+ 0 & 50 & 78 & 0 \\
+ 0 & 0 & -\frac{232}{25} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }4) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & 0 \\
+ 0 & 50 & 78 & 0 \\
+ 0 & 0 & -\frac{232}{25} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{25}{232}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & 0 \\
+ 0 & 50 & 78 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }78\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 10 & 0 \\
+ 0 & 50 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 0 & 0 \\
+ 0 & 50 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }50: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 6 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3599.txt

@@ -0,0 +1,227 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ 6 & -2 \\
+ 3 & 3 \\
+ -8 & 8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 8 \\
+ 3 & 3 \\
+ 6 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{8}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -8 & 8 \\
+ 0 & 6 \\
+ 6 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{3}{4}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 8 \\
+ 0 & 6 \\
+ 0 & 4 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{2}{3}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cc}
+ -8 & 8 \\
+ 0 & 6 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }6: \\
+ \left(
+\begin{array}{cc}
+ -8 & 8 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }8\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -8 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-8: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+ 0 & 0 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3700.txt

@@ -0,0 +1,375 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 0 & -1 & 2 & 10 \\
+ -8 & -10 & 7 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 5 & -8 & -8 & 2 \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{8}\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & -\frac{57}{4} & -\frac{29}{8} & \frac{33}{4} \\
+ -3 & 1 & -9 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{3}{8}\, \times \, \text{(row }1) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & -\frac{57}{4} & -\frac{29}{8} & \frac{33}{4} \\
+ 0 & \frac{19}{4} & -\frac{93}{8} & -\frac{11}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{57}{4}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & 0 & -\frac{257}{8} & -\frac{537}{4} \\
+ 0 & \frac{19}{4} & -\frac{93}{8} & -\frac{11}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{19}{4}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & 0 & -\frac{257}{8} & -\frac{537}{4} \\
+ 0 & 0 & -\frac{17}{8} & \frac{179}{4} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{17}{257}\, \times \, \text{(row }3) \text{from }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & 0 & -\frac{257}{8} & -\frac{537}{4} \\
+ 0 & 0 & 0 & \frac{13783}{257} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{257}{13783}: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & 0 & -\frac{257}{8} & -\frac{537}{4} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{537}{4}\, \times \, \text{(row }4) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 10 \\
+ 0 & 0 & -\frac{257}{8} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }4) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 10 \\
+ 0 & -1 & 2 & 0 \\
+ 0 & 0 & -\frac{257}{8} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }4) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 0 \\
+ 0 & -1 & 2 & 0 \\
+ 0 & 0 & -\frac{257}{8} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{8}{257}: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 0 \\
+ 0 & -1 & 2 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 7 & 0 \\
+ 0 & -1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 0 & 0 \\
+ 0 & -1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ -8 & -10 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }10\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -8 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-8: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3745.txt

@@ -0,0 +1,177 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cc}
+ -1 & 2 \\
+ -9 & 10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cc}
+ -1 & 2 \\
+ 0 & -8 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }-8: \\
+ \left(
+\begin{array}{cc}
+ -1 & 2 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cc}
+ -1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cc}
+ 1 & 0 \\
+ 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3747.txt

@@ -0,0 +1,388 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 2 & -2 & 6 & 4 \\
+ 6 & 4 & -2 & -4 \\
+ 1 & -9 & -7 & 6 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 6 & 4 & -2 & -4 \\
+ 2 & -2 & 6 & 4 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 2 & -2 & 6 & 4 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }2\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 16 & 20 & -8 \\
+ -7 & 9 & -7 & 5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }1) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 16 & 20 & -8 \\
+ 0 & -54 & -56 & 47 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{8}{29}\, \times \, \text{(row }2) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & \frac{260}{29} & \frac{88}{29} \\
+ 0 & -54 & -56 & 47 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{27}{29}\, \times \, \text{(row }2) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & \frac{260}{29} & \frac{88}{29} \\
+ 0 & 0 & -\frac{544}{29} & \frac{283}{29} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }3 \text{with }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & -\frac{544}{29} & \frac{283}{29} \\
+ 0 & 0 & \frac{260}{29} & \frac{88}{29} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{65}{136}\, \times \, \text{(row }3) \text{to }\text{row }4: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & -\frac{544}{29} & \frac{283}{29} \\
+ 0 & 0 & 0 & \frac{1047}{136} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }4 \text{by }\frac{136}{1047}: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & -\frac{544}{29} & \frac{283}{29} \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{283}{29}\, \times \, \text{(row }4) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & -40 \\
+ 0 & 0 & -\frac{544}{29} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }40\, \times \, \text{(row }4) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 6 \\
+ 0 & 58 & 40 & 0 \\
+ 0 & 0 & -\frac{544}{29} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }6\, \times \, \text{(row }4) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 0 \\
+ 0 & 58 & 40 & 0 \\
+ 0 & 0 & -\frac{544}{29} & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{29}{544}: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 0 \\
+ 0 & 58 & 40 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }40\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & -7 & 0 \\
+ 0 & 58 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }3) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & 0 & 0 \\
+ 0 & 58 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }58: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -9 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }9\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & 0 \\
+ 0 & 1 & 0 & 0 \\
+ 0 & 0 & 1 & 0 \\
+ 0 & 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3749.txt

@@ -0,0 +1,253 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 9 & 8 & 5 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }9\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 0 & 71 & -13 & 16 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }71: \\
+ \left(
+\begin{array}{cccc}
+ 1 & -7 & 2 & -2 \\
+ 0 & 1 & -\frac{13}{71} & \frac{16}{71} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }7\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{51}{71} & -\frac{30}{71} \\
+ 0 & 1 & -\frac{13}{71} & \frac{16}{71} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{51}{71} & -\frac{30}{71} \\
+ 0 & 1 & -\frac{13}{71} & \frac{16}{71} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Columns }3 \text{and }4 \text{are }\text{the }\text{only }\text{columns }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_3 \text{and }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{51}{71} & -\frac{30}{71} \\
+ 0 & 1 & -\frac{13}{71} & \frac{16}{71} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & \frac{51}{71} & -\frac{30}{71} \\
+ 0 & 1 & -\frac{13}{71} & \frac{16}{71} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1+\frac{51 x_3}{71}-\frac{30 x_4}{71} \\
+ x_2-\frac{13 x_3}{71}+\frac{16 x_4}{71} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1+\frac{51 x_3}{71}-\frac{30 x_4}{71}=0 \\
+ x_2-\frac{13 x_3}{71}+\frac{16 x_4}{71}=0 \\
+\end{array}
+  \text{for }x_1 \text{and }x_2: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{30 x_4}{71}-\frac{51 x_3}{71} \\
+ x_2=\frac{13 x_3}{71}-\frac{16 x_4}{71} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variables }x_3 \text{and }x_4, \text{and }\text{assign }\text{arbitrary }\text{real }\text{values }\text{of }x \text{and }y \text{to }\text{the }\text{variables}: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{30 x_4}{71}-\frac{51 x_3}{71} \\
+ \frac{13 x_3}{71}-\frac{16 x_4}{71} \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{30 y}{71}-\frac{51 x}{71} \\
+ -\frac{16 y}{71}+\frac{13 x}{71} \\
+ x \\
+ y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }\text{the }\text{variables }x \text{and }y \text{are }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{them }\text{with }71 x \text{and }71 y \text{respectively}: \\
+ \left(
+\begin{array}{c}
+ \frac{30 y}{71}-\frac{51 x}{71} \\
+ -\frac{16 y}{71}+\frac{13 x}{71} \\
+ x \\
+ y \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{30 (71 y)}{71}-\frac{51 (71 x)}{71} \\
+ -\frac{16}{71} (71 y)+\frac{13 (71 x)}{71} \\
+ 71 x \\
+ 71 y \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 30 y-51 x \\
+ -16 y+13 x \\
+ 71 x \\
+ 71 y \\
+\end{array}
+\right)\text{ for }x,y\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 30 y-51 x \\
+ -16 y+13 x \\
+ 71 x \\
+ 71 y \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (30 y-51 x,-16 y+13 x,71 x,71 y)\, \text{$\, $: }x,y\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3805.txt

@@ -0,0 +1,270 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{ccc}
+ 5 & -6 & 5 \\
+ -8 & -4 & 1 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 5 & -6 & 5 \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{5}{8}\, \times \, \text{(row }1) \text{to }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 0 & -\frac{17}{2} & \frac{45}{8} \\
+ -5 & 5 & -5 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{5}{8}\, \times \, \text{(row }1) \text{from }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 0 & -\frac{17}{2} & \frac{45}{8} \\
+ 0 & \frac{15}{2} & -\frac{45}{8} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{15}{17}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 0 & -\frac{17}{2} & \frac{45}{8} \\
+ 0 & 0 & -\frac{45}{68} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{68}{45}: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 0 & -\frac{17}{2} & \frac{45}{8} \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\frac{45}{8}\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 1 \\
+ 0 & -\frac{17}{2} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }\text{row }3 \text{from }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 0 \\
+ 0 & -\frac{17}{2} & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }2 \text{by }-\frac{2}{17}: \\
+ \left(
+\begin{array}{ccc}
+ -8 & -4 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }4\, \times \, \text{(row }2) \text{to }\text{row }1: \\
+ \left(
+\begin{array}{ccc}
+ -8 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }1 \text{by }-8: \\
+ \left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{There }\text{are }\text{no }\text{free }\text{variables} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{only }\text{value }\text{of }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right) \text{that }\text{would }\text{make }\left(
+\begin{array}{ccc}
+ 1 & 0 & 0 \\
+ 0 & 1 & 0 \\
+ 0 & 0 & 1 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \text{is }\text{the }\text{zero }\text{vector }\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{is }\text{the }\text{singleton }\text{set }\text{containing }\text{only }\text{the }\text{zero }\text{vector}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (0,0,0)\, \} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}

pretraining/mathematica/linear_algebra/null_space_w_steps/3827.txt

@@ -0,0 +1,366 @@
+Problem:
+Give a list of vectors that forms a basis for the null space of the following matrix:
+$\left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right)$.
+Answer:
+\begin{array}{l}
+ 
+\begin{array}{l}
+ \text{Find the null space of the matrix }M: \\
+ M=\left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+\hline
+ 
+\begin{array}{l}
+ \text{The }\text{null }\text{space }\text{of }\text{matrix }M=\left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right) \text{is }\text{the }\text{set }\text{of }\text{all }\text{vectors }v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{such }\text{that }M.v=0: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Reduce }\text{the }\text{matrix }\left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right) \text{to }\text{row }\text{echelon }\text{form}: \\
+ \left(
+\begin{array}{cccc}
+ 9 & 8 & 2 & -10 \\
+ -7 & 8 & 3 & -2 \\
+ -1 & 10 & 7 & -2 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }1 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ -7 & 8 & 3 & -2 \\
+ 9 & 8 & 2 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }1) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & -62 & -46 & 12 \\
+ 9 & 8 & 2 & -10 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }9\, \times \, \text{(row }1) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & -62 & -46 & 12 \\
+ 0 & 98 & 65 & -28 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Swap }\text{row }2 \text{with }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & 98 & 65 & -28 \\
+ 0 & -62 & -46 & 12 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Add }\frac{31}{49}\, \times \, \text{(row }2) \text{to }\text{row }3: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & 98 & 65 & -28 \\
+ 0 & 0 & -\frac{239}{49} & -\frac{40}{7} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }3 \text{by }-\frac{49}{239}: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & 98 & 65 & -28 \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }65\, \times \, \text{(row }3) \text{from }\text{row }2: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 7 & -2 \\
+ 0 & 98 & 0 & -\frac{24892}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }7\, \times \, \text{(row }3) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 0 & -\frac{2438}{239} \\
+ 0 & 98 & 0 & -\frac{24892}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Divide }\text{row }2 \text{by }98: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 10 & 0 & -\frac{2438}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Subtract }10\, \times \, \text{(row }2) \text{from }\text{row }1: \\
+ \left(
+\begin{array}{cccc}
+ -1 & 0 & 0 & \frac{102}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{row }1 \text{by }-1: \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{102}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Free }\text{variables }\text{in }\text{the }\text{null }\text{space }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right) \text{correspond }\text{to }\text{the }\text{columns }\text{in }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{102}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \text{which }\text{have }\text{no }\text{pivot.} \\
+ \text{Column }4 \text{is }\text{the }\text{only }\text{column }\text{with }\text{no }\text{pivot, }\text{so }\text{we }\text{may }\text{take }x_4 \text{to }\text{be }\text{the }\text{only }\text{free }\text{variable} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Multiply }\text{out }\text{the }\text{reduced }\text{matrix }\left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{102}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right) \text{with }\text{the }\text{proposed }\text{solution }\text{vector }\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right): \\
+ \left(
+\begin{array}{cccc}
+ 1 & 0 & 0 & -\frac{102}{239} \\
+ 0 & 1 & 0 & -\frac{254}{239} \\
+ 0 & 0 & 1 & \frac{280}{239} \\
+\end{array}
+\right).\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ x_1-\frac{102 x_4}{239} \\
+ x_2-\frac{254 x_4}{239} \\
+ x_3+\frac{280 x_4}{239} \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 0 \\
+ 0 \\
+ 0 \\
+\end{array}
+\right) \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Solve }\text{the }\text{equations }\{
+\begin{array}{l}
+ x_1-\frac{102 x_4}{239}=0 \\
+ x_2-\frac{254 x_4}{239}=0 \\
+ x_3+\frac{280 x_4}{239}=0 \\
+\end{array}
+  \text{for }x_1,x_2 \text{and }x_3: \\
+ \{
+\begin{array}{l}
+ x_1=\frac{102 x_4}{239} \\
+ x_2=\frac{254 x_4}{239} \\
+ x_3=-\frac{280 x_4}{239} \\
+\end{array}
+ \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }v \text{in }\text{terms }\text{of }\text{the }\text{free }\text{variable }x_4, \text{and }\text{assign }\text{it }\text{an }\text{arbitrary }\text{real }\text{value }\text{of }x: \\
+ v=\left(
+\begin{array}{c}
+ x_1 \\
+ x_2 \\
+ x_3 \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{102 x_4}{239} \\
+ \frac{254 x_4}{239} \\
+ -\frac{280 x_4}{239} \\
+ x_4 \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ \frac{102 x}{239} \\
+ \frac{254 x}{239} \\
+ -\frac{280 x}{239} \\
+ x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Since }x \text{is }\text{taken }\text{from }\mathbb{R}, \text{we }\text{can }\text{replace }\text{it }\text{with }239 x: \\
+ \left(
+\begin{array}{c}
+ \frac{102 x}{239} \\
+ \frac{254 x}{239} \\
+ -\frac{280 x}{239} \\
+ x \\
+\end{array}
+\right)\, \rightarrow \, \left(
+\begin{array}{c}
+ \frac{102 (239 x)}{239} \\
+ \frac{254 (239 x)}{239} \\
+ -\frac{280}{239} (239 x) \\
+ 239 x \\
+\end{array}
+\right)=\left(
+\begin{array}{c}
+ 102 x \\
+ 254 x \\
+ -280 x \\
+ 239 x \\
+\end{array}
+\right)\text{ for }x\in \mathbb{R} \\
+\end{array}
+ \\
+ 
+\begin{array}{l}
+ \text{Rewrite }\text{the }\text{solution }\text{vector }v=\left(
+\begin{array}{c}
+ 102 x \\
+ 254 x \\
+ -280 x \\
+ 239 x \\
+\end{array}
+\right) \text{in }\text{set }\text{notation}: \\
+ \fbox{$
+\begin{array}{ll}
+ \text{Answer:} &  \\
+ \text{} & \{\, (102 x,254 x,-280 x,239 x)\, \text{$\, $: }x\in \mathbb{R}\} \\
+\end{array}
+$} \\
+\end{array}
+ \\
+\end{array}