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|---|---|---|---|---|---|---|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.3.10
|
A person sets off on a random walk with three possible locations. The Markov matrix of probabilities $A=\left[a_{i j}\right]$ is given by
$$
\left[\begin{array}{rrr}
0.5 & 0.1 & 0.6 \\
0.2 & 0.9 & 0.2 \\
0.3 & 0 & 0.2
\end{array}\right]
$$
It is unknown where the walker starts, but the probability of starting in each location is given by
$$
X_{0}=\left[\begin{array}{r}
0.2 \\
0.25 \\
0.55
\end{array}\right]
$$
What is the probability of the walker being in location 1 at time $n=2$ ?
|
$$
X_{2}=\left[\begin{array}{l}
0.367 \\
0.4625 \\
0.1705
\end{array}\right]
$$
Therefore the probability of ending up in location 1 is 0.367 .
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
X_{2}=\left[\begin{array}{l}
0.367 \\
0.4625 \\
0.1705
\end{array}\right]
$$
Therefore the probability of ending up in location 1 is 0.367 .
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.2.8
|
Consider the matrix
$$
A=\left[\begin{array}{rrr}
1 & t & t^{2} \\
0 & 1 & 2 t \\
t & 0 & 2
\end{array}\right]
$$
Does there exist a value of t for which this matrix fails to have an inverse? Explain.
|
$$
\operatorname{det}\left[\begin{array}{ccc}
1 & t & t^{2} \\
0 & 1 & 2 t \\
t & 0 & 2
\end{array}\right]=t^{3}+2
$$
and so it has no inverse when $t=-\sqrt[3]{2}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\operatorname{det}\left[\begin{array}{ccc}
1 & t & t^{2} \\
0 & 1 & 2 t \\
t & 0 & 2
\end{array}\right]=t^{3}+2
$$
and so it has no inverse when $t=-\sqrt[3]{2}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.2.1.37
|
Let
$$
A=\left[\begin{array}{ll}
2 & 1 \\
3 & 0
\end{array}\right]
$$
Find $A^{-1}$ if possible. If $A^{-1}$ does not exist, explain why.
|
$\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]^{-1}=\left[\begin{array}{cc}0 & \frac{1}{3} \\ 1 & -\frac{2}{3}\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{ll}2 & 1 \\ 3 & 0\end{array}\right]^{-1}=\left[\begin{array}{cc}0 & \frac{1}{3} \\ 1 & -\frac{2}{3}\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.5.8
|
Explain why an $n \times n$ matrix $A$ is both one to one and onto if and only if its rank is $n$.
|
The rank is $n$ is the same as saying the columns are independent which is the same as saying $A$ is one to one which is the same as saying the columns are a basis. Thus the span of the columns of $A$ is all of $\mathbb{R}^{n}$ and so $A$ is onto. If $A$ is onto, then the columns must be linearly independent since otherwise the span of these columns would have dimension less than $n$ and so the dimension of $\mathbb{R}^{n}$ would be less than $n$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The rank is $n$ is the same as saying the columns are independent which is the same as saying $A$ is one to one which is the same as saying the columns are a basis. Thus the span of the columns of $A$ is all of $\mathbb{R}^{n}$ and so $A$ is onto. If $A$ is onto, then the columns must be linearly independent since otherwise the span of these columns would have dimension less than $n$ and so the dimension of $\mathbb{R}^{n}$ would be less than $n$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.7.11
|
Find $\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1 \\ 2 \\ -2\end{array}\right]$ and $\vec{v}=\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right]$.
|
$\overrightarrow{\vec{u} \bullet \vec{v}} \overrightarrow{\vec{u}} \vec{u}=\frac{-5}{10}\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right]=\left[\begin{array}{r}-\frac{1}{2} \\ 0 \\ -\frac{3}{2}\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\overrightarrow{\vec{u} \bullet \vec{v}} \overrightarrow{\vec{u}} \vec{u}=\frac{-5}{10}\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right]=\left[\begin{array}{r}-\frac{1}{2} \\ 0 \\ -\frac{3}{2}\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.21
|
How much work does it take to slide a crate 20 meters along a loading dock by pulling on it with a 200 Newton force at an angle of $30^{\circ}$ from the horizontal? Express your answer in Newton meters.
|
$200\left(\cos \left(\frac{\pi}{6}\right)\right) 20=3464.1$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
200\left(\cos \left(\frac{\pi}{6}\right)\right) 20=3464.1
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.10
|
An airplane is flying due north at 150.0 miles per hour but it is not actually going due North because there is a wind which is pushing the airplane due east at 40.0 miles per hour. After one hour, the plane starts flying $30^{\circ}$ East of North. Assuming the plane starts at $(0,0)$, where is it after 2 hours? Let North be the direction of the positive y axis and let East be the direction of the positive $x$ axis.
|
After two hours it is then at $(40,150)+150\left[\begin{array}{ll}\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]+\left[\begin{array}{ll}40 & 0\end{array}\right]=\left[\begin{array}{ll}155 & 75 \sqrt{3}+150\end{array}\right]=\left[\begin{array}{ll}155.0 & 279.9\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
After two hours it is then at $(40,150)+150\left[\begin{array}{ll}\frac{1}{2} & \frac{\sqrt{3}}{2}\end{array}\right]+\left[\begin{array}{ll}40 & 0\end{array}\right]=\left[\begin{array}{ll}155 & 75 \sqrt{3}+150\end{array}\right]=\left[\begin{array}{ll}155.0 & 279.9\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.3
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{rrr}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right]
$$
Hint: One eigenvalue is -2.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right],\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow-2 \\
{\left[\begin{array}{ccc}
\sqrt{3} / 3 & -\sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & \sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & 0 & \frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]^{T}\left[\begin{array}{rrr}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right]} \\
.\left[\begin{array}{ccc}
\sqrt{3} / 3 & -\sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & \sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & 0 & \frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right] \\
=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array}\right]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right],\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow-2 \\
{\left[\begin{array}{ccc}
\sqrt{3} / 3 & -\sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & \sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & 0 & \frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]^{T}\left[\begin{array}{rrr}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right]} \\
.\left[\begin{array}{ccc}
\sqrt{3} / 3 & -\sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & \sqrt{2} / 2 & -\sqrt{6} / 6 \\
\sqrt{3} / 3 & 0 & \frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right] \\
=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & -2
\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.4
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{rrr}
17 & -7 & -4 \\
-7 & 17 & -4 \\
-4 & -4 & 14
\end{array}\right]
$$
Hint: Two eigenvalues are 18 and 24.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 18,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right]\right\} \leftrightarrow 24
$$
The matrix $U$ has these as its columns.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 18,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right]\right\} \leftrightarrow 24
$$
The matrix $U$ has these as its columns.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.11
|
City $A$ is located at the origin $(0,0)$ while city $B$ is located at $(300,500)$ where distances are in miles. An airplane flies at 250 miles per hour in still air. This airplane wants to fly from city A to city $B$ but the wind is blowing in the direction of the positive y axis at a speed of 50 miles per hour. Find a unit vector such that if the plane heads in this direction, it will end up at city B having flown the shortest possible distance. How long will it take to get there?
|
Therefore, it takes $\frac{583.1}{291.55}=2$ hours.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Therefore, it takes $\frac{583.1}{291.55}=2$ hours.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.6
|
Consider the following augmented matrix in which $*$ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{ccccc|c}
\mathbf{\square} & * & * & * & * & * \\
0 & \mathbf{\square} & * & * & 0 & * \\
0 & 0 & \mathbf{\square} & * & * & * \\
0 & 0 & 0 & 0 & \mathbf{\square} & *
\end{array}\right]
$$
|
The solution exists but is not unique.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The solution exists but is not unique.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.20
|
A large dog drags a sled for 300 feet along the ground by pulling on a rope which is 45 degrees from the horizontal with a force of 20 pounds. How much work does this force do?
|
$20\left(\cos \frac{\pi}{4}\right) 300=4242.6$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
20\left(\cos \frac{\pi}{4}\right) 300=4242.6
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.6
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{ccc}
-\frac{5}{3} & \frac{1}{15} \sqrt{6} \sqrt{5} & \frac{8}{15} \sqrt{5} \\
\frac{1}{15} \sqrt{6} \sqrt{5} & -\frac{14}{5} & -\frac{1}{15} \sqrt{6} \\
\frac{8}{15} \sqrt{5} & -\frac{1}{15} \sqrt{6} & \frac{7}{15}
\end{array}\right]
$$
Hint: The eigenvalues are $-3,-2,1$.
|
eigenvectors:
$$
\left\{\left[\begin{array}{c}
\frac{1}{6} \sqrt{6} \\
0 \\
\frac{1}{6} \sqrt{5} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{2} \sqrt{3} \\
-\frac{1}{5} \sqrt{5} \\
\frac{1}{15} \sqrt{2} \sqrt{15}
\end{array}\right]\right\} \leftrightarrow-2,\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
\frac{2}{5} \sqrt{5} \\
\frac{1}{30} \sqrt{30}
\end{array}\right]\right\} \leftrightarrow-3
$$
These vectors are the columns of $U$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
eigenvectors:
$$
\left\{\left[\begin{array}{c}
\frac{1}{6} \sqrt{6} \\
0 \\
\frac{1}{6} \sqrt{5} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{2} \sqrt{3} \\
-\frac{1}{5} \sqrt{5} \\
\frac{1}{15} \sqrt{2} \sqrt{15}
\end{array}\right]\right\} \leftrightarrow-2,\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
\frac{2}{5} \sqrt{5} \\
\frac{1}{30} \sqrt{30}
\end{array}\right]\right\} \leftrightarrow-3
$$
These vectors are the columns of $U$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.8
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 5 & 1 \\
0 & 1 & 5
\end{array}\right]
$$
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
1 \\
0 \\
0
\end{array}\right]\right\} \leftrightarrow 2,\left\{\left[\begin{array}{c}
0 \\
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2}
\end{array}\right]\right\} \leftrightarrow 4,\left\{\left[\begin{array}{c}
0 \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2}
\end{array}\right]\right\} \leftrightarrow 6
$$
These vectors are the columns of $U$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
1 \\
0 \\
0
\end{array}\right]\right\} \leftrightarrow 2,\left\{\left[\begin{array}{c}
0 \\
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2}
\end{array}\right]\right\} \leftrightarrow 4,\left\{\left[\begin{array}{c}
0 \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2}
\end{array}\right]\right\} \leftrightarrow 6
$$
These vectors are the columns of $U$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.35
|
Find the solution to the system of equations, $65 x+84 y+16 z=546,81 x+105 y+20 z=682$, and $84 x+110 y+21 z=713$.
|
Solution is: $[x=2, y=4, z=5]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=2, y=4, z=5]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.2.4
|
Decide whether $\vec{v}=\left[\begin{array}{l}4 \\ 4 \\ 4\end{array}\right]$ is a linear combination of the vectors $\vec{u}_{1}=\left[\begin{array}{r}3 \\ 1 \\ -1\end{array}\right]$ and $\vec{u}_{2}=\left[\begin{array}{r}2 \\ -2 \\ 1\end{array}\right]$.
|
The system
$$
\left[\begin{array}{l}
4 \\
4 \\
4
\end{array}\right]=a_{1}\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]+a_{2}\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$
has no solution.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The system
$$
\left[\begin{array}{l}
4 \\
4 \\
4
\end{array}\right]=a_{1}\left[\begin{array}{r}
3 \\
1 \\
-1
\end{array}\right]+a_{2}\left[\begin{array}{r}
2 \\
-2 \\
1
\end{array}\right]
$$
has no solution.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.3.9
|
A person sets off on a random walk with three possible locations. The Markov matrix of probabilities $A=\left[a_{i j}\right]$ is given by
$$
\left[\begin{array}{lll}
0.1 & 0.3 & 0.7 \\
0.1 & 0.3 & 0.2 \\
0.8 & 0.4 & 0.1
\end{array}\right]
$$
If the walker starts in location 2 , what is the probability of ending back in location 2 at time $n=3$ ?
|
$$
X_{3}=\left[\begin{array}{l}
0.38 \\
0.18 \\
0.44
\end{array}\right]
$$
Therefore the probability of ending up back in location 2 is 0.18 .
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
X_{3}=\left[\begin{array}{l}
0.38 \\
0.18 \\
0.44
\end{array}\right]
$$
Therefore the probability of ending up back in location 2 is 0.18 .
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.19
|
A girl drags a sled for 200 feet along the ground by pulling on a rope which is 30 degrees from the horizontal with a force of 20 pounds. How much work does this force do?
|
$20 \cos \left(\frac{\pi}{6}\right) 200=3464.1$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
20 \cos \left(\frac{\pi}{6}\right) 200=3464.1
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.4
|
Find the area of the triangle determined by the three points, $(1,2,3),(2,3,4)$ and $(3,4,5)$. Did something interesting happen here? What does it mean geometrically?
|
$\left[\begin{array}{lll}1 & 1 & 1\end{array}\right] \times\left[\begin{array}{lll}2 & 2 & 2\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$. The area is 0. It means the three points are on the same line.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{lll}1 & 1 & 1\end{array}\right] \times\left[\begin{array}{lll}2 & 2 & 2\end{array}\right]=\left[\begin{array}{lll}0 & 0 & 0\end{array}\right]$. The area is 0. It means the three points are on the same line.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.11
|
Suppose $\vec{u}, \vec{v}$, and $\vec{w}$ are three vectors whose components are all integers. Can you conclude the volume of the parallelepiped determined from these three vectors will always be an integer?
|
Yes. It will involve the sum of product of integers and so it will be an integer.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Yes. It will involve the sum of product of integers and so it will be an integer.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.2.14
|
Find the inverse, if it exists, of the matrix
$$
A=\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]
$$
|
$$
\begin{aligned}
& {\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]^{-1} } \\
= & {\left[\begin{array}{ccc}
\frac{1}{2} e^{-t} & 0 & \frac{1}{2} e^{-t} \\
\frac{1}{2} \cos t+\frac{1}{2} \sin t & -\sin t & \frac{1}{2} \sin t-\frac{1}{2} \cos t \\
\frac{1}{2} \sin t-\frac{1}{2} \cos t & \cos t & -\frac{1}{2} \cos t-\frac{1}{2} \sin t
\end{array}\right] }
\end{aligned}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\begin{aligned}
& {\left[\begin{array}{ccc}
e^{t} & \cos t & \sin t \\
e^{t} & -\sin t & \cos t \\
e^{t} & -\cos t & -\sin t
\end{array}\right]^{-1} } \\
= & {\left[\begin{array}{ccc}
\frac{1}{2} e^{-t} & 0 & \frac{1}{2} e^{-t} \\
\frac{1}{2} \cos t+\frac{1}{2} \sin t & -\sin t & \frac{1}{2} \sin t-\frac{1}{2} \cos t \\
\frac{1}{2} \sin t-\frac{1}{2} \cos t & \cos t & -\frac{1}{2} \cos t-\frac{1}{2} \sin t
\end{array}\right] }
\end{aligned}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.16
|
Simplify $\|\vec{u} \times \vec{v}\|^{2}+(\vec{u} \bullet \vec{v})^{2}-\|\vec{u}\|^{2}\|\vec{v}\|^{2}$.
|
$$
\begin{aligned}
\|\vec{u} \times \vec{v}\|^{2} & =\varepsilon_{i j k} u_{j} v_{k} \varepsilon_{i r s} u_{r} v_{s}=\left(\delta_{j r} \delta_{k s}-\delta_{k r} \delta_{j s}\right) u_{r} v_{s} u_{j} v_{k} \\
& =u_{j} v_{k} u_{j} v_{k}-u_{k} v_{j} u_{j} v_{k}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}-(\vec{u} \bullet \vec{v})^{2}
\end{aligned}
$$
It follows that the expression reduces to 0. You can also do the following.
$$
\begin{aligned}
\|\vec{u} \times \vec{v}\|^{2} & =\|\vec{u}\|^{2}\|\vec{v}\|^{2} \sin ^{2} \theta \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}\left(1-\cos ^{2} \theta\right) \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}-\|\vec{u}\|^{2}\|\vec{v}\|^{2} \cos ^{2} \theta \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}-(\vec{u} \bullet \vec{v})^{2}
\end{aligned}
$$
which implies the expression equals 0.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\begin{aligned}
\|\vec{u} \times \vec{v}\|^{2} & =\varepsilon_{i j k} u_{j} v_{k} \varepsilon_{i r s} u_{r} v_{s}=\left(\delta_{j r} \delta_{k s}-\delta_{k r} \delta_{j s}\right) u_{r} v_{s} u_{j} v_{k} \\
& =u_{j} v_{k} u_{j} v_{k}-u_{k} v_{j} u_{j} v_{k}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}-(\vec{u} \bullet \vec{v})^{2}
\end{aligned}
$$
It follows that the expression reduces to 0. You can also do the following.
$$
\begin{aligned}
\|\vec{u} \times \vec{v}\|^{2} & =\|\vec{u}\|^{2}\|\vec{v}\|^{2} \sin ^{2} \theta \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}\left(1-\cos ^{2} \theta\right) \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}-\|\vec{u}\|^{2}\|\vec{v}\|^{2} \cos ^{2} \theta \\
& =\|\vec{u}\|^{2}\|\vec{v}\|^{2}-(\vec{u} \bullet \vec{v})^{2}
\end{aligned}
$$
which implies the expression equals 0.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.2.1
|
Find $-3\left[\begin{array}{r}5 \\ -1 \\ 2 \\ -3\end{array}\right]+5\left[\begin{array}{r}-8 \\ 2 \\ -3 \\ 6\end{array}\right]$.
|
$\left[\begin{array}{r}-55 \\ 13 \\ -21 \\ 39\end{array}\right]$
|
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|
college_math.linear_algebra
|
\left[\begin{array}{r}-55 \\ 13 \\ -21 \\ 39\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.2.1.41
|
Let
$$
A=\left[\begin{array}{lll}
1 & 0 & 3 \\
2 & 3 & 4 \\
1 & 0 & 2
\end{array}\right]
$$
Find $A^{-1}$ if possible. If $A^{-1}$ does not exist, explain why.
|
$\left[\begin{array}{lll}1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2\end{array}\right]^{-1}=\left[\begin{array}{rrr}-2 & 0 & 3 \\ 0 & \frac{1}{3} & -\frac{2}{3} \\ 1 & 0 & -1\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{lll}1 & 0 & 3 \\ 2 & 3 & 4 \\ 1 & 0 & 2\end{array}\right]^{-1}=\left[\begin{array}{rrr}-2 & 0 & 3 \\ 0 & \frac{1}{3} & -\frac{2}{3} \\ 1 & 0 & -1\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.5
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{rrr}
13 & 1 & 4 \\
1 & 13 & 4 \\
4 & 4 & 10
\end{array}\right]
$$
Hint: Two eigenvalues are 12 and 18.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right]\right\} \leftrightarrow 12,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 18 .
$$
The matrix $U$ has these as its columns.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
-\frac{1}{6} \sqrt{6} \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 6,\left\{\left[\begin{array}{c}
-\frac{1}{2} \sqrt{2} \\
\frac{1}{2} \sqrt{2} \\
0
\end{array}\right]\right\} \leftrightarrow 12,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 18 .
$$
The matrix $U$ has these as its columns.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.10
|
An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \rightarrow \cdots \rightarrow\left[\begin{array}{ll}
c & d \\
a & b
\end{array}\right]
$$
|
In this case two rows were switched and so the resulting determinant is -1 times the first.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
In this case two rows were switched and so the resulting determinant is -1 times the first.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.6
|
Find the following determinant by expanding along the second row and first column.
$$
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|
$$
|
$$
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|=6
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left|\begin{array}{lll}
1 & 2 & 1 \\
2 & 1 & 3 \\
2 & 1 & 1
\end{array}\right|=6
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.9
|
Consider the following augmented matrix in which $*$ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{lllll|l}
\mathbf{\square} & * & * & * & * & * \\
0 & \mathbf{\square} & * & * & 0 & * \\
0 & 0 & 0 & 0 & \mathbf{\square} & 0 \\
0 & 0 & 0 & 0 & * & \mathbf{\square}
\end{array}\right]
$$
|
There might be a solution. If so, there are infinitely many.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
There might be a solution. If so, there are infinitely many.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.11
|
An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \rightarrow \cdots \rightarrow\left[\begin{array}{cc}
a & b \\
a+c & b+d
\end{array}\right]
$$
|
The determinant is unchanged. It was just the first row added to the second.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The determinant is unchanged. It was just the first row added to the second.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.9
|
Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{rrrr}
1 & 0 & 1 & 1 \\
1 & -1 & 1 & 0 \\
3 & -1 & 3 & 2 \\
3 & 3 & 0 & 3
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]
$$
|
Solution is: $\left[\begin{array}{c}0 \\ -\hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right], \hat{t} \in \mathbb{R}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{c}0 \\ -\hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right], \hat{t} \in \mathbb{R}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.7.4
|
Let $V=\mathbb{R}^{3}$ and let
$$
W=\operatorname{span}\left\{\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],\left[\begin{array}{r}
-1 \\
2 \\
-1
\end{array}\right]\right\}
$$
Extend this basis of $W$ to a basis of $V$.
|
There are many possible such extensions, one is (how do we know?):
$$
\left\{\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],\left[\begin{array}{r}
-1 \\
2 \\
-1
\end{array}\right],\left[\begin{array}{l}
0 \\
0 \\
1
\end{array}\right]\right\}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
There are many possible such extensions, one is (how do we know?):
$$
\left\{\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right],\left[\begin{array}{r}
-1 \\
2 \\
-1
\end{array}\right],\left[\begin{array}{l}
0 \\
0 \\
1
\end{array}\right]\right\}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.38
|
Find the solution to the system of equations, $3 x-y-2 z=3, y-4 z=0$, and $-2 x+y=-2$.
|
Solution is: $[x=2 t+1, y=4 t, z=t]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=2 t+1, y=4 t, z=t]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.12
|
An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \rightarrow \cdots \rightarrow\left[\begin{array}{cc}
a & b \\
2 c & 2 d
\end{array}\right]
$$
|
The second row was multiplied by 2 so the determinant of the result is 2 times the original determinant.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The second row was multiplied by 2 so the determinant of the result is 2 times the original determinant.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.7
|
Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{rrr}
0 & -1 & 2 \\
1 & 0 & 1 \\
1 & -2 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
$$
|
Solution is: $\left[\begin{array}{c}-\hat{t} \\ 2 \hat{t} \\ \hat{t}\end{array}\right], \hat{t} \in \mathbb{R}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{c}-\hat{t} \\ 2 \hat{t} \\ \hat{t}\end{array}\right], \hat{t} \in \mathbb{R}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.3.30
|
If you have 6 vectors in $\mathbb{R}^{5}$, is it possible they are linearly independent? Explain.
|
No. They can't be.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
No. They can't be.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.25
|
Find the determinant using row operations to first simplify.
$$
\left|\begin{array}{rrrr}
1 & 4 & 1 & 2 \\
3 & 2 & -2 & 3 \\
-1 & 0 & 3 & 3 \\
2 & 1 & 2 & -2
\end{array}\right|
$$
|
One can row reduce this using only row operation 3 to
$$
\left[\begin{array}{rrrr}
1 & 4 & 1 & 2 \\
0 & -10 & -5 & -3 \\
0 & 0 & 2 & \frac{19}{5} \\
0 & 0 & 0 & -\frac{211}{20}
\end{array}\right]
$$
Thus the determinant is given by
$$
\left|\begin{array}{rrrr}
1 & 4 & 1 & 2 \\
3 & 2 & -2 & 3 \\
-1 & 0 & 3 & 3 \\
2 & 1 & 2 & -2
\end{array}\right|=211
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
One can row reduce this using only row operation 3 to
$$
\left[\begin{array}{rrrr}
1 & 4 & 1 & 2 \\
0 & -10 & -5 & -3 \\
0 & 0 & 2 & \frac{19}{5} \\
0 & 0 & 0 & -\frac{211}{20}
\end{array}\right]
$$
Thus the determinant is given by
$$
\left|\begin{array}{rrrr}
1 & 4 & 1 & 2 \\
3 & 2 & -2 & 3 \\
-1 & 0 & 3 & 3 \\
2 & 1 & 2 & -2
\end{array}\right|=211
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.6.3.9
|
If $n$ is an integer, is it always true that $(\cos \theta-i \sin \theta)^{n}=\cos (n \theta)-i \sin (n \theta)$ ? Explain.
|
Yes, this is true.
$$
\begin{aligned}
(\cos \theta-i \sin \theta)^{n} & =(\cos (-\theta)+i \sin (-\theta))^{n} \\
& =\cos (-n \theta)+i \sin (-n \theta) \\
& =\cos (n \theta)-i \sin (n \theta)
\end{aligned}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Yes, this is true.
$$
\begin{aligned}
(\cos \theta-i \sin \theta)^{n} & =(\cos (-\theta)+i \sin (-\theta))^{n} \\
& =\cos (-n \theta)+i \sin (-n \theta) \\
& =\cos (n \theta)-i \sin (n \theta)
\end{aligned}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.6
|
Find the area of the parallelogram determined by the vectors $\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right],\left[\begin{array}{r}4 \\ -2 \\ 1\end{array}\right]$.
|
$\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right] \times\left[\begin{array}{r}4 \\ -2 \\ 1\end{array}\right]=\left[\begin{array}{r}6 \\ 11 \\ -2\end{array}\right]$. The area is $\sqrt{36+121+4}=\sqrt{161}$
|
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|
college_math.linear_algebra
|
\left[\begin{array}{l}1 \\ 0 \\ 3\end{array}\right] \times\left[\begin{array}{r}4 \\ -2 \\ 1\end{array}\right]=\left[\begin{array}{r}6 \\ 11 \\ -2\end{array}\right]$. The area is $\sqrt{36+121+4}=\sqrt{161}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.2.12
|
Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & 5 & 3\end{array}\right]^{T}$.
|
$$
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.17
|
Suppose $A \vec{x}=\vec{b}$ has a solution. Explain why the solution is unique precisely when $A \vec{x}=\overrightarrow{0}$ has only the trivial solution.
|
If not, then there would be a infintely many solutions to $A \vec{x}=\overrightarrow{0}$ and each of these added to a solution to $A \vec{x}=\vec{b}$ would be a solution to $A \vec{x}=\vec{b}$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
If not, then there would be a infintely many solutions to $A \vec{x}=\overrightarrow{0}$ and each of these added to a solution to $A \vec{x}=\vec{b}$ would be a solution to $A \vec{x}=\vec{b}$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.2
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$.
$$
A=\left[\begin{array}{rrr}
4 & 1 & -2 \\
1 & 4 & -2 \\
-2 & -2 & 7
\end{array}\right]
$$
Hint: One eigenvalue is 3.
|
The eigenvectors and eigenvalues are:
$$
\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right], \frac{1}{\sqrt{3}}\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\right\} \leftrightarrow 3,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}
-1 \\
-1 \\
2
\end{array}\right]\right\} \leftrightarrow 9
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right], \frac{1}{\sqrt{3}}\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\right\} \leftrightarrow 3,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}
-1 \\
-1 \\
2
\end{array}\right]\right\} \leftrightarrow 9
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.1
|
Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{lll}
1 & -1 & 2 \\
1 & -2 & 1 \\
3 & -4 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
$$
|
Solution is: $\left[\begin{array}{r}-3 \hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right], \hat{t}_{3} \in \mathbb{R}$. A basis for the solution space is $\left[\begin{array}{r}-3 \\ -1 \\ 1\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{r}-3 \hat{t} \\ -\hat{t} \\ \hat{t}\end{array}\right], \hat{t}_{3} \in \mathbb{R}$. A basis for the solution space is $\left[\begin{array}{r}-3 \\ -1 \\ 1\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.1.1
|
If $A$ is an invertible $n \times n$ matrix, compare the eigenvalues of $A$ and $A^{-1}$. More generally, for $m$ an arbitrary integer, compare the eigenvalues of $A$ and $A^{m}$.
|
$A^{m} X=\lambda^{m} X$ for any integer. In the case of $-1, A^{-1} \lambda X=A A^{-1} X=X$ so $A^{-1} X=\lambda^{-1} X$. Thus the eigenvalues of $A^{-1}$ are just $\lambda^{-1}$ where $\lambda$ is an eigenvalue of $A$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
A^{m} X=\lambda^{m} X$ for any integer. In the case of $-1, A^{-1} \lambda X=A A^{-1} X=X$ so $A^{-1} X=\lambda^{-1} X$. Thus the eigenvalues of $A^{-1}$ are just $\lambda^{-1}$ where $\lambda$ is an eigenvalue of $A$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.12
|
What does it mean geometrically if the box product of three vectors gives zero?
|
It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
It means that if you place them so that they all have their tails at the same point, the three will lie in the same plane.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.10
|
Find the volume of the parallelepiped determined by the vectors $\left[\begin{array}{r}1 \\ -7 \\ -5\end{array}\right]$,
$\left[\begin{array}{r}1 \\ -2 \\ -6\end{array}\right]$, and $\left[\begin{array}{l}3 \\ 2 \\ 3\end{array}\right]$
|
$\left|\begin{array}{rrr}1 & -7 & -5\end{array}\right|=113$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left|\begin{array}{rrr}1 & -7 & -5\end{array}\right|=113
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.9
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{ccc}
\frac{4}{3} & \frac{1}{3} \sqrt{3} \sqrt{2} & \frac{1}{3} \sqrt{2} \\
\frac{1}{3} \sqrt{3} \sqrt{2} & 1 & -\frac{1}{3} \sqrt{3} \\
\frac{1}{3} \sqrt{2} & -\frac{1}{3} \sqrt{3} & \frac{5}{3}
\end{array}\right]
$$
Hint: The eigenvalues are 0,2,2 where 2 is listed twice because it is a root of multiplicity 2.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{5} \sqrt{2} \sqrt{5} \\
\frac{1}{5} \sqrt{3} \sqrt{5} \\
\frac{1}{5} \sqrt{5}
\end{array}\right]\right\} \leftrightarrow 0,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right],\left[\begin{array}{c}
\frac{1}{5} \sqrt{2} \sqrt{5} \\
\frac{1}{5} \sqrt{3} \sqrt{5} \\
-\frac{1}{5} \sqrt{5}
\end{array}\right]\right\} \leftrightarrow 2
$$
The columns are these vectors.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{5} \sqrt{2} \sqrt{5} \\
\frac{1}{5} \sqrt{3} \sqrt{5} \\
\frac{1}{5} \sqrt{5}
\end{array}\right]\right\} \leftrightarrow 0,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right],\left[\begin{array}{c}
\frac{1}{5} \sqrt{2} \sqrt{5} \\
\frac{1}{5} \sqrt{3} \sqrt{5} \\
-\frac{1}{5} \sqrt{5}
\end{array}\right]\right\} \leftrightarrow 2
$$
The columns are these vectors.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.2.13
|
Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & 0 & 3\end{array}\right]^{T}$.
|
$$
\frac{1}{10}\left[\begin{array}{lll}
1 & 0 & 3 \\
0 & 0 & 0 \\
3 & 0 & 9
\end{array}\right]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\frac{1}{10}\left[\begin{array}{lll}
1 & 0 & 3 \\
0 & 0 & 0 \\
3 & 0 & 9
\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.1.21
|
Consider functions defined on $\{1,2, \cdots, n\}$ having values in $\mathbb{R}$. Explain how, if $V$ is the set of all such functions, $V$ can be considered as $\mathbb{R}^{n}$.
|
Let $f(i)$ be the $i^{t h}$ component of a vector $\vec{x} \in \mathbb{R}^{n}$. Thus a typical element in $\mathbb{R}^{n}$ is $(f(1), \cdots, f(n))$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Let $f(i)$ be the $i^{t h}$ component of a vector $\vec{x} \in \mathbb{R}^{n}$. Thus a typical element in $\mathbb{R}^{n}$ is $(f(1), \cdots, f(n))$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.45
|
The steady state temperature, $u$, of a plate solves Laplace's equation, $\Delta u=0$. One way to approximate the solution is to divide the plate into a square mesh and require the temperature at each node to equal the average of the temperature at the four adjacent nodes. In the following picture, the numbers represent the observed temperature at the indicated nodes. Find the temperature at the interior nodes, indicated by $x, y, z$, and $w$. One of the equations is $z=\frac{1}{4}(10+0+w+x)$.
|
Solution is: $[w=15, x=15, y=20, z=10]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[w=15, x=15, y=20, z=10]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.7.12
|
Find $\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1 \\ 2 \\ -2 \\ 1\end{array}\right]$ and $\vec{v}=\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 0\end{array}\right]$.
|
$\overrightarrow{\vec{u} \bullet \vec{v}} \overrightarrow{\vec{u}} \vec{u}=\frac{\left[\begin{array}{llll}1 & 2 & -2 & 1\end{array}\right]^{T} \bullet\left[\begin{array}{llll}1 & 2 & 3 & 0\end{array}\right]^{T}}{1+4+9}\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 0\end{array}\right]=\left[\begin{array}{r}-\frac{1}{14} \\ -\frac{1}{7} \\ -\frac{3}{14} \\ 0\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\overrightarrow{\vec{u} \bullet \vec{v}} \overrightarrow{\vec{u}} \vec{u}=\frac{\left[\begin{array}{llll}1 & 2 & -2 & 1\end{array}\right]^{T} \bullet\left[\begin{array}{llll}1 & 2 & 3 & 0\end{array}\right]^{T}}{1+4+9}\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 0\end{array}\right]=\left[\begin{array}{r}-\frac{1}{14} \\ -\frac{1}{7} \\ -\frac{3}{14} \\ 0\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.24
|
Find the determinant using row operations to first simplify.
$$
\left|\begin{array}{rrrr}
1 & 2 & 1 & 2 \\
3 & 1 & -2 & 3 \\
-1 & 0 & 3 & 1 \\
2 & 3 & 2 & -2
\end{array}\right|
$$
|
One can row reduce this using only row operation 3 to
$$
\left[\begin{array}{rrrr}
1 & 2 & 1 & 2 \\
0 & -5 & -5 & -3 \\
0 & 0 & 2 & \frac{9}{5} \\
0 & 0 & 0 & -\frac{63}{10}
\end{array}\right]
$$
and therefore, the determinant is -63.
$$
\left|\begin{array}{rrrr}
1 & 2 & 1 & 2 \\
3 & 1 & -2 & 3 \\
-1 & 0 & 3 & 1 \\
2 & 3 & 2 & -2
\end{array}\right|=63
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
One can row reduce this using only row operation 3 to
$$
\left[\begin{array}{rrrr}
1 & 2 & 1 & 2 \\
0 & -5 & -5 & -3 \\
0 & 0 & 2 & \frac{9}{5} \\
0 & 0 & 0 & -\frac{63}{10}
\end{array}\right]
$$
and therefore, the determinant is -63.
$$
\left|\begin{array}{rrrr}
1 & 2 & 1 & 2 \\
3 & 1 & -2 & 3 \\
-1 & 0 & 3 & 1 \\
2 & 3 & 2 & -2
\end{array}\right|=63
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.7
|
Is $\vec{u} \times(\vec{v} \times \vec{w})=(\vec{u} \times \vec{v}) \times \vec{w}$ ? What is the meaning of $\vec{u} \times \vec{v} \times \vec{w}$ ? Explain. Hint: Try $(\vec{i} \times \vec{j}) \times \vec{k}$
|
$(\vec{i} \times \vec{j}) \times \vec{j}=\vec{k} \times \vec{j}=-\vec{i}$. However, $\vec{i} \times(\vec{j} \times \vec{j})=\overrightarrow{0}$ and so the cross product is not associative.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
(\vec{i} \times \vec{j}) \times \vec{j}=\vec{k} \times \vec{j}=-\vec{i}$. However, $\vec{i} \times(\vec{j} \times \vec{j})=\overrightarrow{0}$ and so the cross product is not associative.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.5
|
Find the following determinant by expanding along the first column and third row.
$$
\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|
$$
|
$$
\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|=2
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left|\begin{array}{lll}
1 & 2 & 1 \\
1 & 0 & 1 \\
2 & 1 & 1
\end{array}\right|=2
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.9.12
|
Find the matrix for $T(\vec{w})=\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{v}=\left[\begin{array}{lll}1 & 5 & 3\end{array}\right]^{T}$.
|
$$
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\frac{1}{35}\left[\begin{array}{rrr}
1 & 5 & 3 \\
5 & 25 & 15 \\
3 & 15 & 9
\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.29
|
Find the solution of the system whose augmented matrix is
$$
\left[\begin{array}{lll|l}
1 & 1 & 0 & 1 \\
1 & 0 & 4 & 2
\end{array}\right]
$$
|
The solution is $z=t, y=4 t, x=2-4 t$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The solution is $z=t, y=4 t, x=2-4 t$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.3
|
Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{lll}
0 & -1 & 2 \\
1 & -2 & 1 \\
1 & -4 & 5
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0
\end{array}\right]
$$
|
Solution is: $\left[\begin{array}{c}3 \hat{t} \\ 2 \hat{t} \\ \hat{t}\end{array}\right]$, A basis is $\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{c}3 \hat{t} \\ 2 \hat{t} \\ \hat{t}\end{array}\right]$, A basis is $\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.36
|
Find the solution to the system of equations, $8 x+2 y+3 z=-3,8 x+3 y+3 z=-1$, and $4 x+y+3 z=-9$.
|
Solution is: $[x=1, y=2, z=-5]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=1, y=2, z=-5]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.7.1
|
Find $\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \bullet\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]$.
|
$\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \bullet\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]=17$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{l}1 \\ 2 \\ 3 \\ 4\end{array}\right] \bullet\left[\begin{array}{l}2 \\ 0 \\ 1 \\ 3\end{array}\right]=17
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.18
|
A boy drags a sled for 100 feet along the ground by pulling on a rope which is 20 degrees from the horizontal with a force of 40 pounds. How much work does this force do?
|
$40 \cos \left(\frac{20}{180} \pi\right) 100=3758.8$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
40 \cos \left(\frac{20}{180} \pi\right) 100=3758.8
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.12
|
Find $h$ such that
$$
\left[\begin{array}{ll|l}
2 & h & 4 \\
3 & 6 & 7
\end{array}\right]
$$
is the augmented matrix of an inconsistent system.
|
$h=4$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
h=4
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.2.1.40
|
Let
$$
A=\left[\begin{array}{lll}
1 & 2 & 3 \\
2 & 1 & 4 \\
1 & 0 & 2
\end{array}\right]
$$
Find $A^{-1}$ if possible. If $A^{-1}$ does not exist, explain why.
|
$\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2\end{array}\right]^{-1}=\left[\begin{array}{rrr}-2 & 4 & -5 \\ 0 & 1 & -2 \\ 1 & -2 & 3\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 1 & 4 \\ 1 & 0 & 2\end{array}\right]^{-1}=\left[\begin{array}{rrr}-2 & 4 & -5 \\ 0 & 1 & -2 \\ 1 & -2 & 3\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.7
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{ccc}
3 & 0 & 0 \\
0 & \frac{3}{2} & \frac{1}{2} \\
0 & \frac{1}{2} & \frac{3}{2}
\end{array}\right]
$$
|
The eigenvectors and eigenvalues are: $\left\{\left[\begin{array}{c}0 \\ -\frac{1}{2} \sqrt{2} \\ \frac{1}{2} \sqrt{2}\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}0 \\ \frac{1}{2} \sqrt{2} \\ \frac{1}{2} \sqrt{2}\end{array}\right]\right\} \leftrightarrow 2,\left\{\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\right\} \leftrightarrow 3$.
These vectors are the columns of the matrix $U$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are: $\left\{\left[\begin{array}{c}0 \\ -\frac{1}{2} \sqrt{2} \\ \frac{1}{2} \sqrt{2}\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}0 \\ \frac{1}{2} \sqrt{2} \\ \frac{1}{2} \sqrt{2}\end{array}\right]\right\} \leftrightarrow 2,\left\{\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]\right\} \leftrightarrow 3$.
These vectors are the columns of the matrix $U$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.22
|
Find the determinant using row operations to first simplify.
$$
\left|\begin{array}{rrr}
1 & 2 & 1 \\
2 & 3 & 2 \\
-4 & 1 & 2
\end{array}\right|
$$
|
$$
\left|\begin{array}{rrr}
1 & 2 & 1 \\
2 & 3 & 2 \\
-4 & 1 & 2
\end{array}\right|=-6
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left|\begin{array}{rrr}
1 & 2 & 1 \\
2 & 3 & 2 \\
-4 & 1 & 2
\end{array}\right|=-6
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.23
|
Find the determinant using row operations to first simplify.
$$
\left|\begin{array}{rrr}
2 & 1 & 3 \\
2 & 4 & 2 \\
1 & 4 & -5
\end{array}\right|
$$
|
$$
\left|\begin{array}{rrr}
2 & 1 & 3 \\
2 & 4 & 2 \\
1 & 4 & -5
\end{array}\right|=-32
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left|\begin{array}{rrr}
2 & 1 & 3 \\
2 & 4 & 2 \\
1 & 4 & -5
\end{array}\right|=-32
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.33
|
Find the solution to the system of equations, $3 x-y+4 z=6, y+8 z=0$, and $-2 x+y=-4$.
|
Solution is: $[x=2-4 t, y=-8 t, z=t]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=2-4 t, y=-8 t, z=t]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.2.7
|
Consider the matrix
$$
A=\left[\begin{array}{ccc}
1 & 0 & 0 \\
0 & \cos t & -\sin t \\
0 & \sin t & \cos t
\end{array}\right]
$$
Does there exist a value of t for which this matrix fails to have an inverse? Explain.
|
No. It has a nonzero determinant for all $t$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
No. It has a nonzero determinant for all $t
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.11
|
Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix
$$
A=\left[\begin{array}{ccc}
\frac{1}{3} & \frac{1}{6} \sqrt{3} \sqrt{2} & \frac{1}{6} \sqrt{3} \sqrt{6} \\
\frac{1}{6} \sqrt{3} \sqrt{2} & \frac{3}{2} & \frac{1}{12} \sqrt{2} \sqrt{6} \\
\frac{1}{6} \sqrt{3} \sqrt{6} & \frac{1}{12} \sqrt{2} \sqrt{6} & \frac{7}{6}
\end{array}\right]
$$
Hint: The eigenvalues are 1,2,-2.
|
eigenvectors:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow-2,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
-\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 2 .
$$
Then the columns of $U$ are these vectors
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
eigenvectors:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow-2,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
-\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 2 .
$$
Then the columns of $U$ are these vectors
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.34
|
Find the solution to the system of equations, $9 x-2 y+4 z=-17,13 x-3 y+6 z=-25$, and $-2 x-z=3$.
|
Solution is: $[x=-1, y=2, z=-1]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=-1, y=2, z=-1]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.6.3.1
|
Give the complete solution to $x^{4}+16=0$.
|
Solution is:
$$
(1-i) \sqrt{2},-(1+i) \sqrt{2},-(1-i) \sqrt{2},(1+i) \sqrt{2}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is:
$$
(1-i) \sqrt{2},-(1+i) \sqrt{2},-(1-i) \sqrt{2},(1+i) \sqrt{2}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.1.2
|
If $A$ is an $n \times n$ matrix and $c$ is a nonzero constant, compare the eigenvalues of $A$ and $c A$.
|
Say $A X=\lambda X$. Then $c A X=c \lambda X$ and so the eigenvalues of $c A$ are just $c \lambda$ where $\lambda$ is an eigenvalue of $A$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Say $A X=\lambda X$. Then $c A X=c \lambda X$ and so the eigenvalues of $c A$ are just $c \lambda$ where $\lambda$ is an eigenvalue of $A$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.10
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$. Diagonalize $A$ by finding an orthogonal matrix $U$ and a diagonal matrix $D$ such that $U^{T} A U=D$.
$$
A=\left[\begin{array}{ccc}
1 & \frac{1}{6} \sqrt{3} \sqrt{2} & \frac{1}{6} \sqrt{3} \sqrt{6} \\
\frac{1}{6} \sqrt{3} \sqrt{2} & \frac{3}{2} & \frac{1}{12} \sqrt{2} \sqrt{6} \\
\frac{1}{6} \sqrt{3} \sqrt{6} & \frac{1}{12} \sqrt{2} \sqrt{6} & \frac{1}{2}
\end{array}\right]
$$
Hint: The eigenvalues are 2, 1,0.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 0,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
-\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 2
$$
The columns are these vectors.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{3} \sqrt{3} \\
0 \\
\frac{1}{3} \sqrt{2} \sqrt{3}
\end{array}\right]\right\} \leftrightarrow 0,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
-\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
\frac{1}{3} \sqrt{3} \\
\frac{1}{2} \sqrt{2} \\
\frac{1}{6} \sqrt{6}
\end{array}\right]\right\} \leftrightarrow 2
$$
The columns are these vectors.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.40
|
Find the solution to the system of equations, $-19 x+8 y=-108,-71 x+30 y=-404$, $-2 x+y=-12,4 x+z=14$.
|
Solution is: $[x=4, y=-4, z=-2]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=4, y=-4, z=-2]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.2.2
|
Find the eigenvalues and eigenvectors of the matrix
$$
\left[\begin{array}{rrr}
-13 & -28 & 28 \\
4 & 9 & -8 \\
-4 & -8 & 9
\end{array}\right]
$$
One eigenvalue is 3. Diagonalize if possible.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-2 \\
1 \\
0
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
7 \\
-2 \\
2
\end{array}\right]\right\} \leftrightarrow 3
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
2 & -2 & 7 \\
0 & 1 & -2 \\
1 & 0 & 2
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 3
\end{array}\right]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{l}
2 \\
0 \\
1
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
-2 \\
1 \\
0
\end{array}\right]\right\} \leftrightarrow 1,\left\{\left[\begin{array}{c}
7 \\
-2 \\
2
\end{array}\right]\right\} \leftrightarrow 3
$$
The matrix $P$ needed to diagonalize the above matrix is
$$
\left[\begin{array}{rrr}
2 & -2 & 7 \\
0 & 1 & -2 \\
1 & 0 & 2
\end{array}\right]
$$
and the diagonal matrix $D$ is
$$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 3
\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.3.1.13
|
An operation is done to get from the first matrix to the second. Identify what was done and tell how it will affect the value of the determinant.
$$
\left[\begin{array}{ll}
a & b \\
c & d
\end{array}\right] \rightarrow \cdots \rightarrow\left[\begin{array}{ll}
b & a \\
d & c
\end{array}\right]
$$
|
In this case the two columns were switched so the determinant of the second is -1 times the determinant of the first.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
In this case the two columns were switched so the determinant of the second is -1 times the determinant of the first.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.7
|
The wind blows from West to East at a speed of 50 miles per hour and an airplane which travels at 400 miles per hour in still air is heading North West. What is the velocity of the airplane relative to the ground? What is the component of this velocity in the direction North?
|
The velocity is the sum of two vectors. $50 \vec{i}+\frac{300}{\sqrt{2}}(\vec{i}+\vec{j})=\left(50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}} \vec{j}$. The component in the direction of North is then $\frac{300}{\sqrt{2}}=150 \sqrt{2}$ and the velocity relative to the ground is $\left(50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}} \vec{j}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The velocity is the sum of two vectors. $50 \vec{i}+\frac{300}{\sqrt{2}}(\vec{i}+\vec{j})=\left(50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}} \vec{j}$. The component in the direction of North is then $\frac{300}{\sqrt{2}}=150 \sqrt{2}$ and the velocity relative to the ground is $\left(50+\frac{300}{\sqrt{2}}\right) \vec{i}+\frac{300}{\sqrt{2}} \vec{j}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.1
|
Find the eigenvalues and an orthonormal basis of eigenvectors for $A$.
$$
A=\left[\begin{array}{rrr}
11 & -1 & -4 \\
-1 & 11 & -4 \\
-4 & -4 & 14
\end{array}\right]
$$
Hint: Two eigenvalues are 12 and 18.
|
The eigenvectors and eigenvalues are:
$$
\left\{\frac{1}{\sqrt{3}}\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\right\} \leftrightarrow 6,\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right]\right\} \leftrightarrow 12,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}
-1 \\
-1 \\
2
\end{array}\right]\right\} \leftrightarrow 18
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\frac{1}{\sqrt{3}}\left[\begin{array}{l}
1 \\
1 \\
1
\end{array}\right]\right\} \leftrightarrow 6,\left\{\frac{1}{\sqrt{2}}\left[\begin{array}{r}
-1 \\
1 \\
0
\end{array}\right]\right\} \leftrightarrow 12,\left\{\frac{1}{\sqrt{6}}\left[\begin{array}{r}
-1 \\
-1 \\
2
\end{array}\right]\right\} \leftrightarrow 18
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.41
|
Suppose a system of equations has fewer equations than variables and you have found a solution to this system of equations. Is it possible that your solution is the only one? Explain.
|
No. Consider $x+y+z=2$ and $x+y+z=1$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
No. Consider $x+y+z=2$ and $x+y+z=1$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.3.29
|
If you have 5 vectors in $\mathbb{R}^{5}$ and the vectors are linearly independent, can it always be concluded they span $\mathbb{R}^{5}$ ?
|
Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Yes. If not, there would exist a vector not in the span. But then you could add in this vector and obtain a linearly independent set of vectors with more vectors than a basis.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.9.11
|
Write the solution set of the following system as a linear combination of vectors
$$
\left[\begin{array}{llll}
1 & 1 & 0 & 1 \\
2 & 1 & 1 & 2 \\
1 & 0 & 1 & 1 \\
0 & 0 & 0 & 0
\end{array}\right]\left[\begin{array}{l}
x \\
y \\
z \\
w
\end{array}\right]=\left[\begin{array}{l}
0 \\
0 \\
0 \\
0
\end{array}\right]
$$
|
Solution is: $\left[\begin{array}{c}-s-t \\ s \\ s \\ t\end{array}\right], s, t \in \mathbb{R}$. A basis is
$$
\left\{\left[\begin{array}{r}
-1 \\
1 \\
1 \\
0
\end{array}\right],\left[\begin{array}{r}
-1 \\
0 \\
0 \\
1
\end{array}\right]\right\}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{c}-s-t \\ s \\ s \\ t\end{array}\right], s, t \in \mathbb{R}$. A basis is
$$
\left\{\left[\begin{array}{r}
-1 \\
1 \\
1 \\
0
\end{array}\right],\left[\begin{array}{r}
-1 \\
0 \\
0 \\
1
\end{array}\right]\right\}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.1
|
Find the point $\left(x_{1}, y_{1}\right)$ which lies on both lines, $x+3 y=1$ and $4 x-y=3$.
|
Solution is: $\left[x=\frac{10}{13}, y=\frac{1}{13}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[x=\frac{10}{13}, y=\frac{1}{13}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.6.3.3
|
Find the four fourth roots of 16.
|
Solution is:
$$
(1-i) \sqrt{2},-(1+i) \sqrt{2},-(1-i) \sqrt{2},(1+i) \sqrt{2}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is:
$$
(1-i) \sqrt{2},-(1+i) \sqrt{2},-(1-i) \sqrt{2},(1+i) \sqrt{2}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.4.2
|
Let $M=\left\{\vec{u}=\left(u_{1}, u_{2}, u_{3}, u_{4}\right) \in \mathbb{R}^{4}: \sin \left(u_{1}\right)=1\right\}$. Is $M$ a subspace of $\mathbb{R}^{4}$ ?
|
This is not a subspace.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
This is not a subspace.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.3
|
Find the area of the triangle determined by the three points, $(1,0,3),(4,1,0)$ and $(-3,1,1)$.
|
$\left[\begin{array}{r}3 \\ 1 \\ -3\end{array}\right] \times\left[\begin{array}{r}-4 \\ 1 \\ -2\end{array}\right]=\left[\begin{array}{c}1 \\ 18 \\ 7\end{array}\right]$. The area is given by
$$
\frac{1}{2} \sqrt{1+(18)^{2}+49}=\frac{1}{2} \sqrt{374}
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{r}3 \\ 1 \\ -3\end{array}\right] \times\left[\begin{array}{r}-4 \\ 1 \\ -2\end{array}\right]=\left[\begin{array}{c}1 \\ 18 \\ 7\end{array}\right]$. The area is given by
$$
\frac{1}{2} \sqrt{1+(18)^{2}+49}=\frac{1}{2} \sqrt{374}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.8.1
|
Let $V=\mathbb{R}^{3}$ and let
$$
W=\operatorname{span}(S), \text { where } S=\left\{\left[\begin{array}{r}
1 \\
-1 \\
1
\end{array}\right],\left[\begin{array}{r}
-2 \\
2 \\
-2
\end{array}\right],\left[\begin{array}{r}
-1 \\
1 \\
1
\end{array}\right],\left[\begin{array}{r}
1 \\
-1 \\
3
\end{array}\right]\right\}
$$
Find a basis of $W$ consisting of vectors in $S$.
|
In this case $\operatorname{dim}(W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (nonzero) vector from $S$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
In this case $\operatorname{dim}(W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (nonzero) vector from $S$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.9.9.15
|
Let $B=\left\{\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right],\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right],\left[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\right]\right\}$ be a basis of $\mathbb{R}^{3}$ and let $\vec{x}=\left[\begin{array}{r}5 \\ -1 \\ 4\end{array}\right]$ be a vector in $\mathbb{R}^{2}$. Find $C_{B}(\vec{x})$.
|
$C_{B}(\vec{x})=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
C_{B}(\vec{x})=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.11.21
|
Find the least squares solution to the following system.
$$
\begin{gathered}
x+2 y=1 \\
2 x+3 y=2 \\
3 x+5 y=4
\end{gathered}
$$
|
Solution is: $\left[\begin{array}{c}\frac{2}{3} \\ \frac{1}{3}\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $\left[\begin{array}{c}\frac{2}{3} \\ \frac{1}{3}\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.7.8
|
Find the angle between the vectors
$$
\vec{u}=\left[\begin{array}{r}
3 \\
-1 \\
-1
\end{array}\right], \vec{v}=\left[\begin{array}{l}
1 \\
4 \\
2
\end{array}\right]
$$
|
$\frac{\left[\begin{array}{lll}3 & -1 & -1\end{array}\right]^{T} \cdot\left[\begin{array}{lll}1 & 4 & 2\end{array}\right]^{T}}{\sqrt{9+1+1} \sqrt{1+16+4}}=\frac{-3}{\sqrt{11} \sqrt{21}}=-0.19739=\cos \theta$
Therefore we need to solve
$$
-0.19739=\cos \theta
$$
Thus $\theta=1.7695$ radians.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\frac{\left[\begin{array}{lll}3 & -1 & -1\end{array}\right]^{T} \cdot\left[\begin{array}{lll}1 & 4 & 2\end{array}\right]^{T}}{\sqrt{9+1+1} \sqrt{1+16+4}}=\frac{-3}{\sqrt{11} \sqrt{21}}=-0.19739=\cos \theta$
Therefore we need to solve
$$
-0.19739=\cos \theta
$$
Thus $\theta=1.7695$ radians.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.7.1
|
Let $V=\mathbb{R}^{3}$ and let
$$
W=\operatorname{span}(S), \text { where } S=\left\{\left[\begin{array}{r}
1 \\
-1 \\
1
\end{array}\right],\left[\begin{array}{r}
-2 \\
2 \\
-2
\end{array}\right],\left[\begin{array}{r}
-1 \\
1 \\
1
\end{array}\right],\left[\begin{array}{r}
1 \\
-1 \\
3
\end{array}\right]\right\}
$$
Find a basis of $W$ consisting of vectors in $S$.
|
In this case $\operatorname{dim}(W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (nonzero) vector from $S$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
In this case $\operatorname{dim}(W)=1$ and a basis for $W$ consisting of vectors in $S$ can be obtained by taking any (nonzero) vector from $S$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.15
|
Simplify $(\vec{u} \times \vec{v}) \bullet(\vec{v} \times \vec{w}) \times(\vec{w} \times \vec{z})$.
|
Here $[\vec{v}, \vec{w}, \vec{z}]$ denotes the box product. Consider the cross product term. From the above,
$$
\begin{aligned}
(\vec{v} \times \vec{w}) \times(\vec{w} \times \vec{z}) & =[\vec{v}, \vec{w}, \vec{z}] \vec{w}-[\vec{w}, \vec{w}, \vec{z}] \vec{v} \\
& =[\vec{v}, \vec{w}, \vec{z}] \vec{w}
\end{aligned}
$$
Thus it reduces to
$$
(\vec{u} \times \vec{v}) \bullet[\vec{v}, \vec{w}, \vec{z}] \vec{w}=[\vec{v}, \vec{w}, \vec{z}][\vec{u}, \vec{v}, \vec{w}]
$$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Here $[\vec{v}, \vec{w}, \vec{z}]$ denotes the box product. Consider the cross product term. From the above,
$$
\begin{aligned}
(\vec{v} \times \vec{w}) \times(\vec{w} \times \vec{z}) & =[\vec{v}, \vec{w}, \vec{z}] \vec{w}-[\vec{w}, \vec{w}, \vec{z}] \vec{v} \\
& =[\vec{v}, \vec{w}, \vec{z}] \vec{w}
\end{aligned}
$$
Thus it reduces to
$$
(\vec{u} \times \vec{v}) \bullet[\vec{v}, \vec{w}, \vec{z}] \vec{w}=[\vec{v}, \vec{w}, \vec{z}][\vec{u}, \vec{v}, \vec{w}]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.5.8.2
|
Let $B=\left\{\left[\begin{array}{r}1 \\ -1 \\ 2\end{array}\right],\left[\begin{array}{l}2 \\ 1 \\ 2\end{array}\right],\left[\begin{array}{r}-1 \\ 0 \\ 2\end{array}\right]\right\}$ be a basis of $\mathbb{R}^{3}$ and let $\vec{x}=\left[\begin{array}{r}5 \\ -1 \\ 4\end{array}\right]$ be a vector in $\mathbb{R}^{2}$. Find $C_{B}(\vec{x})$.
|
$C_{B}(\vec{x})=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
C_{B}(\vec{x})=\left[\begin{array}{r}2 \\ 1 \\ -1\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.59
|
Consider the system $-5 x+2 y-z=0$ and $-5 x-2 y-z=0$. Both equations equal zero and so $-5 x+2 y-z=-5 x-2 y-z$ which is equivalent to $y=0$. Does it follow that $x$ and $z$ can equal anything? Notice that when $x=1, z=-4$, and $y=0$ are plugged in to the equations, the equations do not equal 0 . Why?
|
These are not legitimate row operations. They do not preserve the solution set of the system.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
These are not legitimate row operations. They do not preserve the solution set of the system.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.7.10
|
Find $\operatorname{proj}_{\vec{v}}(\vec{w})$ where $\vec{w}=\left[\begin{array}{r}1 \\ 0 \\ -2\end{array}\right]$ and $\vec{v}=\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$.
|
$\frac{\vec{u} \bullet \vec{v}}{\vec{u} \bullet \vec{u}} \vec{u}=\frac{-5}{14}\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]=\left[\begin{array}{r}-\frac{5}{14} \\ -\frac{5}{7} \\ -\frac{15}{14}\end{array}\right]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\frac{\vec{u} \bullet \vec{v}}{\vec{u} \bullet \vec{u}} \vec{u}=\frac{-5}{14}\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]=\left[\begin{array}{r}-\frac{5}{14} \\ -\frac{5}{7} \\ -\frac{15}{14}\end{array}\right]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.7.4.12
|
Find the eigenvalues and an orthonormal basis of eigenvectors for the matrix
$$
A=\left[\begin{array}{ccc}
-\frac{1}{2} & -\frac{1}{5} \sqrt{6} \sqrt{5} & \frac{1}{10} \sqrt{5} \\
-\frac{1}{5} \sqrt{6} \sqrt{5} & \frac{7}{5} & -\frac{1}{5} \sqrt{6} \\
\frac{1}{10} \sqrt{5} & -\frac{1}{5} \sqrt{6} & -\frac{9}{10}
\end{array}\right]
$$
Hint: The eigenvalues are $-1,2,-1$ where -1 is listed twice because it has multiplicity 2 as a zero of the characteristic equation.
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
0 \\
\frac{1}{6} \sqrt{5} \sqrt{6}
\end{array}\right],\left[\begin{array}{c}
\frac{1}{3} \sqrt{2} \sqrt{3} \\
\frac{1}{5} \sqrt{5} \\
\frac{1}{15} \sqrt{2} \sqrt{15}
\end{array}\right]\right\} \leftrightarrow-1,\left\{\left[\begin{array}{c}
\frac{1}{6} \sqrt{6} \\
-\frac{2}{5} \sqrt{5} \\
\frac{1}{30} \sqrt{30}
\end{array}\right]\right\} \leftrightarrow 2 .
$$
The columns of $U$ are these vectors.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
The eigenvectors and eigenvalues are:
$$
\left\{\left[\begin{array}{c}
-\frac{1}{6} \sqrt{6} \\
0 \\
\frac{1}{6} \sqrt{5} \sqrt{6}
\end{array}\right],\left[\begin{array}{c}
\frac{1}{3} \sqrt{2} \sqrt{3} \\
\frac{1}{5} \sqrt{5} \\
\frac{1}{15} \sqrt{2} \sqrt{15}
\end{array}\right]\right\} \leftrightarrow-1,\left\{\left[\begin{array}{c}
\frac{1}{6} \sqrt{6} \\
-\frac{2}{5} \sqrt{5} \\
\frac{1}{30} \sqrt{30}
\end{array}\right]\right\} \leftrightarrow 2 .
$$
The columns of $U$ are these vectors.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.9.5
|
Find the area of the parallelogram determined by the vectors $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right],\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]$.
|
$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \times\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]=\left[\begin{array}{r}8 \\ 8 \\ -8\end{array}\right]$. The area is $8 \sqrt{3}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right] \times\left[\begin{array}{r}3 \\ -2 \\ 1\end{array}\right]=\left[\begin{array}{r}8 \\ 8 \\ -8\end{array}\right]$. The area is $8 \sqrt{3}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.4.12.24
|
An object moves 20 meters in the direction of $\vec{k}+\vec{j}$. There are two forces acting on this object, $\vec{F}_{1}=\vec{i}+\vec{j}+2 \vec{k}$, and $\vec{F}_{2}=\vec{i}+2 \vec{j}-6 \vec{k}$. Find the total work done on the object by the two forces. Hint: You can take the work done by the resultant of the two forces or you can add the work done by each force. Why?
|
$\left[\begin{array}{r}2 \\ 3 \\ -4\end{array}\right] \bullet\left[\begin{array}{c}0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right] 20=-10 \sqrt{2}$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
\left[\begin{array}{r}2 \\ 3 \\ -4\end{array}\right] \bullet\left[\begin{array}{c}0 \\ \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right] 20=-10 \sqrt{2}
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.2.1.32
|
Suppose $A B=A C$ and $A$ is an invertible $n \times n$ matrix. Does it follow that $B=C$ ? Explain why or why not.
|
Yes $B=C$. Multiply $A B=A C$ on the left by $A^{-1}$.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Yes $B=C$. Multiply $A B=A C$ on the left by $A^{-1}$.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.2.1.14
|
Let $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right], B=\left[\begin{array}{ll}1 & 2 \\ 3 & k\end{array}\right]$. Is it possible to choose $k$ such that $A B=B A$ ? If so, what should $k$ equal?
|
Solution is: $[k=4]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[k=4]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.32
|
Find the solution to the system of equations, $7 x+14 y+15 z=22,2 x+4 y+3 z=5$, and $3 x+6 y+10 z=13$.
|
Solution is: $[x=1-2 t, z=1, y=t]$
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
Solution is: $[x=1-2 t, z=1, y=t]
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.7
|
Consider the following augmented matrix in which $*$ denotes an arbitrary number and denotes a nonzero number. Determine whether the given augmented matrix is consistent. If consistent, is the solution unique?
$$
\left[\begin{array}{ccc|c}
\mathbf{0} & * & * & * \\
0 & \mathbf{\square} & * & * \\
0 & 0 & \mathbf{\square} & *
\end{array}\right]
$$
|
A solution exists and is unique.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
A solution exists and is unique.
|
college_math.A_First_Course_in_Linear_Algebra
|
exercise.1.2.17
|
Determine if the system is consistent. If so, is the solution unique?
$$
\begin{gathered}
x+2 y+z-w=2 \\
x-y+z+w=1 \\
2 x+y-z=1 \\
4 x+2 y+z=5
\end{gathered}
$$
|
There is no solution.
|
Creative Commons License (CC BY)
|
college_math.linear_algebra
|
There is no solution.
|
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