Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
\\( - x = 7 \\) . | Since \\( - - x \\) is actually \\( - 1 \\cdot x \\) and \\( \\left( {-1}\\right) \\left( {-1}\\right) = 1 \\), we can isolate \\( x \\) by multiplying both sides of the\n\nequation by -1 .\n\n\\[ \n\\left( {-1}\\right) \\left( {-x}\\right) = - 1 \\cdot 7 \n\\]\n\n\\[ \nx = - 7 \n\\]\n\nCheck: When \\( x = 7 \\) ,\n\n\... | No |
[{6x} - 4 = - {16}] | -4 is associated with \( x \) by subtraction. Undo the association by adding 4 to both sides.\n\n\[{6x} - 4 + 4 = - {16} + 4\]\n\n\[{6x} = - {12}\]\n\n6 is associated with \( x \) by multiplication. Undo the association by dividing both sides by 6\n\n\[\frac{6x}{6} = \frac{-{12}}{6}\]\n\n\[x = - 2\] | Yes |
\( {5m} - 6 - {4m} = {4m} - 8 + {3m} \) | Begin by solving this equation by combining like terms. \( m - 6 = {7m} - 8 \) Choose a side on which to isolate \( m \). Since 7 is greater than 1, we’ll isolate \( m \) on the right side.\n\nSubtract \( m \) from both sides.\n\n\[ \n- m - 6 - m = {7m} - 8 - m \n\]\n\n\[ \n- 6 = {6m} - 8 \n\]\n\n8 is associated with \... | Yes |
\( \frac{8x}{7} = - 2 \) | 7 is associated with \( x \) by division. Undo the association by multiplying both sides by 7 .\n\n\[ \overset{―}{)7} \cdot \frac{8x}{\overset{―}{)7}} = 7\left( { - 2}\right) \]\n\n\[ 7 \cdot \frac{8x}{7} = - {14} \]\n\n\( {8x} = - {14} \)\n\n8 is associated with \( x \) by multiplication. Undo the association by divid... | Yes |
[ \underset{9}{\underbrace{\text{ Nine }}}\underset{ + }{\underbrace{\text{ more than }}}\underset{x}{\underbrace{\text{ some number }}} ] | Translation: \( 9 + x \) . | Yes |
[
Example 11.36
\[
\underset{18}{\underbrace{\text{ Eighteen }}}\underset{ - }{\underbrace{\text{ minus }}}\underset{x}{\underbrace{\text{ a number }}}
\]
] | [
Translation: \( {18} - x \) .
] | Yes |
\[ \underset{y}{\underbrace{\text{ A quantity less five. }}} \] | Translation: \( y - 5 \) . | Yes |
\( \underset{4}{\underbrace{\text{ Four }}}\underset{.}{\underbrace{\text{ times }}}\underset{x}{\underbrace{\text{ a number }}}\underset{ = }{\underbrace{\text{ is }}}\underset{16}{\underbrace{\text{ sixteen. }}} \) | Translation: \( {4x} = {16} \) | Yes |
[ \underset{\frac{1}{5}}{\underbrace{\text{ One fifth }}}\underset{\text{. }}{\underbrace{\text{ of }}}\underset{n}{\underbrace{\text{ a number }}}\underset{ = }{\underbrace{\text{ is }}}\underset{30}{\underbrace{\text{ thirty. }}} ] | Translation: \( \frac{1}{5}n = {30} \), or \( \frac{n}{5} = {30} \) | Yes |
\[ \underset{5}{\underbrace{\text{ Five }}}\underset{x}{\underbrace{\text{ times }}}\underset{x}{\underbrace{\text{ a number }}}\underset{ = }{\underbrace{\text{ is }}}\underset{2}{\underbrace{\text{ two }}}\underset{ + }{\underbrace{\text{ more than }}}\underset{2 \cdot }{\underbrace{\text{ twice }}}\underset{x}{\unde... | Translation: \( {5x} = 2 + {2x} \) | Yes |
Sometimes the structure of the sentence indicates the use of grouping symbols. We'll be alert for commas. They set off terms. | Translation: \( \frac{x}{4} - 6 = {12} \) | No |
What number decreased by six is five? | Step 1: Let \( n \) represent the unknown number.\n\nStep 2: Translate the words to mathematical symbols and construct an equation. Read phrases. What number: \( n \)\ndecreased by: -\nsix: \( \;6\} n - 6 = 5 \)\nis: \( \; = \)\nfive: \( \;5 \)\n\nStep 3: Solve this equation.\n\n\( n - 6 = 5 \) Add 6 to both sides.\n\n... | Yes |
The sum of three consecutive odd integers is equal to one less than twice the first odd integer. Find the three integers. | Let \( \;n = \) the first odd integer. Then,\nStep 1. \( \;n + 2 = \) the second odd integer, and\n\( n + 4 = \) the third odd integer.\nStep 2. Translate the words to mathematical symbols and construct an equation. Read phrases. The sum of: \( \; \) add some numbers\nthree consecutive odd integers: \( n, n + 2, n + 4 ... | Yes |
The perimeter (length around) of a rectangle is 20 meters. If the length is 4 meters longer than the width, find the length and width of the rectangle. | Step 1: Let \( x = \) the width of the rectangle. Then, \( x + 4 = \) the length of the rectangle. Step 2: The length around the rectangle is \( x + x + 4 + x + x + 4 = 20 \). Step 3: \( 4x + 8 = 20 \) Subtract 8 from both sides. \( 4x = 12 \) Divide both sides by 4. \( x = 3 \) Then, \( x + 4 = 3 + 4 = 7 \). Step 4: C... | Yes |
If we had only one processor working on this task, it is easy to determine the finishing time; just add up the individual times. We assume one person can't work on two tasks at the same time, ignore things like drying times during which someone could work on another task. Scheduling with one processor, a possible sched... | <table><tr><td>Time:</td><td>0</td><td>1</td><td>2</td><td>3</td><td>4</td><td>5</td><td>6</td><td>7</td><td>8</td><td>9</td><td>10</td></tr><tr><td>\( {\mathrm{P}}_{1} \)</td><td>\( {\mathrm{T}}_{1} \)</td><td>|</td><td>\( {\mathrm{T}}_{3} \)</td><td>1</td><td></td><td>\( {\mathbf{T}}_{5} \)</td><td>|</td><td>\( {\mat... | Yes |
Suppose that we have collected weights from 100 male subjects as part of a nutrition study. For our weight data, we have values ranging from a low of 121 pounds to a high of 263 pounds, giving a total span of \( {263} - {121} = {142} \) . We could create 7 intervals with a width of around 20, 14 intervals with a width ... | <table><thead><tr><th>Interval</th><th>Frequency</th></tr></thead><tr><td>120 - 134</td><td>4</td></tr><tr><td>135 – 149</td><td>14</td></tr><tr><td>150 – 164</td><td>16</td></tr><tr><td>165 – 179</td><td>28</td></tr><tr><td>180 – 194</td><td>12</td></tr><tr><td>195 – 209</td><td>8</td></tr><tr><td>210 – 224</td><td>7<... | Yes |
The number of touchdown (TD) passes thrown by each of the 31 teams in the National Football League in the 2000 season are shown below.\n\n37 33 33 32 29 28 28 23 22 22 22 21 21 21 20\n\n201919181818181615141414121296 | Adding these values, we get 634 total TDs. Dividing by 31, the number of data values, we get \( {634}/{31} = {20.4516} \) . It would be appropriate to round this to 20.5 .\n\nIt would be most correct for us to report that \ | No |
Find the median of these quiz scores: 5 10 8 6 4 8 2 5 7 7 | We start by listing the data in order: 2 4 5 5 6 7 7 8 8 10\n\nSince there are 10 data values, an even number, there is no one middle number. So we find the mean of the two middle numbers,6 and 7, and get \( \left( {6 + 7}\right) /2 = {6.5} \) .\n\nThe median quiz score was 6.5. | No |
The box plot below is based on the household income data with 5 number summary: \( {15},{27.5},{35},{40},{50} \) |  | No |
Use a Venn diagram to illustrate \( {\left( H \cap F\right) }^{c} \cap W \) | We’ll start by identifying everything in the set \( H \cap F \)\n\n\n\nNow, \( {\left( H \cap F\right) }^{c} \cap W \) will contain everything not in the set identified above that is also in set \( W \).\n\n![60d71e5... | Yes |
Multiply: \( \left( {2 + {5i}}\right) \left( {4 + i}\right) \) . | \n\n\( \left( {2 + {5i}}\right) \left( {4 + i}\right) \) Expand\n\n\( = 8 + {20i} + {2i} + 5{i}^{2} \) Since \( i = \sqrt{-1},{i}^{2} = - 1 \)\n\n\( = 8 + {20i} + {2i} + 5\left( {-1}\right) \) Simplify\n\n\( = 3 + {22i} \) | Yes |
Visualize the product \( i\left( {1 + {2i}}\right) \) | Multiplying, we'd get\n\n\( i \cdot 1 + i \cdot {2i} \)\n\n\[ \n= i + 2{i}^{2} \n\]\n\n\[ \n= i + 2\left( {-1}\right) \n\]\n\n\[ \n= - 2 + i \n\]\n\nIn this case, the distance from the origin has not changed, but the point has been rotated about the origin, \( {90}^{ \circ } \) counter-clockwise. | Yes |
If I go to the mall, then I'll buy new jeans. If I buy new jeans, I'll buy a shirt to go with it. Conclusion: If I go to the mall, I'll buy a shirt. | Let \( m = \mathrm{I} \) go to the mall, \( j = \mathrm{I} \) buy jeans, and \( s = \mathrm{I} \) buy a shirt. The premises and conclusion can be stated as: Premise: \( \;m \rightarrow j \) Premise: \( \;j \rightarrow s \) Conclusion: \( \;m \rightarrow s \) We can construct a truth table for \( \left\lbrack {\left( {m... | Yes |
If I drop my phone into the swimming pool, my phone will be ruined. | If we let \( d = \mathrm{I} \) drop the phone in the pool and \( r = \) the phone is ruined, then we can represent the argument this way: Premise \( \;d \rightarrow r \) Premise \( \; \sim r \) Conclusion: \( \sim d \) The form of this argument matches what we need to invoke the law of contraposition, so it is a valid ... | Yes |
Example 39\n\nPremise: : I can either drive or take the train.\n\nPremise: I refuse to drive.\n\nConclusion: I will take the train.\n\nIf we let \( d = \mathrm{I} \) drive and \( t = \mathrm{I} \) take the train, then the symbolic representation of the argument\n\nis:\n\nPremise \( \;d \vee t \)\n\nPremise \( \; \sim d... | This argument is valid because it has the form of a disjunctive syllogism. I have two choices, and one of them is not going to happen, so the other one must happen. | Yes |
If I don't buy a boat, I must not have worked hard. | If we let \( h = \) working hard, \( r = \) getting a raise, and \( b = \) buying a boat, then we can represent our argument symbolically: Premise \( \;h \rightarrow r \) Premise \( \;r \rightarrow b \) Conclusion: \( \; \sim b \rightarrow \sim h \) Using the transitive property with the two premises, we can conclude t... | Yes |
Proposition 1.3.10 0 and 1 are unique. Also \( - x \) is unique and \( {x}^{-1} \) is unique. Furthermore, \( {0x} = {x0} = 0 \) and \( - x = \left( {-1}\right) x \) . | Proof: Suppose \( {0}^{\prime } \) is another additive identity. Then\n\n\[ \n{0}^{\prime } = {0}^{\prime } + 0 = 0.\n\]\n\nThus 0 is unique. Say \( {1}^{\prime } \) is another multiplicative identity. Then\n\n\[ \n1 = {1}^{\prime }1 = {1}^{\prime }.\n\]\n\nNow suppose \( y \) acts like the additive inverse of \( x \) ... | Yes |
1. If \( x < y \) and \( y < z \), then \( x < z \) . | First consider \( ▱ \), called the transitive law. Suppose that \( x < y \) and \( y < z \) . Then from the axioms, \( x + y < y + z \) and so, adding \( - y \) to both sides, it follows\n\n\[ x < z \] | No |
Theorem 1.5.4 Let \( r > 0 \) be given. Then if \( n \) is a positive integer,\n\n\[{\left\lbrack r\left( \cos t + i\sin t\right) \right\rbrack }^{n} = {r}^{n}\left( {\cos {nt} + i\sin {nt}}\right) .\] | Proof: It is clear the formula holds if \( n = 1 \) . Suppose it is true for \( n \).\n\n\[{\left\lbrack r\left( \cos t + i\sin t\right) \right\rbrack }^{n + 1} = {\left\lbrack r\left( \cos t + i\sin t\right) \right\rbrack }^{n}\left\lbrack {r\left( {\cos t + i\sin t}\right) }\right\rbrack\]\n\nwhich by induction equal... | Yes |
Corollary 1.5.5 Let \( z \) be a non zero complex number. Then there are always exactly \( k{k}^{th} \) roots of \( z \) in \( \mathbb{C} \) . | Proof: Let \( z = x + {iy} \) and let \( z = \left| z\right| \left( {\cos t + i\sin t}\right) \) be the polar form of the complex number. By De Moivre's theorem, a complex number,\n\n\[ r\left( {\cos \alpha + i\sin \alpha }\right) \]\n\nis a \( {k}^{\text{th }} \) root of \( z \) if and only if\n\n\[ {r}^{k}\left( {\co... | Yes |
Example 1.5.6 Find the three cube roots of \( i \) . | First note that \( i = 1\left( {\cos \left( \frac{\pi }{2}\right) + i\sin \left( \frac{\pi }{2}\right) }\right) \). Using the formula in the proof of the above corollary, the cube roots of \( i \) are\n\n\[ 1\left( {\cos \left( \frac{\left( {\pi /2}\right) + {2l\pi }}{3}\right) + i\sin \left( \frac{\left( {\pi /2}\righ... | Yes |
Example 1.5.7 Factor the polynomial \( {x}^{3} - {27} \) . | First find the cube roots of 27. By the above procedure using De Moivre's theorem, these cube roots are \( 3,3\left( {\frac{-1}{2} + i\frac{\sqrt{3}}{2}}\right) \), and \( 3\left( {\frac{-1}{2} - i\frac{\sqrt{3}}{2}}\right) \). Therefore, \( {x}^{3} + {27} = \n\n\[ \n\left( {x - 3}\right) \left( {x - 3\left( {\frac{-1}... | Yes |
Proposition 1.7.3 Let \( S \) be a nonempty set and suppose \( \sup \left( S\right) \) exists. Then for every \( \delta > 0 \)\n\n\[ S \cap (\sup \left( S\right) - \delta ,\sup \left( S\right) \rbrack \neq \varnothing . \]\n\nIf \( \inf \left( S\right) \) exists, then for every \( \delta > 0 \) ,\n\n\[ S \cap \lbrack \... | Proof: Consider the first claim. If the indicated set equals \( \varnothing \), then \( \sup \left( S\right) - \delta \) is an upper bound for \( S \) which is smaller than \( \sup \left( S\right) \), contrary to the definition of \( \sup \left( S\right) \) as the least upper bound. In the second claim, if the indicate... | Yes |
Theorem 1.8.3 (Mathematical induction) A set \( S \subseteq \mathbb{Z} \), having the property that \( a \in S \) and \( n + 1 \in S \) whenever \( n \in S \) contains all integers \( x \in \mathbb{Z} \) such that \( x \geq a \) . | Proof: Let \( T \equiv \left( {\lbrack a,\infty }\right) \cap \mathbb{Z}) \smallsetminus S \) . Thus \( T \) consists of all integers larger than or equal to \( a \) which are not in \( S \) . The theorem will be proved if \( T = \varnothing \) . If \( T \neq \varnothing \) then by the well ordering principle, there wo... | Yes |
Show that for all \( n \in \mathbb{N},\frac{1}{2} \cdot \frac{3}{4}\cdots \frac{{2n} - 1}{2n} < \frac{1}{\sqrt{{2n} + 1}} \) . | If \( n = 1 \) this reduces to the statement that \( \frac{1}{2} < \frac{1}{\sqrt{3}} \) which is obviously true. Suppose then that the inequality holds for \( n \) . Then\n\n\[ \frac{1}{2} \cdot \frac{3}{4}\cdots \frac{{2n} - 1}{2n} \cdot \frac{{2n} + 1}{{2n} + 2} < \frac{1}{\sqrt{{2n} + 1}}\frac{{2n} + 1}{{2n} + 2} \... | Yes |
Proposition 1.8.6 \( \mathbb{R} \) has the Archimedean property. | Proof: Suppose it is not true. Then there exists \( x \in \mathbb{R} \) and \( a > 0 \) such that \( {na} \leq x \) for all \( n \in \mathbb{N} \) . Let \( S = \{ {na} : n \in \mathbb{N}\} \) . By assumption, this is bounded above by \( x \) . By completeness, it has a least upper bound \( y \) . By Proposition 1.7.3 t... | Yes |
Theorem 1.8.7 Suppose \( x < y \) and \( y - x > 1 \) . Then there exists an integer \( l \in \mathbb{Z} \), such that \( x < l < y \) . If \( x \) is an integer, there is no integer \( y \) satisfying \( x < y < x + 1 \) . | Proof: Let \( x \) be the smallest positive integer. Not surprisingly, \( x = 1 \) but this can be proved. If \( x < 1 \) then \( {x}^{2} < x \) contradicting the assertion that \( x \) is the smallest natural number. Therefore,1 is the smallest natural number. This shows there is no integer, \( y \) , satisfying \( x ... | No |
Theorem 1.8.8 If \( x < y \) then there exists a rational number \( r \) such that \( x < r < y \) . | Proof: Let \( n \in \mathbb{N} \) be large enough that\n\n\[ n\left( {y - x}\right) > 1 \]\n\nThus \( \left( {y - x}\right) \) added to itself \( n \) times is larger than 1 . Therefore,\n\n\[ n\left( {y - x}\right) = {ny} + n\left( {-x}\right) = {ny} - {nx} > 1. \]\n\nIt follows from Theorem 1.8.7 there exists \( m \i... | Yes |
Theorem 1.8.10 Suppose \( 0 < a \) and let \( b \geq 0 \) . Then there exists a unique integer \( p \) and real number \( r \) such that \( 0 \leq r < a \) and \( b = {pa} + r \) . | Proof: Let \( S \equiv \{ n \in \mathbb{N} : {an} > b\} \) . By the Archimedean property this set is nonempty. Let \( p + 1 \) be the smallest element of \( S \) . Then \( {pa} \leq b \) because \( p + 1 \) is the smallest in \( S \) . Therefore,\n\n\[ r \equiv b - {pa} \geq 0. \]\n\nIf \( r \geq a \) then \( b - {pa} ... | Yes |
Theorem 1.9.3 Let \( m, n \) be two positive integers and define\n\n\[ S \equiv \{ {xm} + {yn} \in \mathbb{N} : x, y \in \mathbb{Z}\} . \]\n\nThen the smallest number in \( S \) is the greatest common divisor, denoted by \( \left( {m, n}\right) \) . | Proof: First note that both \( m \) and \( n \) are in \( S \) so it is a nonempty set of positive integers. By well ordering, there is a smallest element of \( S \), called \( p = {x}_{0}m + {y}_{0}n \) . Either \( p \) divides \( m \) or it does not. If \( p \) does not divide \( m \), then by Theorem 1.8.10,\n\n\[ m... | Yes |
Example 1.9.5 Find the greatest common divisor of 165 and 385. | Use the Euclidean algorithm to write\n\n\[ \n{385} = 2\left( {165}\right) + {55} \]\n\nThus the next two numbers are 55 and 165 . Then\n\n\[ \n{165} = 3 \times {55} \]\n\nand so the greatest common divisor of the first two numbers is 55 . | Yes |
Find the greatest common divisor of 1237 and 4322. | Use the Euclidean algorithm\n\n\\[ \n{4322} = 3\\left( {1237}\\right) + {611} \n\\]\n\nNow the two new numbers are 1237,611. Then\n\n\\[ \n{1237} = 2\\left( {611}\\right) + {15} \n\\]\n\nThe two new numbers are 15,611 . Then\n\n\\[ \n{611} = {40}\\left( {15}\\right) + {11} \n\\]\n\nThe two new numbers are 15,11. Then\n... | Yes |
Theorem 1.9.7 If \( p \) is a prime and \( p \mid {ab} \) then either \( p \mid a \) or \( p \mid b \) . | Proof: Suppose \( p \) does not divide \( a \) . Then since \( p \) is prime, the only factors of \( p \) are 1 and \( p \) so follows \( \left( {p, a}\right) = 1 \) and therefore, there exists integers, \( x \) and \( y \) such that\n\n\[ 1 = {ax} + {yp} \]\n\nMultiplying this equation by \( b \) yields\n\n\[ b = {abx... | Yes |
Theorem 1.9.8 (Fundamental theorem of arithmetic) Let \( a \in \mathbb{N} \smallsetminus \{ 1\} \) . Then \( a = \mathop{\prod }\limits_{{i = 1}}^{n}{p}_{i} \) where \( {p}_{i} \) are all prime numbers. Furthermore, this prime factorization is unique except for the order of the factors. | Proof: If \( a \) equals a prime number, the prime factorization clearly exists. In particular the prime factorization exists for the prime number 2. Assume this theorem is true for all \( a \leq n - 1 \) . If \( n \) is a prime, then it has a prime factorization. On the other hand, if \( n \) is not a prime, then ther... | Yes |
Find the solutions to the system,\n\n\[ \nx + {3y} + {6z} = {25} \]\n\n\[ \n{2x} + {7y} + {14z} = {58} \]\n\n\[ \n{2y} + {5z} = {19} \]\n | To solve this system replace the second equation by \( \\left( {-2}\\right) \) times the first equation added to the second. This yields. the system\n\n\[ \nx + {3y} + {6z} = {25} \]\n\n\[ \ny + {2z} = 8 \]\n\n\[ \n{2y} + {5z} = {19} \]\n\nNow take \( \\left( {-2}\\right) \) times the second and add to the third. More ... | Yes |
Example 1.10.3 Give the complete solution to the system of equations, \( {5x} + {10y} - {7z} = - 2 \) , \( {2x} + {4y} - {3z} = - 1 \), and \( {3x} + {6y} + {5z} = 9 \) . | The augmented matrix for this system is\n\n\[ \left( \begin{matrix} 2 & 4 & - 3 & - 1 \\ 5 & {10} & - 7 & - 2 \\ 3 & 6 & 5 & 9 \end{matrix}\right) \]\n\nMultiply the second row by 2, the first row by 5, and then take (-1) times the first row and add to the second. Then multiply the first row by \( 1/5 \) . This yields\... | Yes |
Example 1.10.4 Give the complete solution to the system of equations, \( {3x} - y - {5z} = 9 \) , \( y - {10z} = 0 \), and \( - {2x} + y = - 6 \) . | The augmented matrix of this system is\n\n\[ \left( \begin{matrix} 3 & - 1 & - 5 & 9 \\ 0 & 1 & - {10} & 0 \\ - 2 & 1 & 0 & - 6 \end{matrix}\right) \]\n\nReplace the last row with 2 times the top row added to 3 times the bottom row. This gives\n\n\[ \left( \begin{matrix} 3 & - 1 & - 5 & 9 \\ 0 & 1 & - {10} & 0 \\ 0 & 1... | Yes |
Example 1.10.7 Give the complete solution to the system of equations, \( - {41x} + {15y} = {168} \) , \( {109x} - {40y} = - {447}, - {3x} + y = {12} \), and \( {2x} + z = - 1 \) . | The augmented matrix is\n\n\[ \left( \begin{matrix} - {41} & {15} & 0 & {168} \\ {109} & - {40} & 0 & - {447} \\ - 3 & 1 & 0 & {12} \\ 2 & 0 & 1 & - 1 \end{matrix}\right) \]\n\nTo solve this multiply the top row by 109 , the second row by 41 , add the top row to the second row, and multiply the top row by \( 1/{109} \)... | Yes |
Find \( \left( {1,2,0, - 1}\right) \cdot \left( {0, i,2,3}\right) \) . | This equals \( 0 + 2\left( {-i}\right) + 0 + - 3 = - 3 - {2i} \) | Yes |
Theorem 1.15.4 The inner product satisfies the inequality\n\n\[ \left| {\mathbf{a} \cdot \mathbf{b}}\right| \leq \left| \mathbf{a}\right| \left| \mathbf{b}\right| \]\n\nFurthermore equality is obtained if and only if one of \( \mathbf{a} \) or \( \mathbf{b} \) is a scalar multiple of the other. | Proof: First define \( \theta \in \mathbb{C} \) such that\n\n\[ \bar{\theta }\left( {\mathbf{a} \cdot \mathbf{b}}\right) = \left| {\mathbf{a} \cdot \mathbf{b}}\right| ,\left| \theta \right| = 1, \]\n\nand define a function of \( t \in \mathbb{R} \)\n\n\[ f\left( t\right) = \left( {\mathbf{a} + {t\theta }\mathbf{b}}\rig... | Yes |
Theorem 1.15.5 (Triangle inequality) For \( \mathbf{a},\mathbf{b} \in {\mathbb{F}}^{n} \)\n\n\[ \left| {\mathbf{a} + \mathbf{b}}\right| \leq \left| \mathbf{a}\right| + \left| \mathbf{b}\right| \]\n\nand equality holds if and only if one of the vectors is a nonnegative scalar multiple of the other. | Proof: By properties of the inner product and the Cauchy Schwartz inequality,\n\n\[ {\left| \mathbf{a} + \mathbf{b}\right| }^{2} = \left( {\mathbf{a} + \mathbf{b}}\right) \cdot \left( {\mathbf{a} + \mathbf{b}}\right) = \left( {\mathbf{a} \cdot \mathbf{a}}\right) + \left( {\mathbf{a} \cdot \mathbf{b}}\right) + \left( {\... | Yes |
Compute\n\n\[ \left( \begin{matrix} 1 & 2 & 1 & 3 \\ 0 & 2 & 1 & - 2 \\ 2 & 1 & 4 & 1 \end{matrix}\right) \left( \begin{array}{l} 1 \\ 2 \\ 0 \\ 1 \end{array}\right) \] | First of all, this is of the form \( \left( {3 \times 4}\right) \left( {4 \times 1}\right) \) and so the result should be a \( \left( {3 \times 1}\right) \) . Note how the inside numbers cancel. To get the entry in the second row and first and only column, compute\n\n\[ \mathop{\sum }\limits_{{k = 1}}^{4}{a}_{2k}{v}_{k... | No |
Example 2.1.6 Multiply the following.\n\n\\[ \n\\left( \\begin{array}{lll} 1 & 2 & 1 \\\\ 0 & 2 & 1 \\end{array}\\right) \\left( \\begin{matrix} 1 & 2 & 0 \\\\ 0 & 3 & 1 \\\\ - 2 & 1 & 1 \\end{matrix}\\right) \n\\] | The first thing you need to check before doing anything else is whether it is possible to do the multiplication. The first matrix is a \( 2 \times 3 \) and the second matrix is a \( 3 \times 3 \) . Therefore,\n\nis it possible to multiply these matrices. According to the above discussion it should be a \( 2 \times 3 \)... | Yes |
Example 2.1.7 Multiply the following.\n\n\\[ \n\\left( \\begin{matrix} 1 & 2 & 0 \\\\ 0 & 3 & 1 \\\\ - 2 & 1 & 1 \\end{matrix}\\right) \\left( \\begin{array}{lll} 1 & 2 & 1 \\\\ 0 & 2 & 1 \\end{array}\\right) \n\\] | First check if it is possible. This is of the form \\( \\left( {3 \\times 3}\\right) \\left( {2 \\times 3}\\right) \\) . The inside numbers do not match and so you can't do this multiplication. This means that anything you write will be absolute nonsense because it is impossible to multiply these matrices in this order... | Yes |
Example 2.1.9 Multiply if possible \( \left( \begin{array}{ll} 1 & 2 \\ 3 & 1 \\ 2 & 6 \end{array}\right) \left( \begin{array}{lll} 2 & 3 & 1 \\ 7 & 6 & 2 \end{array}\right) \) . | First check to see if this is possible. It is of the form \( \left( {3 \times 2}\right) \left( {2 \times 3}\right) \) and since the inside numbers match, it must be possible to do this and the result should be a \( 3 \times 3 \) matrix. The answer is of the form\n\n\[ \left( {\left( \begin{array}{ll} 1 & 2 \\ 3 & 1 \\ ... | Yes |
Example 2.1.10 Multiply if possible \( \left( \begin{array}{ll} 1 & 2 \\ 3 & 1 \\ 2 & 6 \end{array}\right) \left( \begin{array}{lll} 2 & 3 & 1 \\ 7 & 6 & 2 \\ 0 & 0 & 0 \end{array}\right) \) . | This is not possible because it is of the form \( \left( {3 \times 2}\right) \left( {3 \times 3}\right) \) and the middle numbers don't match. | Yes |
Example 2.1.11 Multiply if possible \( \left( \begin{array}{lll} 2 & 3 & 1 \\ 7 & 6 & 2 \\ 0 & 0 & 0 \end{array}\right) \left( \begin{array}{ll} 1 & 2 \\ 3 & 1 \\ 2 & 6 \end{array}\right) \) . | This is possible because in this case it is of the form \( \left( {3 \times 3}\right) \left( {3 \times 2}\right) \) and the middle numbers do match. When the multiplication is done it equals\n\n\[ \left( \begin{matrix} {13} & {13} \\ {29} & {32} \\ 0 & 0 \end{matrix}\right) \] | Yes |
Example 2.1.12 Multiply if possible \( \left( \begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right) \left( \begin{array}{llll} 1 & 2 & 1 & 0 \end{array}\right) \) . | In this case you are trying to do \( \left( {3 \times 1}\right) \left( {1 \times 4}\right) \) . The inside numbers match so you can do it. Verify\n\n\[ \left( \begin{array}{l} 1 \\ 2 \\ 1 \end{array}\right) \left( \begin{array}{llll} 1 & 2 & 1 & 0 \end{array}\right) = \left( \begin{array}{llll} 1 & 2 & 1 & 0 \\ 2 & 4 &... | Yes |
Write the matrix which is associated with this directed graph and find the number of ways to go from 2 to 4 in three steps. | Here you need to use a \( 4 \times 4 \) matrix. The one you need is\n\n\[ \left( \begin{array}{llll} 0 & 1 & 1 & 0 \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 0 & 1 \\ 0 & 1 & 0 & 1 \end{array}\right) \]\n\nThen to find the answer, you just need to multiply this matrix by itself three times and look at the entry in the second row and ... | Yes |
Compare \( \left( \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right) \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \) and \( \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \left( \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right) \) . | The first product is\n\n\[ \left( \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right) \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) = \left( \begin{array}{ll} 2 & 1 \\ 4 & 3 \end{array}\right) \]\n\nthe second product is\n\n\[ \left( \begin{array}{ll} 0 & 1 \\ 1 & 0 \end{array}\right) \left( \begin{array}{... | Yes |
Proposition 2.1.15 If all multiplications and additions make sense, the following hold for matrices, \( A, B, C \) and \( a, b \) scalars.\n\n\[ A\left( {{aB} + {bC}}\right) = a\left( {AB}\right) + b\left( {AC}\right) \] | Proof: Using the above definition of matrix multiplication,\n\n\[ {\left( A\left( aB + bC\right) \right) }_{ij} = \mathop{\sum }\limits_{k}{A}_{ik}{\left( aB + bC\right) }_{kj} \]\n\n\[ = \mathop{\sum }\limits_{k}{A}_{ik}\left( {a{B}_{kj} + b{C}_{kj}}\right) \]\n\n\[ = a\mathop{\sum }\limits_{k}{A}_{ik}{B}_{kj} + b\mat... | Yes |
Lemma 2.1.17 Let \( A \) be an \( m \times n \) matrix and let \( B \) be a \( n \times p \) matrix. Then\n\n\[ \n{\left( AB\right) }^{T} = {B}^{T}{A}^{T} \n\] | Proof: From the definition,\n\n\[ \n{\left( {\left( AB\right) }^{T}\right) }_{ij} = {\left( AB\right) }_{ji} \n\]\n\n\[ \n= \mathop{\sum }\limits_{k}{A}_{jk}{B}_{ki} \n\]\n\n\[ \n= \mathop{\sum }\limits_{k}{\left( {B}^{T}\right) }_{ik}{\left( {A}^{T}\right) }_{kj} \n\]\n\n\[ \n= {\left( {B}^{T}{A}^{T}\right) }_{ij} \n\... | Yes |
Lemma 2.1.21 Suppose \( A \) is an \( m \times n \) matrix and \( {I}_{n} \) is the \( n \times n \) identity matrix. Then \( A{I}_{n} = A \) . If \( {I}_{m} \) is the \( m \times m \) identity matrix, it also follows that \( {I}_{m}A = A \) . | \[ \n{\left( A{I}_{n}\right) }_{ij} = \mathop{\sum }\limits_{k}{A}_{ik}{\delta }_{kj} \n\] \n\[ \n= {A}_{ij} \n\] \nand so \( A{I}_{n} = A \) . The other case is left as an exercise for you. | No |
Proposition 2.1.23 Suppose \( {AB} = {BA} = I \) . Then \( B = {A}^{-1} \) . | Proof: From the definition \( B \) is an inverse for \( A \) . Could there be another one \( {B}^{\prime } \) ?\n\n\[ \n{B}^{\prime } = {B}^{\prime }I = {B}^{\prime }\left( {AB}\right) = \left( {{B}^{\prime }A}\right) B = {IB} = B.\n\]\n\nThus, the inverse, if it exists, is unique. | Yes |
Example 2.1.24 Let \( A = \left( \begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right) \) . Does \( A \) have an inverse? | One might think \( A \) would have an inverse because it does not equal zero. However, \[ \left( \begin{array}{ll} 1 & 1 \\ 1 & 1 \end{array}\right) \left( \begin{matrix} - 1 \\ 1 \end{matrix}\right) = \left( \begin{array}{l} 0 \\ 0 \end{array}\right) \] and if \( {A}^{-1} \) existed, this could not happen because you ... | Yes |
Example 2.1.27 Let \( A = \left( \begin{matrix} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 3 & 1 & - 1 \end{matrix}\right) \) . Find \( {A}^{-1} \) . | Set up the augmented matrix \( \left( {A \mid I}\right) \)\n\n\[ \left( \begin{matrix} 1 & 2 & 2 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 & 1 & 0 \\ 3 & 1 & - 1 & 0 & 0 & 1 \end{matrix}\right) \]\n\nNext take \( \left( {-1}\right) \) times the first row and add to the second followed by \( \left( {-3}\right) \) times the first row... | Yes |
Example 2.1.28 Let \( A = \left( \begin{array}{lll} 1 & 2 & 2 \\ 1 & 0 & 2 \\ 2 & 2 & 4 \end{array}\right) \) . Find \( {A}^{-1} \) . | Write the augmented matrix \( \left( {A \mid I}\right) \)\n\n\[ \left( \begin{array}{llllll} 1 & 2 & 2 & 1 & 0 & 0 \\ 1 & 0 & 2 & 0 & 1 & 0 \\ 2 & 2 & 4 & 0 & 0 & 1 \end{array}\right) \]\n\nand proceed to do row operations attempting to obtain \( \left( {I \mid {A}^{-1}}\right) \) . Take \( \left( {-1}\right) \) times ... | Yes |
Lemma 2.3.3 Let \( \\mathbf{v} \\in {\\mathbb{F}}^{n} \) . Thus \( \\mathbf{v} \) is a list of numbers arranged vertically, \( {v}_{1},\\cdots ,{v}_{n} \) . Then\n\n\[ \n{\\mathbf{e}}_{i}^{T}\\mathbf{v} = {v}_{i} \n\]\n\n\( \\left( {2.20}\\right) \)\n\nAlso, if \( A \) is an \( m \\times n \) matrix, then letting \( {\... | Proof: First note that \( {\\mathbf{e}}_{i}^{T} \) is a \( 1 \\times n \) matrix and \( \\mathbf{v} \) is an \( n \\times 1 \) matrix so the above multiplication in (2.20) makes perfect sense. It equals\n\n\[ \n\\left( {0,\\cdots ,1,\\cdots 0}\\right) \\left( \\begin{matrix} {v}_{1} \\\\ \\vdots \\\\ {v}_{i} \\\\ \\vdo... | Yes |
Theorem 2.3.4 Let \( L : {\mathbb{F}}^{n} \rightarrow {\mathbb{F}}^{m} \) be a linear transformation. Then there exists a unique \( m \times n \) matrix \( A \) such that\n\n\[ A\mathbf{x} = L\mathbf{x} \]\n\nfor all \( \mathbf{x} \in {\mathbb{F}}^{n} \) . The \( i{k}^{\text{th }} \) entry of this matrix is given by\n\... | Proof: By the lemma,\n\n\[ {\left( L\mathbf{x}\right) }_{i} = {\mathbf{e}}_{i}^{T}L\mathbf{x} = {\mathbf{e}}_{i}^{T}{x}_{k}L{\mathbf{e}}_{k} = \left( {{\mathbf{e}}_{i}^{T}L{\mathbf{e}}_{k}}\right) {x}_{k}. \]\n\nLet \( {A}_{ik} = {\mathbf{e}}_{i}^{T}L{\mathbf{e}}_{k} \), to prove the existence part of the theorem.\n\nT... | Yes |
Corollary 2.3.5 A linear transformation, \( L : {\mathbb{F}}^{n} \rightarrow {\mathbb{F}}^{m} \) is completely determined by the vectors \( \left\{ {L{\mathbf{e}}_{1},\cdots, L{\mathbf{e}}_{n}}\right\} \) . | Proof: This follows immediately from the above theorem. The unique matrix determining the linear transformation which is given in (2.22) depends only on these vectors. | No |
Find the linear transformation, \( L : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{2} \) which has the property that \( L{\mathbf{e}}_{1} = \left( \begin{array}{l} 2 \\ 1 \end{array}\right) \) and \( L{\mathbf{e}}_{2} = \left( \begin{array}{l} 1 \\ 3 \end{array}\right) \). | From the above theorem and corollary, this linear transformation is that determined by matrix multiplication by the matrix \[ \left( \begin{array}{ll} 2 & 1 \\ 1 & 3 \end{array}\right) \] | Yes |
Theorem 2.3.8 Let \( A \) be an \( m \times n \) matrix where \( m < n \) . Then \( N\left( A\right) \) contains nonzero vectors. | Proof: First consider the case where \( A \) is a \( 1 \times n \) matrix for \( n > 1 \) . Say\n\n\[ A = \left( \begin{array}{lll} {a}_{1} & \cdots & {a}_{n} \end{array}\right) \]\n\nIf \( {a}_{1} = 0 \), consider the vector \( \mathbf{x} = {\mathbf{e}}_{1} \) . If \( {a}_{1} \neq 0 \), let\n\n\[ \mathbf{x} = \left( \... | Yes |
Proposition 2.4.2 Let \( V \subseteq {\mathbb{F}}^{n} \) . Then \( V \) is a subspace if and only if it is a vector space itself with respect to the same operations of scalar multiplication and vector addition. | Proof: Suppose first that \( V \) is a subspace. All algebraic properties involving scalar multiplication and vector addition hold for \( V \) because these things hold for \( {\mathbb{F}}^{n} \) . Is \( \mathbf{0} \in V \) ? Yes it is. This is because \( 0\mathbf{v} \in V \) and \( 0\mathbf{v} = \mathbf{0} \) . By ass... | Yes |
Lemma 2.4.3 A set of vectors \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{p}}\right\} \) is linearly independent if and only if none of the vectors can be obtained as a linear combination of the others. | Proof: Suppose first that \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{p}}\right\} \) is linearly independent. If \( {\mathbf{x}}_{k} = \mathop{\sum }\limits_{{j \neq k}}{c}_{j}{\mathbf{x}}_{j} \), then\n\n\[ \mathbf{0} = 1{\mathbf{x}}_{k} + \mathop{\sum }\limits_{{j \neq k}}\left( {-{c}_{j}}\right) {\mathbf{x}}_... | Yes |
Theorem 2.4.4 (Exchange Theorem) Let \( \\left\\{ {{\\mathbf{x}}_{1},\\cdots ,{\\mathbf{x}}_{r}}\\right\\} \) be a linearly independent set of vectors such that each \( {\\mathbf{x}}_{i} \) is in \( \\operatorname{span}\\left( {{\\mathbf{y}}_{1},\\cdots ,{\\mathbf{y}}_{s}}\\right) \) . Then \( r \\leq s \) . | Proof 1: Suppose not. Then \( r > s \) . By assumption, there exist scalars \( {a}_{ji} \) such that\n\n\[ \n{\\mathbf{x}}_{i} = \\mathop{\\sum }\\limits_{{j = 1}}^{s}{a}_{ji}{\\mathbf{y}}_{j} \n\]\n\nThe matrix whose \( j{i}^{th} \) entry is \( {a}_{ji} \) has more columns than rows. Therefore, by Theorem 2.3.8 there ... | Yes |
Corollary 2.4.6 Let \( \left\{ {{\mathbf{x}}_{1},\cdots ,{\mathbf{x}}_{r}}\right\} \) and \( \left\{ {{\mathbf{y}}_{1},\cdots ,{\mathbf{y}}_{s}}\right\} \) be two bases \( {}^{m} \) of \( {\mathbb{F}}^{n} \) . Then \( r = s = n \) . | Proof: From the exchange theorem, \( r \leq s \) and \( s \leq r \) . Now note the vectors,\n\n\[ \n{\mathbf{e}}_{i} = \overset{1\text{ is in the }{i}^{\text{th }}\text{ slot }}{\overbrace{\left( 0,\cdots ,0,1,0\cdots ,0\right) }} \n\]\n\nfor \( i = 1,2,\cdots, n \) are a basis for \( {\mathbb{F}}^{n} \) . ∎ | No |
Lemma 2.4.7 Let \( \\left\\{ {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{r}}\\right\\} \) be a set of vectors. Then \( V \\equiv \\operatorname{span}\\left( {{\\mathbf{v}}_{1},\\cdots ,{\\mathbf{v}}_{r}}\\right\\} \) is a subspace. | Proof: Suppose \( \\alpha ,\\beta \) are two scalars and let \( \\mathop{\\sum }\\limits_{{k = 1}}^{r}{c}_{k}{\\mathbf{v}}_{k} \) and \( \\mathop{\\sum }\\limits_{{k = 1}}^{r}{d}_{k}{\\mathbf{v}}_{k} \) are two elements\n\nof \( V \) . What about\n\[ \n\\alpha \\mathop{\\sum }\\limits_{{k = 1}}^{r}{c}_{k}{\\mathbf{v}}_... | Yes |
Corollary 2.4.9 Let \( \\left\\{ {{\\mathbf{x}}_{1},\\cdots ,{\\mathbf{x}}_{r}} \\right\\} \) and \( \\left\\{ {{\\mathbf{y}}_{1},\\cdots ,{\\mathbf{y}}_{s}} \\right\\} \) be two bases for \( V \) . Then \( r = s \) . | Proof: From the exchange theorem, \( r \\leq s \) and \( s \\leq r \) . | Yes |
Lemma 2.4.11 Suppose \( \mathbf{v} \notin \operatorname{span}\left( {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right) \) and \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k}}\right\} \) is linearly independent. Then \( \left\{ {{\mathbf{u}}_{1},\cdots ,{\mathbf{u}}_{k},\mathbf{v}}\right\} \) is also linearly inde... | Proof: Suppose \( \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{\mathbf{u}}_{i} + d\mathbf{v} = \mathbf{0} \) . It is required to verify that each \( {c}_{i} = 0 \) and that \( d = 0 \) . But if \( d \neq 0 \), then you can solve for \( \mathbf{v} \) as a linear combination of the vectors, \( \left\{ {{\mathbf{u}}_{1},\cd... | Yes |
Theorem 2.4.12 Let \( V \) be a nonzero subspace of \( {\mathbb{F}}^{n} \). Then \( V \) has a basis. | Proof: Let \( {\mathbf{v}}_{1} \in V \) where \( {\mathbf{v}}_{1} \neq 0 \). If \( \operatorname{span}\left\{ {\mathbf{v}}_{1}\right\} = V \), stop. \( \left\{ {\mathbf{v}}_{1}\right\} \) is a basis for \( V \). Otherwise, there exists \( {\mathbf{v}}_{2} \in V \) which is not in span \( \left\{ {\mathbf{v}}_{1}\right\... | Yes |
Corollary 2.4.13 Let \( V \) be a subspace of \( {\mathbb{F}}^{n} \) and let \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) be a linearly independent set of vectors in \( V \) . Then either it is a basis for \( V \) or there exist vectors, \( {\mathbf{v}}_{r + 1},\cdots ,{\mathbf{v}}_{s} \) such that... | Proof: This follows immediately from the proof of Theorem 2.4.12. You do exactly the same argument except you start with \( \left\{ {{\mathbf{v}}_{1},\cdots ,{\mathbf{v}}_{r}}\right\} \) rather than \( \left\{ {\mathbf{v}}_{1}\right\} \) . | No |
Theorem 2.4.14 Let \( V \) be a subspace of \( {\mathbb{F}}^{n} \) and suppose \( \operatorname{span}\left( {{\mathbf{u}}_{1}\cdots ,{\mathbf{u}}_{p}}\right) = V \) where the \( {\mathbf{u}}_{i} \) are nonzero vectors. Then there exist vectors \( \left\{ {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right\} \) such that \... | Proof: Let \( r \) be the smallest positive integer with the property that for some set \( \left\{ {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right\} \subseteq \left\{ {{\mathbf{u}}_{1}\cdots ,{\mathbf{u}}_{p}}\right\} \)\n\n\[ \operatorname{span}\left( {{\mathbf{v}}_{1}\cdots ,{\mathbf{v}}_{r}}\right) = V \]\n\nThen \... | Yes |
Lemma 2.6.2 Let \( A\left( t\right) \) be an \( m \times n \) matrix and let \( B\left( t\right) \) be an \( n \times p \) matrix with the property that all the entries of these matrices are differentiable functions. Then\n\n\[ \n{\left( A\left( t\right) B\left( t\right) \right) }^{\prime } = {A}^{\prime }\left( t\righ... | Proof: This is like the usual proof.\n\n\[ \n\frac{1}{h}\left( {A\left( {t + h}\right) B\left( {t + h}\right) - A\left( t\right) B\left( t\right) }\right) = \n\]\n\n\[ \n\frac{1}{h}\left( {A\left( {t + h}\right) B\left( {t + h}\right) - A\left( {t + h}\right) B\left( t\right) }\right) + \frac{1}{h}\left( {A\left( {t + ... | Yes |
Theorem 2.6.4 Let \( \mathbf{i}\left( t\right) ,\mathbf{j}\left( t\right) ,\mathbf{k}\left( t\right) \) be as described. Then there exists a unique vector \( \mathbf{\Omega }\left( t\right) \) such that if \( \mathbf{u}\left( t\right) \) is a vector whose components are constant with respect to \( \mathbf{i}\left( t\ri... | Proof: It only remains to prove uniqueness. Suppose \( {\mathbf{\Omega }}_{1} \) also works. Then \( \mathbf{u}\left( t\right) = Q\left( t\right) \mathbf{u} \) and so \( {\mathbf{u}}^{\prime }\left( t\right) = {Q}^{\prime }\left( t\right) \mathbf{u} \) and\n\n\[{Q}^{\prime }\left( t\right) \mathbf{u} = \mathbf{\Omega }... | Yes |
Example 2.6.5 Suppose a rock is dropped from a tall building. Where will it strike? | Assume \( \mathbf{a} = - g\mathbf{k} \) and the \( \mathbf{j} \) component of \( {\mathbf{a}}_{B} \) is approximately\n\n\[- {2\omega }\left( {{x}^{\prime }\cos \phi + {z}^{\prime }\sin \phi }\right) .\n\]\n\nThe dominant term in this expression is clearly the second one because \( {x}^{\prime } \) will be small. Also,... | Yes |
Example 3.1.2 Find \( \\det \\left( \\begin{matrix} 2 & 4 \\\\ - 1 & 6 \\end{matrix}\\right) \) . | From the definition this is just \( \\left( 2\\right) \\left( 6\\right) - \\left( {-1}\\right) \\left( 4\\right) = {16} \) . | Yes |
Theorem 3.1.4 Let \( A \) be an \( n \times n \) matrix where \( n \geq 2 \). Then\n\n\[ \det \left( A\right) = \mathop{\sum }\limits_{{j = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij} = \mathop{\sum }\limits_{{i = 1}}^{n}{a}_{ij}\operatorname{cof}{\left( A\right) }_{ij}. \] | The first formula consists of expanding the determinant along the \( {i}^{\text{th }} \) row and the second expands the determinant along the \( {j}^{\text{th }} \) column.\n\nNote that for a \( n \times n \) matrix, you will need \( n \) ! terms to evaluate the determinant in this way. If \( n = {10} \), this is \( {1... | No |
Corollary 3.1.6 Let \( M \) be an upper (lower) triangular matrix. Then \( \det \left( M\right) \) is obtained by taking the product of the entries on the main diagonal. | Proof: The corollary is true if the matrix is one to one. Suppose it is \( n \times n \) . Then the\n\nmatrix is of the form\n\[\n\left( \begin{matrix} {m}_{11} & \mathbf{a} \\ \mathbf{0} & {M}_{1} \end{matrix}\right)\n\]\n\nwhere \( {M}_{1} \) is \( \left( {n - 1}\right) \times \left( {n - 1}\right) \) . Then expandin... | Yes |
Find \( \det \left( A\right) \) . | From the above corollary, this is -6 . | No |
Find the determinant of the matrix\n\n\[ A = \left( \begin{matrix} 1 & 2 & 3 & 4 \\ 5 & 1 & 2 & 3 \\ 4 & 5 & 4 & 3 \\ 2 & 2 & - 4 & 5 \end{matrix}\right) \] | Replace the second row by \( \left( {-5}\right) \) times the first row added to it. Then replace the third row by \( \left( {-4}\right) \) times the first row added to it. Finally, replace the fourth row by \( \left( {-2}\right) \) times the first row added to it. This yields the matrix\n\n\[ B = \left( \begin{matrix} ... | Yes |
Find the inverse of the matrix\n\n\[ A = \left( \begin{array}{lll} 1 & 2 & 3 \\ 3 & 0 & 1 \\ 1 & 2 & 1 \end{array}\right) \] | First find the determinant of this matrix. This is seen to be 12 . The cofactor matrix of\n\n\( A \) is\n\[ \left( \begin{matrix} - 2 & - 2 & 6 \\ 4 & - 2 & 0 \\ 2 & 8 & - 6 \end{matrix}\right) \]\n\nEach entry of \( A \) was replaced by its cofactor. Therefore, from the above theorem, the inverse of \( A \) should equ... | Yes |
Suppose\n\n\[ A\left( t\right) = \left( \begin{matrix} {e}^{t} & 0 & 0 \\ 0 & \cos t & \sin t \\ 0 & - \sin t & \cos t \end{matrix}\right) \]\n\nFind \( A{\left( t\right) }^{-1} \) . | First note \( \det \left( {A\left( t\right) }\right) = {e}^{t} \) . A routine computation using the above theorem shows that this inverse is\n\n\[ \frac{1}{{e}^{t}}{\left( \begin{matrix} 1 & 0 & 0 \\ 0 & {e}^{t}\cos t & {e}^{t}\sin t \\ 0 & - {e}^{t}\sin t & {e}^{t}\cos t \end{matrix}\right) }^{T} = \left( \begin{matri... | Yes |
Lemma 3.3.1 There exists a unique function, \( {\operatorname{sgn}}_{n} \) which maps each ordered list of numbers from \( \{ 1,\cdots, n\} \) to one of the three numbers, \( 0,1 \), or -1 which also has the following properties.\n\n\[ \n{\operatorname{sgn}}_{n}\left( {1,\cdots, n}\right) = 1 \n\]\n\n(3.2)\n\n\[ \n{\op... | Proof: To begin with, it is necessary to show the existence of such a function. This is clearly true if \( n = 1 \) . Define \( {\operatorname{sgn}}_{1}\left( 1\right) \equiv 1 \) and observe that it works. No switching is possible. In the case where \( n = 2 \), it is also clearly true. Let \( {\operatorname{sgn}}_{2}... | Yes |
Lemma 3.3.2 Every ordered list of \( \{ 1,2,\cdots, n\} \) can be obtained from every other ordered list by a finite number of switches. Also, sgn is unique. | Proof: This is obvious if \( n = 1 \) or 2 . Suppose then that it is true for sets of \( n - 1 \) elements. Take two ordered lists of numbers, \( {P}_{1},{P}_{2} \) . To get from \( {P}_{1} \) to \( {P}_{2} \) using switches, first make a switch to obtain the last element in the list coinciding with the last element of... | Yes |
Proposition 3.3.6 Let \( \left( {{r}_{1},\cdots ,{r}_{n}}\right) \) be an ordered list of numbers from \( \{ 1,\cdots, n\} \) . Then\n\n\[ \operatorname{sgn}\left( {{r}_{1},\cdots ,{r}_{n}}\right) \det \left( A\right) = \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{k}_{1},\cdo... | Proof: Let \( \left( {1,\cdots, n}\right) = \left( {1,\cdots, r,\cdots s,\cdots, n}\right) \) so \( r < s \) .\n\n\[ \det \left( {A\left( {1,\cdots, r,\cdots, s,\cdots, n}\right) }\right) = \]\n\n\( \left( {3.10}\right) \)\n\n\[ \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{k}... | Yes |
The following formula for \( \det \left( A\right) \) is valid. | From Proposition 3.3.6, if the \( {r}_{i} \) are distinct,\n\n\[ \det \left( A\right) = \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{r}_{1},\cdots ,{r}_{n}}\right) \operatorname{sgn}\left( {{k}_{1},\cdots ,{k}_{n}}\right) {a}_{{r}_{1}{k}_{1}}\cdots {a}_{{r}_{n}{k}_{n}}. \]\n\... | Yes |
If two rows or two columns in an \( n \times n \) matrix \( A \), are switched, the determinant of the resulting matrix equals \( \left( {-1}\right) \) times the determinant of the original matrix. If \( A \) is an \( n \times n \) matrix in which two rows are equal or two columns are equal then \( \det \left( A\right)... | Proof: By Proposition 3.3.6 when two rows are switched, the determinant of the resulting matrix is \( \left( {-1}\right) \) times the determinant of the original matrix. By Corollary 3.3.8 the same holds for columns because the columns of the matrix equal the rows of the transposed matrix. Thus if \( {A}_{1} \) is the ... | Yes |
Corollary 3.3.11 Suppose \( A \) is an \( n \times n \) matrix and some column (row) is a linear combination of \( r \) other columns (rows). Then \( \det \left( A\right) = 0 \) . | Proof: Let \( A = \left( \begin{array}{lll} {\mathbf{a}}_{1} & \cdots & {\mathbf{a}}_{n} \end{array}\right) \) be the columns of \( A \) and suppose the condition that one column is a linear combination of \( r \) of the others is satisfied. Then by using Corollary 3.3. you may rearrange the columns to have the \( {n}^... | Yes |
Theorem 3.3.13 Let \( A \) and \( B \) be \( n \times n \) matrices. Then\n\n\[ \det \left( {AB}\right) = \det \left( A\right) \det \left( B\right) . \] | Proof: Let \( {c}_{ij} \) be the \( i{j}^{th} \) entry of \( {AB} \) . Then by Proposition 3.3.6,\n\n\[ \det \left( {AB}\right) = \mathop{\sum }\limits_{\left( {k}_{1},\cdots ,{k}_{n}\right) }\operatorname{sgn}\left( {{k}_{1},\cdots ,{k}_{n}}\right) {c}_{1{k}_{1}}\cdots {c}_{n{k}_{n}} \]\n\n\[ = \mathop{\sum }\limits_{... | Yes |
Theorem 3.3.14 Let \( A \) be an \( n \times m \) matrix with \( n \geq m \) and let \( B \) be a \( m \times n \) matrix. Also let \( {A}_{i} \) be the \( m \times m \) submatrices of \( A \) which are obtained by deleting \( n - m \) rows and let \( {B}_{i} \) be the \( m \times m \) submatrices of \( B \) which are ... | Proof: This follows from a computation. By Corollary 3.3.8 on Page SZ, \( \det \left( {BA}\right) = \)\[ \frac{1}{m!}\mathop{\sum }\limits_{\left( {i}_{1}\cdots {i}_{m}\right) }\mathop{\sum }\limits_{\left( {j}_{1}\cdots {j}_{m}\right) }\operatorname{sgn}\left( {{i}_{1}\cdots {i}_{m}}\right) \operatorname{sgn}\left( {{... | Yes |
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