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Simplify the following matrix expressions.\n\n1. \( A + B \) | 1. \( A + B = \left\lbrack \begin{array}{lll} 3 & 6 & 9 \\ 0 & 4 & 3 \\ 4 & 5 & 9 \end{array}\right\rbrack \) . | Yes |
Find the matrix \( X \) such that\n\n\[ {2A} + {3X} = - {4A}\text{.} \] | Solution We can use basic algebra techniques to manipulate this equation for \( X \) ; first, let’s subtract \( {2A} \) from both sides. This gives us\n\n\[ {3X} = - {6A}\text{.} \]\n\nNow divide both sides by 3 to get\n\n\[ X = - {2A}\text{.} \]\n\nNow we just need to compute \( - {2A} \) ; we find that\n\n\[ X = \lef... | Yes |
Find the following products.\n\n1. \( \overrightarrow{u}\overrightarrow{x} \) 3. \( \overrightarrow{u}\overrightarrow{y} \) \n\n5. \( \overrightarrow{x}\overrightarrow{u} \) \n\n2. \( \overrightarrow{vy} \) 4. \( \overrightarrow{u}\overrightarrow{v} \) | \[ \text{1.}\overrightarrow{u}\overrightarrow{x} = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \end{array}\right\rbrack \left\lbrack \begin{matrix} - 2 \\ 4 \\ 3 \end{matrix}\right\rbrack = 1\left( {-2}\right) + 2\left( 4\right) + 3\left( 3\right) = {15} \] | No |
multiply\n\n\\[ \n\\left\\lbrack \\begin{array}{ll} 1 & 2 \\\\ 3 & 4 \\end{array}\\right\\rbrack \\left\\lbrack \\begin{matrix} 1 & - 1 & 0 \\\\ 2 & 2 & - 1 \\end{matrix}\\right\\rbrack \n\\] | Solution Let’s call our first matrix \\( A \\) and the second \\( B \\) . We should first check to see that we can actually perform this multiplication. Matrix \\( A \\) is \\( 2 \\times 2 \\) and \\( B \\) is \\( 2 \\times 3 \\) . The \ | No |
Example 26 Multiply\n\n\\[ \n\\left\\lbrack \\begin{matrix} 1 & - 1 \\\\ 5 & 2 \\\\ - 2 & 3 \\end{matrix}\\right\\rbrack \\left\\lbrack \\begin{array}{llll} 1 & 1 & 1 & 1 \\\\ 2 & 6 & 7 & 9 \\end{array}\\right\\rbrack \n\\] | Solution Let's first check to make sure this product is defined. Again calling the first matrix \\( A \\) and the second \\( B \\), we see that \\( A \\) is a \\( 3 \\times 2 \\) matrix and \\( B \\) is a \\( 2 \\times 4 \\) matrix; the inner dimensions match so the product is defined, and the product will be a \\( 3 \... | No |
Example 27 Multiply, if possible,\n\n\[ \n\left\lbrack \begin{array}{lll} 2 & 3 & 4 \\ 9 & 8 & 7 \end{array}\right\rbrack \left\lbrack \begin{matrix} 3 & 6 \\ 5 & - 1 \end{matrix}\right\rbrack \n\] | Solution Again, we’ll call the first matrix \( A \) and the second \( B \) . Checking the dimensions of each matrix, we see that \( A \) is a \( 2 \times 3 \) matrix, whereas \( B \) is a \( 2 \times 2 \) matrix. The inner dimensions do not match, therefore this multiplication is not defined. | Yes |
Find \( \overrightarrow{x}\overrightarrow{u} \) where \( \overrightarrow{x} = \left\lbrack \begin{matrix} - 2 \\ 4 \\ 3 \end{matrix}\right\rbrack \) and \( \overrightarrow{u} = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \end{array}\right\rbrack \). | Solution Again, we need to check to make sure the dimensions work correctly (remember that even though we are referring to \( \overrightarrow{u} \) and \( \overrightarrow{x} \) as vectors, they are, in fact, just matrices). The column vector \( \overrightarrow{x} \) has dimensions \( 3 \times 1 \), whereas the row vect... | Yes |
Find the following products.\n\n\[ \text{1. AB 3.}A{\mathbf{0}}_{3 \times 4}\;\text{5. IA 7. BC} \] | 1. \( {AB} = \left\lbrack \begin{matrix} 1 & 2 & 3 \\ 2 & - 7 & 5 \\ - 2 & - 8 & 3 \end{matrix}\right\rbrack \left\lbrack \begin{array}{lll} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{array}\right\rbrack = \left\lbrack \begin{matrix} 6 & 6 & 6 \\ 0 & 0 & 0 \\ - 7 & - 7 & - 7 \end{matrix}\right\rbrack \) | No |
Find the following:\n\n1. \( A\left( {B + C}\right) \) | \[ A\left( {B + C}\right) = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack \left( {\left\lbrack \begin{matrix} 1 & 1 \\ 1 & - 1 \end{matrix}\right\rbrack + \left\lbrack \begin{array}{ll} 2 & 1 \\ 1 & 2 \end{array}\right\rbrack }\right) \]\n\n\[ = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \en... | Yes |
Example 31 Let\n\n\\[ \n\\overrightarrow{x} = \\left\\lbrack \\begin{matrix} 1 \\ - 1 \\end{matrix}\\right\\rbrack \\;\\overrightarrow{y} = \\left\\lbrack \\begin{array}{l} 2 \\ 3 \\end{array}\\right\\rbrack \\;\\text{ and }\\;\\overrightarrow{z} = \\left\\lbrack \\begin{matrix} - 3 \\ 2 \\end{matrix}\\right\\rbrack .\... | Solution To draw \\( \\overrightarrow{x} \\), start at the point \\( \\left( {0, - 1}\\right) \\) as directed, then move to the right one unit and down one unit and draw the tip. Thus the arrow \ | No |
Example 32 Let\n\n\\[ \n\\overrightarrow{x} = \\left\\lbrack \\begin{array}{l} 1 \\\\ 1 \\end{array}\\right\\rbrack \\;\\text{ and }\\;\\overrightarrow{y} = \\left\\lbrack \\begin{array}{l} 3 \\\\ 1 \\end{array}\\right\\rbrack \n\\]\n\nSketch \\( \\overrightarrow{x},\\overrightarrow{y} \\) and \\( \\overrightarrow{x} +... | Solution A starting point for drawing each vector was not given; by default, we’ll start at the origin. (This is in many ways nice; this means that the vector \\( \\left\\lbrack \\begin{array}{l} 3 \\\\ 1 \\end{array}\\right\\rbrack \\) \ | No |
Example 33 Consider the vectors \( \overrightarrow{x} \) and \( \overrightarrow{y} \) as drawn in Figure 2.5. Sketch the vector \( \overrightarrow{x} + \overrightarrow{y} \) . | Solution\n\n\n\nFigure 2.5: Vectors \( \overrightarrow{x} \) and \( \overrightarrow{y} \) in Example 33\n\nWe’ll apply the Parallelogram Law, as given in Key Idea 5. As before, we draw \( \overrightarrow{x} + \overrigh... | Yes |
Sketch \( \overrightarrow{x},\overrightarrow{y},3\overrightarrow{x} \) and \( - 1\overrightarrow{y} \) . | Solution We begin by computing \( 3\overrightarrow{x} \) and \( - \overrightarrow{y} \) :\n\n\[ 3\overrightarrow{x} = \left\lbrack \begin{array}{l} 3 \\ 3 \end{array}\right\rbrack \;\text{ and }\; - \overrightarrow{y} = \left\lbrack \begin{matrix} 2 \\ - 1 \end{matrix}\right\rbrack .\n\]\n\nAll four vectors are sketche... | Yes |
Example 36 Let vectors \( \overrightarrow{x} \) and \( \overrightarrow{y} \) be as in Figure 2.10. Draw \( \overrightarrow{x} - \overrightarrow{y} \) . | Solution ION TO draw \( \overrightarrow{x} - \overrightarrow{y} \), we will first draw \( - \overrightarrow{y} \) and then apply the Parallelogram Law to add \( \overrightarrow{x} \) to \( - \overrightarrow{y} \) . See Figure 2.11. | No |
Example 37 7 Let \( \overrightarrow{x} \) and \( \overrightarrow{y} \) be as in Figure ?? (a). Draw \( \overrightarrow{x} - \overrightarrow{y} \) . | Solution We simply apply Key Idea 6: we draw an arrow from \( \overrightarrow{y} \) to \( \overrightarrow{x} \) . We do so in Figure 2.13; \( \overrightarrow{x} - \overrightarrow{y} \) is dashed. | Yes |
Find the length of each of the vectors given below. | Solution We apply Definition 16 to each vector.\n\n\[ \begin{Vmatrix}\overrightarrow{{x}_{1}}\end{Vmatrix} = \sqrt{{1}^{2} + {1}^{2}} = \sqrt{2} \]\n\n\[ \begin{Vmatrix}\overrightarrow{{x}_{2}}\end{Vmatrix} = \sqrt{{2}^{2} + {\left( -3\right) }^{2}} = \sqrt{13}. \]\n\n\[ \begin{Vmatrix}\overrightarrow{{x}_{3}}\end{Vmat... | Yes |
Example 39 Let \( \overrightarrow{x} = \left\lbrack \begin{matrix} 2 \\ - 1 \end{matrix}\right\rbrack \) . Compute \( \parallel \overrightarrow{x}\parallel ,\parallel 3\overrightarrow{x}\parallel ,\parallel - 2\overrightarrow{x}\parallel \), and \( \parallel c\overrightarrow{x}\parallel \), where \( c \) is a scalar. | Solution We apply Definition 16 to each of the vectors. \[ \parallel \overrightarrow{x}\parallel = \sqrt{4 + 1} = \sqrt{5} \] Before computing the length of \( \parallel 3\overrightarrow{x}\parallel \), we note that \( 3\overrightarrow{x} = \left\lbrack \begin{matrix} 6 \\ - 3 \end{matrix}\right\rbrack \) . \( \left| \... | Yes |
Example 40 Let \( A \) be a matrix, and \( \overrightarrow{x},\overrightarrow{y} \), and \( \overrightarrow{z} \) be vectors as given below.\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 3 \end{array}\right\rbrack ,\;\overrightarrow{x} = \left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack ,\;\overrigh... | Solution\n\n---\n\n\n\nFigure 2.15: Multiplying vectors by a matrix in Example 40.\n\nIt is straightforward to compute:\n\n\[ A\overrightarrow{x} = \left\lbrack \begin{array}{l} 5 \\ 5 \end{array}\right\rbrack ,\;A\ove... | No |
Example 41 Let \( A \) be a matrix and \( \overrightarrow{x} \) and \( \overrightarrow{y} \) be vectors as given below.\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 1 \\ 1 & 2 \end{array}\right\rbrack ,\;\overrightarrow{x} = \left\lbrack \begin{array}{l} 2 \\ 1 \end{array}\right\rbrack ,\;\overrightarrow{y} = \left\lbr... | It is pretty straightforward to compute:\n\n\[ \overrightarrow{x} + \overrightarrow{y} = \left\lbrack \begin{array}{l} 1 \\ 2 \end{array}\right\rbrack ;\;A\overrightarrow{x} = \left\lbrack \begin{array}{l} 3 \\ 4 \end{array}\right\rbrack ;\;A\overrightarrow{y} = \left\lbrack \begin{array}{l} 0 \\ 1 \end{array}\right\rb... | Yes |
Example 42 Let \( A,\overrightarrow{x},\overrightarrow{y} \), and \( \overrightarrow{z} \) be as given below.\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & - 1 \\ 1 & - 1 \end{array}\right\rbrack ,\;\overrightarrow{x} = \left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack ,\;\overrightarrow{y} = \left\lbrack \b... | Solution\n\n\n\nFigure 2.17: Multiplying vectors by a matrix in Example 42.\n\nIt is straightforward to compute:\n\n\[ A\overrightarrow{x} = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rbrack ,\;A\overrighta... | Yes |
Solve the equation \( A\overrightarrow{x} = \overrightarrow{b} \) for \( \overrightarrow{x} \) where\n\n\[ A = \left\lbrack \begin{matrix} 1 & 2 & 3 \\ - 1 & 2 & 1 \\ 1 & 1 & 0 \end{matrix}\right\rbrack \;\text{ and }\;\left\lbrack \begin{matrix} 5 \\ - 1 \\ 2 \end{matrix}\right\rbrack \] | Solution The solution is rather straightforward, even though we did a lot of work before to find the answer. Form the augmented matrix \( \left\lbrack \begin{array}{ll} A & \overrightarrow{b} \end{array}\right\rbrack \) and interpret its reduced row echelon form.\n\n\[ \left\lbrack \begin{matrix} 1 & 2 & 3 & 5 \\ - 1 &... | Yes |
Solve the linear system \( A\overrightarrow{x} = \overrightarrow{0} \) for \( \overrightarrow{x} \) and write the solution in vector form, where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right\rbrack \;\text{ and }\;\overrightarrow{0} = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rb... | Solution (Note: we didn't really need to specify that\n\n\[ \overrightarrow{0} = \left\lbrack \begin{array}{l} 0 \\ 0 \end{array}\right\rbrack \]\n\nbut we did just to eliminate any uncertainty.)\n\nTo solve this system, put the augmented matrix into reduced row echelon form, which we do below.\n\n\[ \left\lbrack \begi... | Yes |
Example 45 Solve the linear system \( A\overrightarrow{x} = \overrightarrow{0} \) and write the solution in vector form, where\n\n\[ A = \left\lbrack \begin{matrix} 2 & - 3 \\ - 2 & 3 \end{matrix}\right\rbrack \] | 4 Again, to solve this problem, we form the proper augmented matrix and we put it into reduced row echelon form, which we do below.\n\n\[ \left\lbrack \begin{matrix} 2 & - 3 & 0 \\ - 2 & 3 & 0 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{matrix} 1 & - 3/2 & 0 \\ 0 & 0 & 0 \end{m... | Yes |
Solve the linear system \( A\overrightarrow{x} = \overrightarrow{b} \), where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right\rbrack \;\text{ and }\;\overrightarrow{b} = \left\lbrack \begin{array}{l} 3 \\ 6 \end{array}\right\rbrack \] | Solution (Note that this is the same matrix \( A \) that we used in Example 44. This will be important later.)\n\nOur methodology is the same as before; we form the augmented matrix and put it into reduced row echelon form.\n\n\[ \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \end{array}\right\rbrack \;\overrig... | Yes |
Solve the linear systems \( A\overrightarrow{x} = \overrightarrow{0} \) and \( A\overrightarrow{x} = \overrightarrow{b} \) for \( \overrightarrow{x} \), and write the solutions in vector form. | Solution We’ll tackle \( A\overrightarrow{x} = \overrightarrow{0} \) first. We form the associated augmented matrix, put it into reduced row echelon form, and interpret the result.\n\n\[ \left\lbrack \begin{matrix} 1 & - 1 & 1 & 3 & 0 \\ 4 & 2 & 4 & 6 & 0 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}... | No |
Find the solutions to \( A\overrightarrow{x} = \overrightarrow{b} \) and \( A\overrightarrow{x} = \overrightarrow{0} \). | Solution We go through the familiar work of finding the reduced row echelon form of the appropriate augmented matrix and interpreting the solution.\n\n\[ \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{matrix} 1 & 0 & - 1 \\ 0 &... | No |
Example 51 Solve the matrix equation \( {AX} = B \) where\n\n\[ A = \left\lbrack \begin{matrix} 1 & - 1 \\ 5 & 3 \end{matrix}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{matrix} - 8 & - {13} & 1 \\ {32} & - {17} & {21} \end{matrix}\right\rbrack \] | Solution To solve \( {AX} = B \) for \( X \), we form the proper augmented matrix, put it into reduced row echelon form, and interpret the result.\n\n\[ \left\lbrack \begin{matrix} 1 & - 1 & - 8 & - {13} & 1 \\ 5 & 3 & {32} & - {17} & {21} \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \... | Yes |
Example 52 Solve the matrix equation \( {AX} = B \) where\n\n\[ A = \left\lbrack \begin{matrix} 1 & 0 & 2 \\ 0 & - 1 & - 2 \\ 2 & - 1 & 0 \end{matrix}\right\rbrack \;\text{ and }\;B = \left\lbrack \begin{matrix} - 1 & 2 \\ 2 & - 6 \\ 2 & - 4 \end{matrix}\right\rbrack \] | Solution To solve, let's again form the augmented matrix\n\n\[ \left\lbrack \begin{array}{ll} A & B \end{array}\right\rbrack \]\n\nput it into reduced row echelon form, and interpret the result.\n\n\[ \left\lbrack \begin{matrix} 1 & 0 & 2 & - 1 & 2 \\ 0 & - 1 & - 2 & 2 & - 6 \\ 2 & - 1 & 0 & 2 & - 4 \end{matrix}\right\... | Yes |
Find a matrix \( X \) such that \( {AX} = I \) . | Solution We know how to solve this from the previous section: we form the proper augmented matrix, put it into reduced row echelon form and interpret the\n\nresults.\n\[ \left\lbrack \begin{array}{llll} 2 & 1 & 1 & 0 \\ 1 & 1 & 0 & 1 \end{array}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{m... | Yes |
If there is a matrix \( X \) such that \( {AX} = {I}_{n} \), then \( {XA} = {I}_{n} \) . | The second obstacle is easier to address. We want to know if another matrix \( Y \) exists where \( {AY} = I = {YA} \) . Let’s suppose that it does. Consider the expression \( {XAY} \) .\n\nSince matrix multiplication is associative, we can group this any way we choose. We could group this as \( \left( {XA}\right) Y \)... | Yes |
Find the inverse of \( A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 2 & 4 \end{array}\right\rbrack \) . | Solution By solving the equation \( {AX} = I \) for \( X \) will give us the inverse of \( A \) . Forming the appropriate augmented matrix and finding its reduced row echelon form\ngives us\n\n\[ \left\lbrack \begin{array}{llll} 1 & 2 & 1 & 0 \\ 2 & 4 & 0 & 1 \end{array}\right\rbrack \;\overrightarrow{\operatorname{rre... | Yes |
Find the inverse, if it exists, of \( A = \left\lbrack \begin{matrix} 1 & 1 & - 1 \\ 1 & - 1 & 1 \\ 1 & 2 & 3 \end{matrix}\right\rbrack \) . | We’ll try to solve \( {AX} = I \) for \( X \) and see what happens.\n\n\[ \left\lbrack \begin{matrix} 1 & 1 & - 1 & 1 & 0 & 0 \\ 1 & - 1 & 1 & 0 & 1 & 0 \\ 1 & 2 & 3 & 0 & 0 & 1 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{matrix} 1 & 0 & 0 & {0.5} & {0.5} & 0 \\ 0 & 1 & 0 & {0.... | Yes |
Theorem 7 The Inverse of a \( 2 \times 2 \) Matrix\n\nLet\n\n\[ A = \left\lbrack \begin{array}{ll} a & b \\ c & d \end{array}\right\rbrack \]\n\n\( A \) is invertible if and only if \( {ad} - {bc} \neq 0 \) . | If \( {ad} - {bc} \neq 0 \), then\n\n\[ {A}^{-1} = \frac{1}{{ad} - {bc}}\left\lbrack \begin{matrix} d & - b \\ - c & a \end{matrix}\right\rbrack \] | Yes |
Example 56 Use Theorem 7 to find the inverse of\n\n\[ A = \left\lbrack \begin{matrix} 3 & 2 \\ - 1 & 9 \end{matrix}\right\rbrack \]\n\nif it exists. | Since \( {ad} - {bc} = {29} \neq 0,{A}^{-1} \) exists. By the Theorem,\n\n\[ {A}^{-1} = \frac{1}{3\left( 9\right) - 2\left( {-1}\right) }\left\lbrack \begin{matrix} 9 & - 2 \\ 1 & 3 \end{matrix}\right\rbrack \]\n\n\[ = \frac{1}{29}\left\lbrack \begin{matrix} 9 & - 2 \\ 1 & 3 \end{matrix}\right\rbrack \]\n\nWe can leave... | No |
Solve \( A\overrightarrow{x} = \overrightarrow{b} \) by computing \( \overrightarrow{x} = {A}^{-1}\overrightarrow{b} \), where\n\n\[ A = \left\lbrack \begin{matrix} 1 & 0 & - 3 \\ - 3 & - 4 & {10} \\ 4 & - 5 & - {11} \end{matrix}\right\rbrack \text{ and }\overrightarrow{b} = \left\lbrack \begin{matrix} - {15} \\ {57} \... | Solution Without showing our steps, we compute\n\n\[ {A}^{-1} = \left\lbrack \begin{matrix} {94} & {15} & - {12} \\ 7 & 1 & - 1 \\ {31} & 5 & - 4 \end{matrix}\right\rbrack \]\n\nWe then find the solution to \( A\overrightarrow{x} = \overrightarrow{b} \) by computing \( {A}^{-1}\overrightarrow{b} \):\n\n\[ \overrightarr... | Yes |
Theorem 9\n\n## Invertible Matrix Theorem\n\nLet \( A \) be an \( n \times n \) matrix. The following statements are equivalent.\n\n(a) \( A \) is invertible.\n\n(b) There exists a matrix \( B \) such that \( {BA} = I \) .\n\n(c) There exists a matrix \( C \) such that \( {AC} = I \) .\n\n(d) The reduced row echelon fo... | Let’s make note of a few things about the Invertible Matrix Theorem.\n\n1. First, note that the theorem uses the phrase \ | No |
Example 58 Let\n\n\\[ \nA = \\left\\lbrack \\begin{array}{ll} 3 & 2 \\\\ 0 & 1 \\end{array}\\right\\rbrack \\text{ and }B = \\left\\lbrack \\begin{matrix} - 2 & 0 \\\\ 1 & 1 \\end{matrix}\\right\\rbrack \n\\]\n\nFind:\n\n1. \\( {A}^{-1} \\) 3. \\( {\\left( AB\\right) }^{-1} \\) 5. \\( {\\left( A + B\\right) }^{-1} \\)\... | ## Solution\n\n1. Computing \\( {A}^{-1} \\) is straightforward; we’ll use Theorem 7.\n\n\\[ \n{A}^{-1} = \\frac{1}{3}\\left\\lbrack \\begin{matrix} 1 & - 2 \\\\ 0 & 3 \\end{matrix}\\right\\rbrack = \\left\\lbrack \\begin{matrix} 1/3 & - 2/3 \\\\ 0 & 1 \\end{matrix}\\right\\rbrack \n\\]\n\n2. We compute \\( {B}^{-1} \\... | Yes |
Find the inverse of \( A = \left\lbrack \begin{matrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & - 7 \end{matrix}\right\rbrack \) . | We’ll find \( {A}^{-1} \) using Key Idea 10.\n\n\[ \left\lbrack \begin{matrix} 2 & 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 0 & 1 & 0 \\ 0 & 0 & - 7 & 0 & 0 & 1 \end{matrix}\right\rbrack \;\overrightarrow{\operatorname{rref}}\;\left\lbrack \begin{matrix} 1 & 0 & 0 & 1/2 & 0 & 0 \\ 0 & 1 & 0 & 0 & 1/3 & 0 \\ 0 & 0 & 1 & 0 & 0 &... | Yes |
Example 60 Find the transpose of \( A = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\right\rbrack \) . | Solution Note that \( A \) is a \( 2 \times 3 \) matrix, so \( {A}^{T} \) will be a \( 3 \times 2 \) matrix. By the definition, the first column of \( {A}^{T} \) is the first row of \( A \) ; the second column of \( {A}^{T} \) is the second row of \( A \) . Therefore, \[ {A}^{T} = \left\lbrack \begin{array}{ll} 1 & 4 \... | Yes |
Find the transpose of the following matrices. | Solution We find each transpose using the definition without explanation. Make note of the dimensions of the original matrix and the dimensions of its transpose.\n\n\[ \n{A}^{T} = \left\lbrack \begin{matrix} 7 & 2 & - 5 \\ 2 & - 1 & 3 \\ 9 & 3 & 0 \\ 1 & 0 & {11} \end{matrix}\right\rbrack \;{B}^{T} = \left\lbrack \begi... | Yes |
Identify the diagonal of each matrix, and state whether each matrix is diagonal, upper triangular, lower triangular, or none of the above. | Solution We first compute the transpose of each matrix.\n\n\[ \n{A}^{T} = \left\lbrack \begin{array}{lll} 1 & 0 & 0 \\ 2 & 4 & 0 \\ 3 & 5 & 6 \end{array}\right\rbrack \;{B}^{T} = \left\lbrack \begin{matrix} 3 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & - 1 \end{matrix}\right\rbrack \;{C}^{T} = \left\lbrack \begin{array}{llll} 1 & ... | Yes |
Find \( {A}^{T} + {B}^{T} \) and \( {\left( A + B\right) }^{T} \) . | Solution We note that\n\n\[ \n{A}^{T} = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 5 \\ 3 & 6 \end{array}\right\rbrack \text{ and }{B}^{T} = \left\lbrack \begin{matrix} 1 & 3 \\ 2 & - 1 \\ 1 & 0 \end{matrix}\right\rbrack .\n\]\n\nTherefore\n\n\[ \n{A}^{T} + {B}^{T} = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 5 \\ 3 ... | Yes |
Find \( {\left( AB\right) }^{T},{A}^{T}{B}^{T} \) and \( {B}^{T}{A}^{T} \) . | Solution We first note that\n\n\[ {A}^{T} = \left\lbrack \begin{array}{ll} 1 & 3 \\ 2 & 4 \end{array}\right\rbrack \text{ and }{B}^{T} = \left\lbrack \begin{matrix} 1 & 1 \\ 2 & 0 \\ - 1 & 1 \end{matrix}\right\rbrack \]\n\nFind \( {\left( AB\right) }^{T} \) :\n\n\[ {\left( AB\right) }^{T} = {\left( \left\lbrack \begin{... | Yes |
Find \( {\left( {A}^{-1}\right) }^{T} \) and \( {\left( {A}^{T}\right) }^{-1} \) . | Solution We first find \( {A}^{-1} \) and \( {A}^{T} \) :\n\n\[ \n{A}^{-1} = \left\lbrack \begin{matrix} 4 & - 7 \\ - 1 & 2 \end{matrix}\right\rbrack \text{ and }{A}^{T} = \left\lbrack \begin{array}{ll} 2 & 1 \\ 7 & 4 \end{array}\right\rbrack .\n\]\n\nFinding \( {\left( {A}^{-1}\right) }^{T} \) :\n\n\[ \n{\left( {A}^{-... | Yes |
Find \( A{A}^{T}, A + {A}^{T} \) and \( A - {A}^{T} \) . | Solution Finding \( A{A}^{T} \) :\n\n\[ A{A}^{T} = \left\lbrack \begin{matrix} 2 & 1 & 3 \\ 2 & - 1 & 1 \\ 1 & 0 & 1 \end{matrix}\right\rbrack \left\lbrack \begin{matrix} 2 & 2 & 1 \\ 1 & - 1 & 0 \\ 3 & 1 & 1 \end{matrix}\right\rbrack \]\n\n\[ = \left\lbrack \begin{matrix} {14} & 6 & 5 \\ 6 & 4 & 3 \\ 5 & 3 & 2 \end{ma... | Yes |
Find the trace of \( A, B, C \) and \( {I}_{4} \), where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack, B = \left\lbrack \begin{matrix} 1 & 2 & 0 \\ 3 & 8 & 1 \\ - 2 & 7 & - 5 \end{matrix}\right\rbrack \text{ and }C = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \end{array}\... | Solution To find the trace of \( A \), note that the diagonal elements of \( A \) are 1 and 4. Therefore, \( \operatorname{tr}\left( A\right) = 1 + 4 = 5 \) .\n\nWe see that the diagonal elements of \( B \) are 1,8 and -5, so \( \operatorname{tr}\left( B\right) = 1 + 8 - 5 = 4 \) .\n\nThe matrix \( C \) is not a square... | Yes |
Find the determinant of \( A, B \) and \( C \) where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack, B = \left\lbrack \begin{matrix} 3 & - 1 \\ 2 & 7 \end{matrix}\right\rbrack \text{ and }C = \left\lbrack \begin{matrix} 1 & - 3 \\ - 2 & 6 \end{matrix}\right\rbrack . \] | Solution Finding the determinant of \( A \) :\n\n\[ \det \left( A\right) = \left| \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right| \]\n\n\[ = 1\left( 4\right) - 2\left( 3\right) \]\n\n\[ = - 2\text{.} \]\n\nSimilar computations show that \( \det \left( B\right) = 3\left( 7\right) - \left( {-1}\right) \left( 2\right)... | No |
Find the cofactor expansions along the second row and down the first column. | Solution 1 By the definition, the cofactor expansion along the second row is\nthe sum\n\n\[{a}_{2,1}{C}_{2,1} + {a}_{2,2}{C}_{2,2} + {a}_{2,3}{C}_{2,3}\]\n\n(Be sure to compare the above line to the definition of cofactor expansion, and see how the \ | No |
\[ A = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{array}\right\rbrack \] | Solution Notice that this is the matrix from Example 70. The cofactor expansion along the first row is\n\n\[ \det \left( A\right) = {a}_{1,1}{C}_{1,1} + {a}_{1,2}{C}_{1,2} + {a}_{1,3}{C}_{1,3} \]\n\nWe’ll compute each cofactor first then take the appropriate sum.\n\n\[ {C}_{1,1} = {\left( -1\right) }^{1 + 1}{A}_{1,1} \... | Yes |
Example 72 Find the determinant of\n\n\[ A = \left\lbrack \begin{matrix} 3 & 6 & 7 \\ 0 & 2 & - 1 \\ 3 & - 1 & 1 \end{matrix}\right\rbrack \] | Solution We'll compute each cofactor first then find the determinant.\n\n\[ {C}_{1,1} = {\left( -1\right) }^{1 + 1}{A}_{1,1} \]\n\n\[ {C}_{1,2} = {\left( -1\right) }^{1 + 2}{A}_{1,2} \]\n\[{C}_{1,3} = {\left( -1\right) }^{1 + 3}{A}_{1,3}\]\n\n\[ = 1 \cdot \left| \begin{matrix} 2 & - 1 \\ - 1 & 1 \end{matrix}\right| \] ... | Yes |
Example 74 Find the determinant of\n\n\[ A = \left\lbrack \begin{matrix} 1 & 2 & 0 & 9 \\ 2 & - 3 & 0 & 5 \\ 7 & 2 & 3 & 8 \\ - 4 & 1 & 0 & 2 \end{matrix}\right\rbrack \] | Solution Our first reaction may well be \ | No |
Find the determinant of\n\n\[ A = \left\lbrack \begin{matrix} 1 & 2 & 3 & 4 & 5 \\ 0 & 6 & 7 & 8 & 9 \\ 0 & 0 & {10} & {11} & {12} \\ 0 & 0 & 0 & {13} & {14} \\ 0 & 0 & 0 & 0 & {15} \end{matrix}\right\rbrack \] | Solution At first glance, we think \ | No |
Find \( \det \left( A\right) \) as well as \( \det \left( B\right) \) for each of the row operations above. | Solution It is straightforward to compute \( \det \left( A\right) = - 2 \) . Let \( B \) be formed by performing the row operation in 1) on \( A \) ; thus \[ B = \left\lbrack \begin{array}{ll} 1 & 2 \\ 5 & 8 \end{array}\right\rbrack \] It is clear that \( \det \left( B\right) = - 2 \), the same as \( \det \left( A\righ... | Yes |
Compute \( \det \left( A\right) \), then find the determinants of the following matrices by inspection using Theorem 15. | Solution Computing \( \det \left( A\right) \) by cofactor expansion down the first column or along the second row seems like the best choice, utilizing the one zero in the matrix. We can quickly confirm that \( \det \left( A\right) = 1 \) . To compute \( \det \left( B\right) \), notice that the rows of \( A \) were rea... | Yes |
Find the determinant of \( A \) by first putting \( A \) into a triangular form, where \( A = \left\lbrack \begin{matrix} 2 & 4 & - 2 \\ - 1 & - 2 & 5 \\ 3 & 2 & 1 \end{matrix}\right\rbrack \) | In putting \( A \) into a triangular form, we need not worry about getting leading \( 1\mathrm{\;s} \), but it does tend to make our life easier as we work out a problem by hand. So let’s scale the first row by \( 1/2 \) : \( \left\lbrack \begin{matrix} 1 & 2 & - 1 \\ - 1 & - 2 & 5 \\ 3 & 2 & 1 \end{matrix}\right\rbrac... | Yes |
Find the determinants of the matrices \( A, B, A + B,{3A},{AB},{A}^{T},{A}^{-1} \), and compare the determinant of these matrices to their trace. | Solution We can quickly compute that \( \det \left( A\right) = - 2 \) and that \( \det \left( B\right) = 7 \) . \n\n\[ \det \left( {A - B}\right) = \det \left( {\left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack - \left\lbrack \begin{array}{ll} 2 & 1 \\ 3 & 5 \end{array}\right\rbrack }\right) \]\n\n... | No |
Example 81 Find the determinant of \( A \) using the previously described shortcut, where \[ A = \left\lbrack \begin{matrix} 1 & 3 & 9 \\ - 2 & 3 & 4 \\ - 5 & 7 & 2 \end{matrix}\right\rbrack \] | Rewriting the first 2 columns, drawing the proper diagonals, and multiplying, we get:  Summing the numbers on the right and subtracting the sum of the numbers on the left, we get \[ \det \left( A\right) = \left( {6 -... | Yes |
Example 82 Use Cramer’s Rule to solve the linear system \( A\overrightarrow{x} = \overrightarrow{b} \) where\n\n\[ A = \left\lbrack \begin{matrix} 1 & 5 & - 3 \\ 1 & 4 & 2 \\ 2 & - 1 & 0 \end{matrix}\right\rbrack \text{ and }\overrightarrow{b} = \left\lbrack \begin{matrix} - {36} \\ - {11} \\ 7 \end{matrix}\right\rbrac... | Solution We first compute the determinant of \( A \) to see if we can apply Cramer’s Rule.\n\n\[ \det \left( A\right) = \left| \begin{matrix} 1 & 5 & - 3 \\ 1 & 4 & 2 \\ 2 & - 1 & 0 \end{matrix}\right| = {49} \]\n\nSince \( \det \left( A\right) \neq 0 \), we can apply Cramer’s Rule. Following Theorem 18, we compute \( ... | Yes |
Example 83 Use Cramer’s Rule to solve the linear system \( A\overrightarrow{x} = \overrightarrow{b} \) where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right\rbrack \text{ and }\overrightarrow{b} = \left\lbrack \begin{matrix} - 1 \\ 1 \end{matrix}\right\rbrack \] | Solution The determinant of \( A \) is -2, so we can apply Cramer’s Rule.\n\n\[ \det \left( {{A}_{1}\left( \overrightarrow{b}\right) }\right) = \left| \begin{matrix} - 1 & 2 \\ 1 & 4 \end{matrix}\right| = - 6 \]\n\n\[ \det \left( {{A}_{2}\left( \overrightarrow{b}\right) }\right) = \left| \begin{matrix} 1 & - 1 \\ 3 & 1... | Yes |
Find the eigenvalues of \( A \), that is, find \( \lambda \) such that \( \det \left( {A - {\lambda I}}\right) = 0 \) | where\n\n\[ A = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 3 \end{array}\right\rbrack \]\n\n1 (Note that this is the matrix we used at the beginning of this section.) First, we write out what \( A - {\lambda I} \) is:\n\n\[ A - {\lambda I} = \left\lbrack \begin{array}{ll} 1 & 4 \\ 2 & 3 \end{array}\right\rbrack - \lam... | Yes |
Example 85 Find \( \\overrightarrow{x} \) such that \( A\\overrightarrow{x} = 5\\overrightarrow{x} \), where\n\n\[ A = \\left\\lbrack \\begin{array}{ll} 1 & 4 \\\\ 2 & 3 \\end{array}\\right\\rbrack \] | Solution Recall that our algebra from before showed that if\n\n\[ A\\overrightarrow{x} = \\lambda \\overrightarrow{x}\\;\\text{ then }\\;\\left( {A - {\\lambda I}}\\right) \\overrightarrow{x} = \\overrightarrow{0}. \]\n\nTherefore, we need to solve the equation \( \\left( {A - {\\lambda I}}\\right) \\overrightarrow{x} ... | Yes |
Find the eigenvalues of \( A \), and for each eigenvalue, find an eigenvector where \( A = \left\lbrack \begin{matrix} - 3 & {15} \\ 3 & 9 \end{matrix}\right\rbrack \) | Solution To find the eigenvalues, we must compute det \( \left( {A - {\lambda I}}\right) \) and set it equal to 0 . \[ \det \left( {A - {\lambda I}}\right) = \left| \begin{matrix} - 3 - \lambda & {15} \\ 3 & 9 - \lambda \end{matrix}\right| \] \[ = \left( {-3 - \lambda }\right) \left( {9 - \lambda }\right) - {45} \] \[ ... | No |
Example 87 Let \( A = \left\lbrack \begin{matrix} - 3 & 0 \\ 5 & 1 \end{matrix}\right\rbrack \) . Find the eigenvalues of \( A \) and an eigenvector for each eigenvalue. | We first compute the characteristic polynomial, set it equal to 0 , then solve for \( \lambda \) .\n\n\[ \det \left( {A - {\lambda I}}\right) = \left| \begin{matrix} - 3 - \lambda & 0 \\ 5 & 1 - \lambda \end{matrix}\right| \]\n\n\[ = \left( {-3 - \lambda }\right) \left( {1 - \lambda }\right) \]\n\nFrom this, we see tha... | Yes |
Find the eigenvalues of \( A \), and for each eigenvalue, give one eigenvector, where\n\n\[ A = \left\lbrack \begin{matrix} - 7 & - 2 & {10} \\ - 3 & 2 & 3 \\ - 6 & - 2 & 9 \end{matrix}\right\rbrack \] | Solution We first compute the characteristic polynomial, set it equal to 0 , then solve for \( \lambda \) . A warning: this process is rather long. We’ll use cofactor expansion along the first row; don't get bogged down with the arithmetic that comes from each step;\n\n\[ \det \left( {A - {\lambda I}}\right) = \left| \... | Yes |
Example 90 Let \( A = \left\lbrack \begin{array}{lll} 1 & 2 & 3 \\ 0 & 4 & 5 \\ 0 & 0 & 6 \end{array}\right\rbrack \) . Find the eigenvalues of \( A \) . | Solution To find the eigenvalues, we compute \( \det \left( {A - {\lambda I}}\right) \) :\n\n\[ \det \left( {A - {\lambda I}}\right) = \left| \begin{matrix} 1 - \lambda & 2 & 3 \\ 0 & 4 - \lambda & 5 \\ 0 & 0 & 6 - \lambda \end{matrix}\right| \]\n\n\[ = \left( {1 - \lambda }\right) \left( {4 - \lambda }\right) \left( {... | Yes |
Find the eigenvalues and eigenvectors of \( A \) and \( B \) | For \( A \), we have eigenvalues \( \lambda = - 6 \) and 12, with eigenvectors\n\n\[ \overrightarrow{x} = {x}_{2}\left\lbrack \begin{matrix} - 5 \\ 1 \end{matrix}\right\rbrack \text{ and }{x}_{2}\left\lbrack \begin{array}{l} 1 \\ 1 \end{array}\right\rbrack \text{, respectively. }\n\nFor \( B \), we have eigenvalues \( ... | Yes |
Find the eigenvalues and eigenvectors of the matrix \( A = \left\lbrack \begin{array}{ll} 1 & 2 \\ 1 & 2 \end{array}\right\rbrack \) . | Solution To find the eigenvalues, we compute \( \det \left( {A - {\lambda I}}\right) \) :\n\n\[ \det \left( {A - {\lambda I}}\right) = \left| \begin{matrix} 1 - \lambda & 2 \\ 1 & 2 - \lambda \end{matrix}\right| \]\n\n\[ = \left( {1 - \lambda }\right) \left( {2 - \lambda }\right) - 2 \]\n\n\[ = {\lambda }^{2} - {3\lamb... | Yes |
Example 94 Plot the transformed unit square after it has been transformed by \( A \), where \[ A = \left\lbrack \begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}\right\rbrack \] | Solution We’ll put the vectors that correspond to each corner in a matrix \( B \) as before and then multiply it on the left by \( A \) . Doing so gives: \[ {AB} = \left\lbrack \begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix}\right\rbrack \left\lbrack \begin{array}{llll} 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 1 \end{array}\right\rbr... | Yes |
First, find the matrix \( A \) that transforms the Cartesian plane by stretching it vertically by a factor of 1.5 , then stretches it horizontally by a factor of 0.5 , then rotates it clockwise about the origin \( {90}^{ \circ } \) . Secondly, using the new locations of \( \overrightarrow{{e}_{1}} \) and \( \overrighta... | Solution To find \( A \), first consider the new location of \( \overrightarrow{{e}_{1}} \) . Stretching the plane vertically does not affect \( \overrightarrow{{e}_{1}} \) ; stretching the plane horizontally by a factor of 0.5 changes \( \overrightarrow{{e}_{1}} \) to \( \left\lbrack \begin{matrix} 1/2 \\ 0 \end{matri... | Yes |
Find the matrix \( A \) that transforms the Cartesian plane by performing the following operations in order:\n1. Vertical shear by a factor of 3. Horizontal stretch by a factor of 2\n2. Counterclockwise rotation about the origin by an angle of \( \theta = {30}^{ \circ } \)\n3. Diagonal reflection across the line \( y =... | Solution Wow! We already know how to do this - sort of. We know we can find the columns of \( A \) by tracing where \( \overrightarrow{{e}_{1}} \) and \( \overrightarrow{{e}_{2}} \) end up, but this also seems difficult. There is so much that is going on. Fortunately, we can accomplish what we need without much difficu... | No |
Determine whether or not the transformation \( T : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{3} \) is a linear transformation, where\n\n\[ T\left( \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \end{array}\right\rbrack \right) = \left\lbrack \begin{matrix} {x}_{1}^{2} \\ 2{x}_{1} \\ {x}_{1}{x}_{2} \end{matrix}\right... | We’ll arbitrarily pick two vectors \( \overrightarrow{x} \) and \( \overrightarrow{y} \) :\n\n\[ \overrightarrow{x} = \left\lbrack \begin{matrix} 3 \\ - 2 \end{matrix}\right\rbrack \;\text{ and }\;\overrightarrow{y} = \left\lbrack \begin{array}{l} 1 \\ 5 \end{array}\right\rbrack \]\n\nLet’s check to see if \( T \) is l... | Yes |
Let \( T : {\mathbb{R}}^{n} \rightarrow {\mathbb{R}}^{m} \) be a linear transformation. Then \( \left\lbrack T\right\rbrack \) is the \( m \times n \) matrix: | \[ \left\lbrack T\right\rbrack = \left\lbrack \begin{array}{llll} T\left( \overrightarrow{{e}_{1}}\right) & T\left( \overrightarrow{{e}_{2}}\right) & \cdots & T\left( \overrightarrow{{e}_{n}}\right) \end{array}\right\rbrack . \] | Yes |
Define \( T : {\mathbb{R}}^{3} \rightarrow {\mathbb{R}}^{4} \) to be the linear transformation where\n\n\[ T\left( \left\lbrack \begin{array}{l} {x}_{1} \\ {x}_{2} \\ {x}_{3} \end{array}\right\rbrack \right) = \left\lbrack \begin{matrix} {x}_{1} + {x}_{2} \\ 3{x}_{1} - {x}_{3} \\ 2{x}_{2} + 5{x}_{3} \\ 4{x}_{1} + 3{x}_... | TON T takes vectors from \( {\mathbb{R}}^{3} \) into \( {\mathbb{R}}^{4} \), so \( \left\lbrack T\right\rbrack \) is going to be a \( 4 \times 3 \) matrix.\n\nNote that\n\n\[ \overrightarrow{{e}_{1}} = \left\lbrack \begin{array}{l} 1 \\ 0 \\ 0 \end{array}\right\rbrack ,\;\overrightarrow{{e}_{2}} = \left\lbrack \begin{a... | Yes |
Example 102 Using Key Idea 16, determine whether or not each of the following transformations is linear. | Solution \( \;{T}_{1} \) is not linear! This may come as a surprise, but we are not allowed to add constants to the variables. By thinking about this, we can see that this transformation is trying to accomplish the translation that got us started in this section - it adds 1 to all the \( x \) values and leaves the \( y... | No |
Example 103 Sketch the following vectors with their origin at the origin.\n\n\[ \overrightarrow{v} = \left\lbrack \begin{array}{l} 2 \\ 1 \\ 3 \end{array}\right\rbrack \;\text{ and }\;\overrightarrow{u} = \left\lbrack \begin{matrix} 1 \\ 3 \\ - 1 \end{matrix}\right\rbrack \] | Solution We’ll start with \( \overrightarrow{v} \) first. Starting at the origin, move 2 units in the \( x \) direction. This puts us at the point \( \left( {2,0,0}\right) \) on the \( x \) axis. Then, move 1 unit in the \( y \) direction. (In our method of drawing, this means moving 1 unit directly to the right. Of co... | No |
Example 104 Sketch the vector \( \overrightarrow{v} = \left\lbrack \begin{matrix} - 3 \\ - 1 \\ 2 \end{matrix}\right\rbrack \) . | Solution We use the same procedure we used in Example 103. Starting at the origin, we move in the negative \( x \) direction 3 units, then 1 unit in the negative \( y \) direction, and then finally up 2 units in the \( z \) direction to find the point \( \left( {-3, - 1,2}\right) \) . We follow by drawing an arrow. Our... | Yes |
Draw the vectors \( \overrightarrow{v} = \left\lbrack \begin{array}{l} 2 \\ 4 \\ 2 \end{array}\right\rbrack \) and \( \overrightarrow{u} = \left\lbrack \begin{matrix} - 2 \\ 1 \\ - 1 \end{matrix}\right\rbrack \) on the same coordinate system. | Solution We follow the steps we've taken before to sketch these vectors, shown in Figure 5.17. The dashed lines are aides for \( \overrightarrow{v} \) and the dotted lines are aids for \( \overrightarrow{u} \) . We again include the vectors without the dashed and dotted lines; but without these, it is very difficult to... | Yes |
Example 106 Let \( \\overrightarrow{v} = \\left\\lbrack \\begin{array}{l} 2 \\\\ 1 \\\\ 3 \\end{array}\\right\\rbrack \) and \( \\overrightarrow{u} = \\left\\lbrack \\begin{matrix} 1 \\\\ 3 \\\\ - 1 \\end{matrix}\\right\\rbrack \) . Sketch \( \\overrightarrow{v} + \\overrightarrow{u} \) . | Solution We sketched each of these vectors previously in Example 103. We sketch them, along with \( \\overrightarrow{v} + \\overrightarrow{u} = \\left\\lbrack \\begin{array}{l} 3 \\\\ 4 \\\\ 2 \\end{array}\\right\\rbrack \), in Figure 5.19 (a). (We use loosely dashed lines for \( \\overrightarrow{v} + \\overrightarrow{... | Yes |
Example 107 Let \( \\overrightarrow{v} = \\left\\lbrack \\begin{array}{l} 2 \\\\ 1 \\\\ 3 \\end{array}\\right\\rbrack \) and \( \\overrightarrow{u} = \\left\\lbrack \\begin{matrix} 1 \\\\ 3 \\\\ - 1 \\end{matrix}\\right\\rbrack \) . Sketch \( \\overrightarrow{v} - \\overrightarrow{u} \) . | Solution It is simple to compute that \( \\overrightarrow{v} - \\overrightarrow{u} = \\left\\lbrack \\begin{matrix} 1 \\\\ - 2 \\\\ 4 \\end{matrix}\\right\\rbrack \) . All three of these vectors are sketched in Figure 5.20 (a), where again \( \\overrightarrow{v} \) is guided by the dashed, \( \\overrightarrow{u} \) by ... | Yes |
Vectors \( \overrightarrow{v} \) and \( \overrightarrow{u} \) are drawn in Figure 5.21. Using this drawing, sketch the vectors \( \overrightarrow{v} + \overrightarrow{u} \) and \( \overrightarrow{v} - \overrightarrow{u} \) . | Solution Using the Parallelogram Law, we draw \( \overrightarrow{v} + \overrightarrow{u} \) by first drawing a gray version of \( \overrightarrow{u} \) coming from the tip of \( \overrightarrow{v};\overrightarrow{v} + \overrightarrow{u} \) is drawn dashed in Figure 5.22.\n\nTo draw \( \overrightarrow{v} - \overrightarr... | Yes |
Sketch \( \overrightarrow{v},2\overrightarrow{v} \), and \( - \overrightarrow{v} \), where\n\n\[ \overrightarrow{v} = \left\lbrack \begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right\rbrack \] | Solution\n\n\n\nFigure 5.23: Sketching scalar multiples of \( \overrightarrow{v} \) in Example 109.\n\nIt is easy to compute\n\n\[ 2\overrightarrow{v} = \left\lbrack \begin{array}{l} 2 \\ 4 \\ 6 \end{array}\right\rbr... | Yes |
Find the lengths of vectors \( \overrightarrow{v} \) and \( \overrightarrow{u} \), where\n\n\[ \overrightarrow{v} = \left\lbrack \begin{matrix} 2 \\ - 3 \\ 5 \end{matrix}\right\rbrack \;\text{ and }\;\overrightarrow{u} = \left\lbrack \begin{matrix} - 4 \\ 7 \\ 0 \end{matrix}\right\rbrack \] | Solution We apply Definition 32 to each vector:\n\n\[ \parallel \overrightarrow{v}\parallel = \sqrt{{2}^{2} + {\left( -3\right) }^{2} + {5}^{2}} \]\n\n\[ = \sqrt{4 + 9 + {25}} \]\n\n\[ = \sqrt{38} \]\n\n\[ \parallel \overrightarrow{u}\parallel = \sqrt{{\left( -4\right) }^{2} + {7}^{2} + {0}^{2}} \]\n\n\[ = \sqrt{{16} +... | Yes |
Consider the following ODE defined on \( \mathbb{R} \) :\n\n\[ \n{\dot{x}}^{2} + {x}^{2} + {t}^{2} = - 1,\;x \in \mathbb{R}. \n\] | This ODE has no solutions since the left hand side is nonnegative and the right hand side is strictly negative. | Yes |
\[ \dot{x} = {ax},\;x \in {\mathbb{R}}^{n}, \] | The solution is given by \[ x\left( t\right) = c{e}^{at}. \] So we see that there are an infinite number of solutions, depending upon the choice of the constant \( c \). If we evaluate the solution (1.6) at \( t = 0 \) we see that \[ x\left( 0\right) = c. \] Substituting this into the solution (1.6), the solution has t... | Yes |
Consider the following ODE defined on \( \mathbb{R} \) :\n\n\[ \n\dot{x} = 3{x}^{\frac{2}{3}},\;x\left( 0\right) = 0, x \in \mathbb{R}.\n\] | It is easy to see that a solution satisfying \( x\left( 0\right) = 0 \) is \( x = 0 \) . However, one can verify directly by substituting into the equation that the following is also a solution satisfying \( x\left( 0\right) = 0 \) :\n\n\[ \nx\left( t\right) = \left\{ \begin{array}{l} 0,\;t \leq a \\ {\left( t - a\righ... | Yes |
Example 4. An example of an ODE with unique solutions that exists only for a finite time.\n\nConsider the following ODE on \( \\mathbb{R} \) :\n\n\[ \n\\dot{x} = {x}^{2}\\;x\\left( 0\\right) = {x}_{0}, x \\in \\mathbb{R}.\n\]\n\n(1.11) | We can easily integrate this equation (it is separable) to obtain the following solution satisfying the initial condition:\n\n\[ \nx\\left( t\\right) = \\frac{{x}_{0}}{1 - {x}_{0}t}\n\]\n\n(1.12)\n\nThe solution becomes infinite, or \ | No |
Consider the following one dimensional autonomous vector field:\n\n\[ \dot{x} = {\lambda x}, x\left( 0\right) = {x}_{0},\;x \in \mathbb{R},\lambda \in \mathbb{R}. \] | The solution is given by:\n\n\[ x\left( {t,0,{x}_{0}}\right) = x\left( {t,{x}_{0}}\right) = {e}^{\lambda t}{x}_{0}. \]\n\nThe time shifted solution is given by:\n\n\[ x\left( {t + s,{x}_{0}}\right) = {e}^{\lambda \left( {t + s}\right) }{x}_{0} \]\n\nWe see that it is a solution of the ODE with the following calculation... | Yes |
Consider the following one dimensional vector field on \( \mathbb{R} \):\n\n\[ \dot{x} = {\lambda tx},\;x\left( 0\right) = {x}_{0},\;x \in \mathbb{R},\lambda \in \mathbb{R}. \] | This vector field is separable and the solution is easily found to be:\n\n\[ x\left( {t,0,{x}_{0}}\right) = {x}_{0}{e}^{\frac{\lambda }{2}{t}^{2}} \]\n\nThe time shifted \ | Yes |
Consider the following one dimensional nonautonomous vector field:\n\n\[ \n\\dot{x} = {e}^{t},\;x \\in \\mathbb{R}. \n\] | The solution is given by:\n\n\[ \nx\\left( t\\right) = {e}^{t} \n\]\n\nIt is easy to verify that the time-shifted function:\n\n\[ \nx\\left( {t + s}\\right) = {e}^{t + s} \n\]\n\ndoes not satisfy the equation. | No |
Consider the following autonomous vector field on the plane:\n\n\[ \n\\dot{x} = - x \n\]\n\n\[ \n\\dot{y} = {y}^{2}\\left( {1 - {y}^{2}}\\right) \\equiv f\\left( y\\right) ,\\;\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2}.\n\]\n\n(3.10) | The two dimensional vector field (3.10) has equilibrium points at:\n\n\[ \n\\left( {x, y}\\right) = \\left( {0,0}\\right) ,\\;\\left( {0,1}\\right) ,\\;\\left( {0,-1}\\right) .\n\]\n\nIn this example it is easy to identify three invariant horizontal lines (examples of invariant sets). Since \( y = 0 \) implies that \( ... | Yes |
We consider the following autonomous, nonlinear vector field on the plane:\n\n\[ \n\dot{x} = x \n\]\n\n\[ \n\dot{y} = - y + {x}^{2},\;\left( {x, y}\right) \in {\mathbb{R}}^{2}.\n\]\n\n(6.16) | This vector field has an equilibrium point at the origin, \( \left( {x, y}\right) = \left( {0,0}\right) \) . The Jacobian of the vector field evaluated at the origin is given by:\n\n\[ \n\left( \begin{matrix} 1 & 0 \\ 0 & - 1 \end{matrix}\right)\n\]\n\n(6.17)\n\nFrom this calculation we can conclude that the origin is ... | Yes |
Consider the following nonlinear autonomous vector field on through the origin and tangent to the un-the plane: stable subspace at the origin. (6.21) characterizes the unstable manifold in terms of the asymptotic behavior of trajectories (as \( t \rightarrow - \infty \) ) whose initial conditions satisfy a particular c... | \[ \dot{x} = x - {x}^{3} \] \[ \dot{y} = - y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}. \] (6.22) Note that the \( x \) and \( y \) components evolve independently. The equilibrium points and the Jacobians associated with their lineariza-tions are given as follows: \[ \left( {x, y}\right) = \left( {0,0}\right) ;\;\le... | Yes |
In this example we consider the following nonlinear autonomous vector field on the plane:\n\n\[ \dot{x} = - x \]\n\n\[ \dot{y} = {y}^{2}\left( {1 - {y}^{2}}\right) ,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}. \]\n\n(6.28)\n\nNote that the \( x \) and \( y \) components evolve independently.\n\nThe equilibrium points a... | \n\n\( \left( {0,0}\right) \) ;\n\n\[ {W}^{s}\left( \left( {0,0}\right) \right) = \{ \left( {x, y}\right) \mid y = 0\} \]\n\n(6.33)\n\n\[ {W}^{c}\left( \left( {0,0}\right) \right) = \{ \left( {x, y}\right) \mid x = 0, - 1 < y < 1\} \]\n\nThe \( x \) -axis is clearly the global stable manifold for this equilibrium point... | Yes |
We want to classify the linearized stability of the equilibria. | The Jacobian of the vector field is given by:\n\n\[ A = \left( \begin{matrix} 0 & 1 \\ 1 - 3{x}^{2} & - \delta \end{matrix}\right) \]\n\nand the eigenvalues of the Jacobian are:\n\n\[ {\lambda }_{ \pm } = - \frac{\delta }{2} \pm \frac{1}{2}\sqrt{{\delta }^{2} + 4 - {12}{x}^{2}}. \]\n\nWe evaluate this expression for th... | Yes |
Consider the following \( {C}^{r}\left( {r \geq 1}\right) \) autonomous vector field on \( {\mathbb{R}}^{n} \) :\n\n\[ \dot{x} = f\left( x\right) ,\;x \in {\mathbb{R}}^{n}. \] | Let \( x = \bar{x} \) be a fixed point of (7.5) and let \( V : U \rightarrow \mathbb{R} \) be a \( {C}^{1} \) function defined in some neighborhood \( U \) of \( \bar{x} \) such that:\n\n1. \( V\left( \bar{x}\right) = 0 \) and \( V\left( x\right) > 0 \) if \( x \neq \bar{x} \).\n\n2. \( \dot{V}\left( x\right) \leq 0 \)... | Yes |
\[ \dot{x} = y \] \[ \dot{y} = - x - \epsilon {x}^{2}y,\;\left( {x, y}\right) \in {\mathbb{R}}^{2}, \] where \( \epsilon \) is a parameter. It is clear that \( \left( {x, y}\right) = \left( {0,0}\right) \) is an equilibrium point of (7.6) and we want to determine the nature of its stability. | We begin by linearizing (7.6) about this equilibrium point. The matrix associated with this linearization is given by: \[ A = \left( \begin{array}{rr} 0 & 1 \\ - 1 & 0 \end{array}\right) \] and its eigenvalues are \( \pm i \). Hence, the origin is not hyperbolic and therefore the information provided by the linearizati... | Yes |
Consider the following vector field on \( {\mathbb{R}}^{2} \) :\n\n\[ \n\dot{x} = y \]\n\n\[ \n\dot{y} = x - {x}^{3} - {\delta y},\;\left( {x, y}\right) \in {\mathbb{R}}^{2},\;\delta > 0.\n\]\n\n(7.15)\n\nThis vector field has three equilibrium points-a saddle point at \( \left( {x, y}\right) = \left( {0,0}\right) \) a... | Consider the function\n\n\[ \nV\left( {x, y}\right) = \frac{{y}^{2}}{2} - \frac{{x}^{2}}{2} + \frac{{x}^{4}}{4}\n\]\n\n(7.16)\n\nand its level sets:\n\n\[ \nV\left( {x, y}\right) = C\text{.\n\]\n\nWe compute the derivative of \( V \) along trajectories of (7.15):\n\n\[ \n\dot{V}\left( {x, y}\right) = \frac{\partial V}{... | Yes |
We consider the following nonlinear autonomous vector field on the plane:\n\n\[ \n\dot{x} = y \equiv f\left( {x, y}\right) \n\]\n\n\[ \n\dot{y} = x - {x}^{3} - {\delta y} \equiv g\left( {x, y}\right) ,\;\left( {x, y}\right) \in {\mathbb{R}}^{2},\;\delta > 0.\;\left( {7.23}\right) \n\] | Computing (7.22) gives:\n\n\[ \n\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = - \delta \n\]\n\n(7.24)\n\nTherefore this vector field has no periodic orbits for \( \delta \neq 0 \) . | Yes |
We consider the following linear autonomous vector field on the plane:\n\n\[ \n\dot{x} = {ax} + {by} \equiv f\left( {x, y}\right) , \n\]\n\n\[ \n\dot{y} = {cx} + {dy} \equiv g\left( {x, y}\right) ,\;\left( {x, y}\right) \in {\mathbb{R}}^{2},\;a, b, c, d \in \mathbb{R}\text{ (7.25) } \n\] | Computing (7.22) gives:\n\n\[ \n\frac{\partial f}{\partial x} + \frac{\partial g}{\partial y} = a + d \n\]\n\n(7.26)\n\nTherefore for \( a + d \neq 0 \) this vector field has no periodic orbits. | Yes |
We consider the following nonlinear autonomous vector field on the plane:\n\n\[ \n\\dot{x} = y \\equiv f\\left( {x, y}\\right) \n\]\n\n\[ \n\\dot{y} = x - {x}^{3} - {\\delta y} + {x}^{2}y \\equiv g\\left( {x, y}\\right) ,\\;\\left( {x, y}\\right) \\in {\\mathbb{R}}^{2},\\;\\left( {7.27}\\right) \n\]\n\nwhere \( \\delta... | Now we will use Bendixson's criterion and the index theorem to determine regions in the phase plane where periodic orbits may exist. For this example (7.22) is given by:\n\n\[ \n- \\delta + {x}^{2}\\text{.} \n\]\n\n(7.33)\n\nHence the two vertical lines \( x = - \\sqrt{\\delta } \) and \( x = \\sqrt{\\delta } \) divide... | Yes |
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