Chapter 9

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{ll} {1} & {3} \\ {3} & {4} \\ {5} & {6} \end{array}\right], \quad B=\left[\begin{array}{rr} {2} & {-1} \\ {3} & {-2} \\ {0} & {1} \end{array}\right] $$

Write the system of linear equations represented by the augmented matrix. Use \(x, y,\) and \(z,\) or, if necessary, \(w, x, y\) and \(z,\) for the variables. $$ \left[\begin{array}{rrrr|r} {1} & {1} & {4} & {1} & {3} \\ {-1} & {1} & {-1} & {0} & {7} \\ {2} & {0} & {0} & {5} & {11} \\ {0} & {0} & {12} & {4} & {5} \end{array}\right] $$

Use Cramer’s Rule to solve each system. $$ \left\\{\begin{array}{l} {x+y=7} \\ {x-y=3} \end{array}\right. $$

Find the products AB and BA to determine.whether \(B\) is the multiplicative inverse of \(A\). $$ A=\left[\begin{array}{rrrr} {0} & {0} & {-2} & {1} \\ {-1} & {0} & {1} & {1} \\ {0} & {1} & {-1} & {0} \\ {1} & {0} & {0} & {-1} \end{array}\right], \quad B=\left[\begin{array}{rrrr} {1} & {2} & {0} & {3} \\ {0} & {1} & {1} & {1} \\ {0} & {1} & {0} & {1} \\ {1} & {2} & {0} & {2} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} 2 w+x-y &=3 \\ w-3 x+2 y &=-4 \\ 3 w+x-3 y+z &=1 \\ w+2 x-4 y-z &=-2 \end{aligned}\right. $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{rrr} {3} & {1} & {1} \\ {-1} & {2} & {5} \end{array}\right], \quad B=\left[\begin{array}{rrr} {2} & {-3} & {6} \\ {-3} & {1} & {-4} \end{array}\right] $$

Write the system of linear equations represented by the augmented matrix. Use \(x, y,\) and \(z,\) or, if necessary, \(w, x, y\) and \(z,\) for the variables. $$ \left[\begin{array}{rrrr|r} {4} & {1} & {5} & {1} & {6} \\ {1} & {-1} & {0} & {-1} & {8} \\ {3} & {0} & {0} & {7} & {4} \\ {0} & {0} & {11} & {5} & {3} \end{array}\right] $$

Use Cramer’s Rule to solve each system. $$ \left\\{\begin{array}{r} {2 x+y=3} \\ {x-y=3} \end{array}\right. $$

Find the products AB and BA to determine.whether \(B\) is the multiplicative inverse of \(A\). $$ A=\left[\begin{array}{rrrr} {1} & {-2} & {1} & {0} \\ {0} & {1} & {-2} & {1} \\ {0} & {0} & {1} & {-2} \\ {0} & {0} & {0} & {1} \end{array}\right], \quad B=\left[\begin{array}{rrrr} {1} & {2} & {3} & {4} \\ {0} & {1} & {2} & {3} \\ {0} & {0} & {1} & {2} \\ {0} & {0} & {0} & {1} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{ll} {2 w-x+3 y+z=0} \\ {3 w+2 x+4 y-z=0} \\ {5 w-2 x-2 y-z=0} \\ {2 w+3 x-7 y-5 z=0} \end{array}\right. $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{r} {2} \\ {-4} \\ {1} \end{array}\right], \quad B=\left[\begin{array}{r} {-5} \\ {3} \\ {-1} \end{array}\right] $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrr|r} {2} & {-6} & {4} & {10} \\ {1} & {5} & {-5} & {0} \\ {3} & {0} & {4} & {7} \end{array}\right]^{\frac{1}{2} R_{1}} $$

Use Cramer’s Rule to solve each system. $$ \left\\{\begin{array}{r} {12 x+3 y=15} \\ {2 x-3 y=13} \end{array}\right. $$

use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ A=\left[\begin{array}{rr} {2} & {3} \\ {-1} & {2} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} w-3 x+y-4 z &=4 \\ -2 w+x+2 y &=-2 \\ 3 w-2 x+y-6 z &=2 \\ -w+3 x+2 y-z &=-6 \end{aligned}\right. $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{lll} {6} & {2} & {-3} \end{array}\right], B=\left[\begin{array}{lll} {4} & {-2} & {3} \end{array}\right] $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrr|r} {3} & {-12} & {6} & {9} \\ {1} & {-4} & {4} & {0} \\ {2} & {0} & {7} & {4} \end{array}\right]^{\frac{1}{3} R_{1}} $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{aligned}x-2 y &=5 \\\5 x-y &=-2\end{aligned}\right.$$

use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ \text { 14. } A=\left[\begin{array}{rr} {0} & {3} \\ {4} & {-2} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} 3 w+2 x-y+2 z=&-12 \\ 4 w-x+y+2 z=& 1 \\ w+x+y+z=&-2 \\ -2 w+3 x+2 y-3 z=& 10 \end{aligned}\right. $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{rrr} {2} & {-10} & {-2} \\ {14} & {12} & {10} \\ {4} & {-2} & {2} \end{array}\right], \quad B=\left[\begin{array}{rrr} {6} & {10} & {-2} \\ {0} & {-12} & {-4} \\ {-5} & {2} & {-2} \end{array}\right] $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrr|r} {1} & {-3} & {2} & {0} \\ {3} & {1} & {-1} & {7} \\ {2} & {-2} & {1} & {3} \end{array}\right] \quad-3 R_{1}+R_{2} $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{4 x-5 y=17} \\\\{2 x+3 y=3}\end{array}\right.$$

use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ A=\left[\begin{array}{rr} {3} & {-1} \\ {-4} & {2} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{l} {2 x+y-z=2} \\ {3 x+3 y-2 z=3} \end{array}\right. $$

Find the following matrices: a. \(A+B\) b. \(A-B\) c. \(-4 A\) d. \(3 A+2 B\) $$ A=\left[\begin{array}{rrr} {6} & {-3} & {5} \\ {6} & {0} & {-2} \\ {-4} & {2} & {-1} \end{array}\right], \quad B=\left[\begin{array}{rrr} {-3} & {5} & {1} \\ {-1} & {2} & {-6} \\ {2} & {0} & {4} \end{array}\right] $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrr|r} {1} & {-1} & {5} & {-6} \\ {3} & {3} & {-1} & {10} \\ {1} & {3} & {2} & {5} \end{array}\right] \quad-3 R_{1}+R_{2} $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{3 x+2 y=2} \\\\{2 x+2 y=3}\end{array}\right.$$

use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ A=\left[\begin{array}{ll} {2} & {-6} \\ {1} & {-2} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{r} {3 x+2 y-z=5} \\ {x+2 y-z=1} \end{array}\right. $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ X-A=B $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrrr|r} {1} & {-1} & {1} & {1} & {3} \\ {0} & {1} & {-2} & {-1} & {0} \\ {2} & {0} & {3} & {4} & {11} \\ {5} & {1} & {2} & {4} & {6} \end{array}\right] \begin{array}{r} {-2 R_{1}+R_{3}} \\ {-5 R_{1}+R_{4}} \end{array} $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{r}{x+2 y=3} \\\\{3 x-4 y=4}\end{array}\right.$$

Use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ A=\left[\begin{array}{rr} {10} & {-2} \\ {-5} & {1} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{r} {x+2 y+3 z=5} \\ {y-5 z=0} \end{array}\right. $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ X-B=A $$

Perform each matrix row operation and write the new matrix. $$ \left[\begin{array}{rrrr|r} {1} & {-5} & {2} & {-2} & {4} \\ {0} & {1} & {-3} & {-1} & {0} \\ {3} & {0} & {2} & {-1} & {6} \\ {-4} & {1} & {4} & {2} & {-3} \end{array}\right] \begin{array}{r} {-3 R_{1}+R_{3}} \\ {4 R_{1}+R_{4}} \end{array} $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{2 x-9 y=5} \\\\{3 x-3 y=11}\end{array}\right.$$

Use the fact that if \(\boldsymbol{A}=\left[\begin{array}{ll}{\boldsymbol{a}} & {\boldsymbol{b}} \\ {\boldsymbol{c}} & {\boldsymbol{d}}\end{array}\right],\) then \(A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}{d} & {-b} \\ {-c} & {a}\end{array}\right]\) to find the inverse of each matrix, if possible. Check that \(A A^{-1}=I_{2}\) and \(A^{-1} A=I_{2}\) $$ \text { 18. } A=\left[\begin{array}{rr} {6} & {-3} \\ {-2} & {1} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{array}{r} {3 x-y+4 z=8} \\ {y+2 z=1} \end{array}\right. $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ 2 X+A=B $$

A few steps in the process of simplifying the given matrix to row-echelon form, with Is down the diagonal from upper left to lower right, and os below the \(1 s,\) are shown. Fill in the missing numbers in the steps that are shown. $$ \left[\begin{array}{rrr|r} {1} & {-1} & {1} & {8} \\ {2} & {3} & {-1} & {-2} \\ {3} & {-2} & {-9} & {9} \end{array}\right] \rightarrow\left[\begin{array}{rrr|r} {1} & {-1} & {1} & {8} \\ {0} & {5} & {\square} & {\square} \\ {0} & {1} & {\square} & {\square} \end{array}\right] $$ $$ \rightarrow\left[\begin{array}{rrr|r} {1} & {-1} & {1} & {8} \\ {0} & {1} & {\square} & {\square} \\ {0} & {1} & {\square} & {\square} \end{array}\right] $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{3 x-4 y=4} \\\\{2 x+2 y=12}\end{array}\right.$$

Find \(\boldsymbol{A}^{-1}\) by forming \([\boldsymbol{A} | \boldsymbol{I}]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B] .\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$ A=\left[\begin{array}{lll} {2} & {0} & {0} \\ {0} & {4} & {0} \\ {0} & {0} & {6} \end{array}\right] $$

Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} x+y-2 z &=2 \\ 3 x-y-6 z &=-7 \end{aligned}\right. $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ 3 X+A=B $$

A few steps in the process of simplifying the given matrix to row-echelon form, with Is down the diagonal from upper left to lower right, and os below the \(1 s,\) are shown. Fill in the missing numbers in the steps that are shown. $$ \left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {2} & {1} & {-4} & {3} \\ {-3} & {4} & {-1} & {-2} \end{array}\right] \rightarrow\left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {0} & {5} & {\square} & {\square} \\ {0} & {-2} & {\square} & {\square} \end{array}\right] $$ $$ \rightarrow\left[\begin{array}{rrr|r} {1} & {-2} & {3} & {4} \\ {0} & {1} & {\square} & {\square} \\ {0} & {-2} & {\square} & {\square} \end{array}\right] $$

Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{3 x=7 y+1} \\\\{2 x=3 y-1}\end{array}\right.$$

Find \(\boldsymbol{A}^{-1}\) by forming \([\boldsymbol{A} | \boldsymbol{I}]\) and then using row operations to obtain \([I | B],\) where \(A^{-1}=[B] .\) Check that \(A A^{-1}=I\) and \(A^{-1} A=I\) $$ A=\left[\begin{array}{lll} {3} & {0} & {0} \\ {0} & {6} & {0} \\ {0} & {0} & {9} \end{array}\right] $$

Let $$ A=\left[\begin{array}{rr} {-3} & {-7} \\ {2} & {-9} \\ {5} & {0} \end{array}\right] \text { and } B=\left[\begin{array}{rr} {-5} & {-1} \\ {0} & {0} \\ {3} & {-4} \end{array}\right] $$ Solve each matrix equation for X. $$ 3 X+2 A=B $$