Chapter 9
Algebra and Trigonometry · 341 exercises
Problem 21
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x+y-z &=-2 \\ 2 x-y+z &=5 \\ -x+2 y+2 z &=1 \end{aligned}\right. $$
3 step solution
Problem 21
Use Cramer’s Rule to solve each system. $$\left\\{\begin{array}{l}{2 x=3 y+2} \\\\{5 x=51-4 y}\end{array}\right.$$
5 step solution
Problem 21
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrr} {1} & {2} & {-1} \\ {-2} & {0} & {1} \\ {1} & {-1} & {0} \end{array}\right] $$
8 step solution
Problem 21
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} w+x-y+z &=-2 \\ 2 w-x+2 y-z &=7 \\ -w+2 x+y+2 z &=-1 \end{aligned}\right. $$
4 step solution
Problem 22
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. $$ 2 X+5 A=B $$
3 step solution
Problem 22
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x-2 y-z &=2 \\ 2 x-y+z &=4 \\ -x+y-2 z &=-4 \end{aligned}\right. $$
3 step solution
Problem 22
Use Cramer’s Rule to solve each system. $$\left\\{\begin{aligned}y &=-4 x+2 \\\2 x &=3 y+8\end{aligned}\right.$$
4 step solution
Problem 22
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} 2 w-3 x+4 y+z &=7 \\ w-x+3 y-5 z &=10 \\ 3 w+x-2 y-2 z &=6 \end{aligned}\right. $$
3 step solution
Problem 22
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 \(A^{-1} A=I\) $$ \text { 22. } A=\left[\begin{array}{rrr} {1} & {-1} & {1} \\ {0} & {2} & {-1} \\ {2} & {3} & {0} \end{array}\right] $$
4 step solution
Problem 23
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. $$ B-X=4 A $$
3 step solution
Problem 23
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x+3 y &=0 \\ x+y+z &=1 \\ 3 x-y-z &=11 \end{aligned}\right. $$
3 step solution
Problem 23
Evaluate each determinant. $$\left|\begin{array}{rrr}{3} & {0} & {0} \\\\{2} & {1} & {-5} \\\\{2} & {5} & {-1}\end{array}\right|$$
3 step solution
Problem 23
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} w+2 x+3 y-z &=7 \\ 2 x-3 y+z &=4 \\ w-4 x+y &=3 \end{aligned}\right. $$
4 step solution
Problem 23
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrr} {2} & {2} & {-1} \\ {0} & {3} & {-1} \\ {-1} & {-2} & {1} \end{array}\right] $$
3 step solution
Problem 24
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. $$ A-X=4 B $$
4 step solution
Problem 24
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{r} {3 y-z=-1} \\ {x+5 y-z=-4} \\ {-3 x+6 y+2 z=11} \end{array}\right. $$
3 step solution
Problem 24
Evaluate each determinant. $$\left|\begin{array}{rrr}{4} & {0} & {0} \\\\{3} & {-1} & {4} \\\\{2} & {-3} & {5}\end{array}\right|$$
4 step solution
Problem 24
Use Gaussian elimination to find the complete solution to each system of equations, or show that none exists. $$ \left\\{\begin{aligned} w-x &+z=0 \\ w-4 x+y+2 z &=0 \\ 3 w-& y+2 z=0 \end{aligned}\right. $$
3 step solution
Problem 24
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 \(A^{-1} A=I\) $$ \text { 24. } A=\left[\begin{array}{rrr} {2} & {4} & {-4} \\ {1} & {3} & {-4} \\ {2} & {4} & {-3} \end{array}\right] $$
4 step solution
Problem 25
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. $$ 4 A+3 B=-2 X $$
3 step solution
Problem 25
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} 2 x-y-z &=4 \\ x+y-5 z &=-4 \\ x-2 y &=4 \end{aligned}\right. $$
5 step solution
Problem 25
Evaluate each determinant. $$\left|\begin{array}{rrr}{3} & {1} & {0} \\\\{-3} & {4} & {0} \\\\{-1} & {3} & {-5}\end{array}\right|$$
4 step solution
Problem 25
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrr} {5} & {0} & {2} \\ {2} & {2} & {1} \\ {-3} & {1} & {-1} \end{array}\right] $$
6 step solution
Problem 26
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. $$ 4 B+3 A=-2 X $$
3 step solution
Problem 26
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} x &-3 z=-2 \\ 2 x+2 y+z &=4 \\ 3 x+y-2 z &=5 \end{aligned}\right. $$
3 step solution
Problem 26
Evaluate each determinant. $$\left|\begin{array}{rrr}{2} & {-4} & {2} \\\\{-1} & {0} & {5} \\\\{3} & {0} & {4}\end{array}\right|$$
3 step solution
Problem 26
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{lll} {3} & {2} & {6} \\ {1} & {1} & {2} \\ {2} & {2} & {5} \end{array}\right] $$
3 step solution
Problem 27
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{ll} {1} & {3} \\ {5} & {3} \end{array}\right], \quad B=\left[\begin{array}{rr} {3} & {-2} \\ {-1} & {6} \end{array}\right] $$
4 step solution
Problem 27
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{l} {x+y+z=4} \\ {x-y-z=0} \\ {x-y+z=2} \end{array}\right. $$
5 step solution
Problem 27
Evaluate each determinant. $$\left|\begin{array}{rrr}{1} & {1} & {1} \\\\{2} & {2} & {2} \\\\{-3} & {4} & {-5}\end{array}\right|$$
3 step solution
Problem 27
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrrr} {1} & {0} & {0} & {0} \\ {0} & {-1} & {0} & {0} \\ {0} & {0} & {3} & {0} \\ {1} & {0} & {0} & {1} \end{array}\right] $$
4 step solution
Problem 28
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{rr} {3} & {-2} \\ {1} & {5} \end{array}\right], \quad B=\left[\begin{array}{rr} {0} & {0} \\ {5} & {-6} \end{array}\right] $$
2 step solution
Problem 28
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} 3 x+y-z &=0 \\ x+y+2 z &=6 \\ 2 x+2 y+3 z &=10 \end{aligned}\right. $$
4 step solution
Problem 28
Evaluate each determinant. $$\left|\begin{array}{rrr}{1} & {2} & {3} \\\\{2} & {2} & {-3} \\\\{3} & {2} & {1}\end{array}\right|$$
3 step solution
Problem 28
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 \(A^{-1} A=I\) $$ A=\left[\begin{array}{rrrr} {2} & {0} & {0} & {1} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {-1} & {0} \\ {0} & {0} & {0} & {2} \end{array}\right] $$
3 step solution
Problem 29
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{llll} {1} & {2} & {3} & {4} \end{array}\right], \quad B=\left[\begin{array}{l} {1} \\ {2} \\ {3} \\ {4} \end{array}\right] $$
3 step solution
Problem 29
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{l} {x+2 y=z-1} \\ {x=4+y-z} \\ {x+y-3 z=-2} \end{array}\right. $$
3 step solution
Problem 29
Use Cramer's Rule to solve each system. $$\left\\{\begin{aligned}x+y+z &=0 \\\2 x-y+z &=-1 \\\\-x+3 y-z &=-8\end{aligned}\right.$$
5 step solution
Problem 29
Write each linear system as a matrix equation in the form \(A X=B\), where \(A\) is the coefficient matrix and \(B\) is the constant matrix. $$ \left\\{\begin{array}{l} {6 x+5 y=13} \\ {5 x+4 y=10} \end{array}\right. $$
5 step solution
Problem 30
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{l} {-1} \\ {-2} \\ {-3} \end{array}\right], \quad B=\left[\begin{array}{lll} {1} & {2} & {3} \end{array}\right] $$
3 step solution
Problem 30
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{r} {2 x+y=z+1} \\ {2 x=1+3 y-z} \\ {x+y+z=4} \end{array}\right. $$
4 step solution
Problem 30
Use Cramer's Rule to solve each system. $$\left\\{\begin{aligned}x-y+2 z &=3 \\\2 x+3 y+z &=9 \\\\-x-y+3 z &=11\end{aligned}\right.$$
3 step solution
Problem 30
Write each linear system as a matrix equation in the form \(A X=B\), where \(A\) is the coefficient matrix and \(B\) is the constant matrix. $$ \left\\{\begin{array}{l} {7 x+5 y=23} \\ {3 x+2 y=10} \end{array}\right. $$
4 step solution
Problem 31
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{rrr} {1} & {-1} & {4} \\ {4} & {-1} & {3} \\ {2} & {0} & {-2} \end{array}\right], \quad B=\left[\begin{array}{rrr} {1} & {1} & {0} \\ {1} & {2} & {4} \\ {1} & {-1} & {3} \end{array}\right] $$
2 step solution
Problem 31
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{array}{r} {3 a-b-4 c=3} \\ {2 a-b+2 c=-8} \\ {a+2 b-3 c=9} \end{array}\right. $$
4 step solution
Problem 31
Use Cramer's Rule to solve each system. $$\left\\{\begin{aligned}4 x-5 y-6 z &=-1 \\\x-2 y-5 z &=-12 \\\2 x-y &=7\end{aligned}\right.$$
4 step solution
Problem 31
Write each linear system as a matrix equation in the form \(A X=B\), where \(A\) is the coefficient matrix and \(B\) is the constant matrix. $$ \left\\{\begin{array}{l} {x+3 y+4 z=-3} \\ {x+2 y+3 z=-2} \\ {x+4 y+3 z=-6} \end{array}\right. $$
4 step solution
Problem 32
Find (if possible) the following matrices: a. \(A B\) b. \(B A\) $$ A=\left[\begin{array}{rrr} {1} & {-1} & {1} \\ {5} & {0} & {-2} \\ {3} & {-2} & {2} \end{array}\right], \quad B=\left[\begin{array}{rrr} {1} & {1} & {0} \\ {1} & {-4} & {5} \\ {3} & {-1} & {2} \end{array}\right] $$
3 step solution
Problem 32
Solve each system of equations using matrices. Use Gaussian elimination with back-substitution or Gauss-Jordan elimination. $$ \left\\{\begin{aligned} 3 a+b-c &=0 \\ 2 a+3 b-5 c &=1 \\ a-2 b+3 c &=-4 \end{aligned}\right. $$
4 step solution
Problem 32
Write each linear system as a matrix equation in the form \(A X=B\), where \(A\) is the coefficient matrix and \(B\) is the constant matrix. $$ \text { 32. }\left\\{\begin{aligned} x+4 y-z &=3 \\ x+3 y-2 z &=5 \\ 2 x+7 y-5 z &=12 \end{aligned}\right. $$
3 step solution