Chapter 12

Algebra for College Students · 155 exercises

Problem 1

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rrr} 2 & -1 & 4 \\ -2 & 0 & 5 \end{array}\right], \quad B=\left[\begin{array}{rrr} -1 & 4 & -7 \\ 5 & -6 & 2 \end{array}\right] $$

8 step solution

Problem 1

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 5 & 7 \\ 2 & 3 \end{array}\right] $$

6 step solution

Problem 1

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ A+B $$

4 step solution

Problem 1

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} x+y>3 \\ x-y>1 \end{array}\right) $$

3 step solution

Problem 2

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} 3 & -6 \\ 2 & -1 \\ -4 & 5 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & 0 \\ 5 & -7 \\ -6 & 9 \end{array}\right] $$

4 step solution

Problem 2

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 3 & 4 \\ 2 & 3 \end{array}\right] $$

5 step solution

Problem 2

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ B-C $$

4 step solution

Problem 2

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} x-y<2 \\ x+y<1 \end{array}\right) $$

5 step solution

Problem 3

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{llll} 2 & -1 & 4 & 12 \end{array}\right], \quad B=\left[\begin{array}{llll} -3 & -6 & 9 & -5 \end{array}\right] $$

4 step solution

Problem 3

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 3 & 8 \\ 2 & 5 \end{array}\right] $$

3 step solution

Problem 3

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 3 C+D $$

2 step solution

Problem 4

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{r} 3 \\ -9 \\ 7 \end{array}\right], \quad B=\left[\begin{array}{r} -6 \\ 12 \\ 9 \end{array}\right] $$

4 step solution

Problem 4

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 2 & 9 \\ 3 & 13 \end{array}\right] $$

4 step solution

Problem 4

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 2 D-E $$

3 step solution

Problem 4

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} 3 x-y>6 \\ 2 x+y \leq 4 \end{array}\right) $$

4 step solution

Problem 5

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rrr} 3 & -2 & 1 \\ -1 & 4 & -7 \\ 0 & 5 & 9 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -1 & -3 \\ 10 & -2 & 4 \\ 7 & 0 & 12 \end{array}\right] $$

4 step solution

Problem 5

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} -1 & 2 \\ 3 & 4 \end{array}\right] $$

3 step solution

Problem 5

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 4 A-3 B $$

3 step solution

Problem 5

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} 2 x+3 y \leq 6 \\ 3 x-2 y \leq 6 \end{array}\right) $$

5 step solution

Problem 6

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} 7 & -4 \\ -5 & 9 \\ -1 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 12 & 3 \\ -2 & -4 \\ -6 & 7 \end{array}\right] $$

5 step solution

Problem 6

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 1 & -2 \\ 4 & -3 \end{array}\right] $$

4 step solution

Problem 6

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 2 B+3 D $$

3 step solution

Problem 6

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} 4 x+3 y \geq 12 \\ 3 x-4 y \geq 12 \end{array}\right) $$

4 step solution

Problem 7

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rr} -1 & 0 \\ 2 & 3 \\ -5 & -4 \\ -7 & 11 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & 2 \\ -3 & 7 \\ 6 & -5 \\ 9 & -2 \end{array}\right] $$

4 step solution

Problem 7

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} -2 & -3 \\ 4 & 6 \end{array}\right] $$

2 step solution

Problem 7

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ (A-B)-C $$

2 step solution

Problem 7

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{r} 2 x-y \geq 4 \\ x+3 y<3 \end{array}\right) $$

5 step solution

Problem 8

For Problems \(1-8\), find \(A+B, A-B, 2 A+3 B\), and \(4 A-2 B\). $$ A=\left[\begin{array}{rrr} 0 & -1 & -2 \\ 3 & -4 & 6 \\ 5 & 4 & -9 \end{array}\right], \quad B=\left[\begin{array}{rrr} 2 & 1 & -7 \\ -6 & 4 & 5 \\ 3 & -2 & -1 \end{array}\right] $$

4 step solution

Problem 8

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{rr} 5 & -1 \\ 3 & 4 \end{array}\right] $$

4 step solution

Problem 8

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ B-(D-E) $$

2 step solution

Problem 8

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{r} 3 x-y<3 \\ x+y \geq 1 \end{array}\right) $$

4 step solution

Problem 9

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rr} 2 & -1 \\ 0 & -4 \\ -5 & 3 \end{array}\right], \quad B=\left[\begin{array}{rrr} 5 & -2 & 6 \\ -1 & 4 & -2 \end{array}\right] $$

3 step solution

Problem 9

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} -3 & 2 \\ -4 & 5 \end{array}\right] $$

4 step solution

Problem 9

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 2 D-4 E $$

4 step solution

Problem 9

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{r} x+2 y>-2 \\ x-y<-3 \end{array}\right) $$

5 step solution

Problem 10

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} -2 & 3 & -1 \\ 7 & -4 & 5 \end{array}\right], \quad B=\left[\begin{array}{rr} 1 & -1 \\ -2 & 3 \\ -5 & -6 \end{array}\right] $$

5 step solution

Problem 10

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 3 & -4 \\ 6 & -8 \end{array}\right] $$

3 step solution

Problem 10

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ 3 A-4 E $$

3 step solution

Problem 10

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{r} x-3 y<-3 \\ 2 x-3 y>-6 \end{array}\right) $$

3 step solution

Problem 11

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{llr} 2 & -1 & -3 \\ 0 & -4 & 7 \end{array}\right], \quad B=\left[\begin{array}{rrrr} 2 & 1 & -1 & 4 \\ 0 & -2 & 3 & 5 \\ -6 & 4 & -2 & 0 \end{array}\right] $$

12 step solution

Problem 11

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} 0 & 1 \\ 5 & 3 \end{array}\right] $$

4 step solution

Problem 11

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ B-(D+E) $$

2 step solution

Problem 12

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} 3 & -1 & -4 \\ -5 & 2 & 2 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -2 \\ -4 & -1 \end{array}\right] $$

4 step solution

Problem 12

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} -2 & 0 \\ -3 & 5 \end{array}\right] $$

6 step solution

Problem 12

For Problems \(1-12\), compute the indicated matrix by using the following matrices: \(\begin{array}{ll} A=\left[\begin{array}{rr} 1 & -2 \\ 3 & 4 \end{array}\right] & B=\left[\begin{array}{lr} 2 & -3 \\ 5 & -1 \end{array}\right] \\ C=\left[\begin{array}{rr} 0 & 6 \\ -4 & 2 \end{array}\right] & D=\left[\begin{array}{rr} -2 & 3 \\ 5 & -4 \end{array}\right] \\ E=\left[\begin{array}{lr} 2 & 5 \\ 7 & 3 \end{array}\right] & \end{array}\) $$ A-(B+C) $$

2 step solution

Problem 12

For Problems \(1-24\), indicate the solution set for each system of inequalities by graphing the system and shading the appropriate region. $$ \left(\begin{array}{l} y \leq x+2 \\ y \geq x \end{array}\right) $$

3 step solution

Problem 13

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} 1 & -1 & 2 \\ 0 & 1 & -2 \\ 3 & 1 & 4 \end{array}\right], \quad B=\left[\begin{array}{rrr} 2 & 3 & -1 \\ 4 & 0 & 2 \\ -5 & 1 & -1 \end{array}\right] $$

4 step solution

Problem 13

For Problems \(1-18\), find the multiplicative inverse (if one exists) of each matrix. $$ \left[\begin{array}{ll} -2 & -3 \\ -1 & -4 \end{array}\right] $$

5 step solution

Problem 13

For Problems \(13-26\), compute \(A B\) and \(B A\). $$ A=\left[\begin{array}{ll} 1 & -1 \\ 2 & -2 \end{array}\right], \quad B=\left[\begin{array}{rr} 3 & -4 \\ -1 & 2 \end{array}\right] $$

3 step solution

Problem 14

For Problems \(9-20\), find \(A B\) and \(B A\), whenever they exist. $$ A=\left[\begin{array}{rrr} 1 & 0 & 1 \\ 0 & 1 & 1 \\ -1 & 2 & 3 \end{array}\right], \quad B=\left[\begin{array}{rrr} -1 & -1 & 1 \\ 0 & 1 & 0 \\ 2 & -3 & 1 \end{array}\right] $$

3 step solution

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