Chapter 11

Contemporary Precalculus · 151 exercises

Problem 1

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{aligned}x=-1, y &=3 \\\2 x+y &=1 \\\\-3 x+2 y &=9\end{aligned}$$

3 step solution

Problem 1

In Exercises \(1-4,\) write the augmented matrix of the system. $$\begin{array}{rr} 2 x-3 y+4 z= & 1 \\ x+2 y-6 z= & 0 \\ 3 x-7 y+4 z= & -3 \end{array}$$

2 step solution

Problem 1

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{lll} 3 & 6 & 7 \\ 8 & 0 & 1 \end{array}\right), \quad B=\left(\begin{array}{rrrr} 2 & 5 & 9 & 1 \\ 7 & 0 & 0 & 6 \\ -1 & 3 & 8 & 7 \end{array}\right)$$

6 step solution

Problem 2

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{aligned}x=-2, y &=5 \\\2 x+3 y &=11 \\\x-2 y &=-12\end{aligned}$$

3 step solution

Problem 2

In Exercises \(1-4,\) write the augmented matrix of the system. $$\begin{aligned} &4 x+y+z+7 w=4\\\ &x-4 y \quad-3 w=0\\\ &5 x \quad-5 z+10 w=-3 \end{aligned}$$

2 step solution

Problem 2

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{ccc} 0 & 5 & 9 \\ 0 & 5 & 3 \\ 0 & 0 & 0 \end{array}\right), \quad B=\left(\begin{array}{ccc} -8 & 7 & 1 \\ 0 & -3 & 4 \end{array}\right)$$

4 step solution

Problem 3

In Exercises \(1-4,\) write the augmented matrix of the system. $$\begin{aligned} 2 x-\frac{5}{2} y+\frac{2}{3} z &=0 \\ x-\frac{1}{4} y+4 z &=0 \\ -3 y+\frac{1}{2} z &=0 \end{aligned}$$

2 step solution

Problem 3

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{aligned}&x=2, y=-1\\\&\frac{1}{3} x+\frac{1}{2} y=\frac{1}{6}\\\&\frac{1}{2} x+\frac{1}{3}y=\frac{2}{3}\end{aligned}$$

5 step solution

Problem 3

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{ll} 1 & 0 \\ 1 & 1 \\ 0 & 1 \end{array}\right), \quad B=\left(\begin{array}{lll} 5 & 6 & 11 \\ 7 & 8 & 15 \end{array}\right)$$

3 step solution

Problem 4

In Exercises \(1-4,\) write the augmented matrix of the system. $$\begin{aligned} x+7 y-\frac{2}{5} z+\frac{5}{6} w &=0 \\ \frac{1}{8} x-y-8 z &=1 \\ \frac{2}{3} y-5 z+\quad w &=-2 \\ \frac{1}{6} x+4 y+\frac{2}{7} z &=3 \end{aligned}$$

2 step solution

Problem 4

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{array}{r}x=.3, y=.7 \\\4 x-1.2 y=.36 \\\3.1 x+2 y=4.7\end{array}$$

4 step solution

Problem 4

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{rr} -1 & 0 \\ 6 & 8 \end{array}\right), \quad B=\left(\begin{array}{rr} 4 & 0 \\ 1 & -7 \end{array}\right)$$

3 step solution

Problem 5

In Exercises \(5-8,\) the augmented matrix of a system of equations is given. Express the system in equation notation. \left(\begin{array}{rrr} 3 & -5 & 4 \\ 9 & 7 & 2 \end{array}\right)

2 step solution

Problem 5

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{aligned}x=\frac{1}{3}, y=2, z &=-1 \\\3 x-y+2 z &=1 \\\4 y-3 z &=5 \\\3 z &=3\end{aligned}$$

3 step solution

Problem 5

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{rr} -4 & 15 \\ 3 & -7 \\ 2 & 10 \end{array}\right), \quad B=\left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right)$$

4 step solution

Problem 6

In Exercises \(5-8,\) the augmented matrix of a system of equations is given. Express the system in equation notation. $$\left(\begin{array}{rrrr} 2 & -3 & 5 & 0 \\ 7 & 0 & -1 & 5 \end{array}\right)$$

3 step solution

Problem 6

Determine whether the given values of \(x, y,\) and z are a solution of the system of equations. $$\begin{aligned}&x=3, y=-\frac{5}{4}, z=\frac{3}{4}\\\&4 x-6 y+10 z=27\\\&4 x+8 y-28 z=-19\\\&x+4 y-8 z=-8\end{aligned}$$

6 step solution

Problem 6

Determine whether the product \(A B\) or \(B A\) is defined. If a product is defined, state its size ( number of rows and columns). Do not actually calculate any products. $$A=\left(\begin{array}{lll} -1 & 5 & 6 \end{array}\right), \quad B=\left(\begin{array}{r} 7 \\ 8 \\ -1 \end{array}\right)$$

6 step solution

Problem 7

In Exercises \(5-8,\) the augmented matrix of a system of equations is given. Express the system in equation notation. $$\left(\begin{array}{rrrrr} 1 & 0 & 1 & 0 & 1 \\ 1 & -1 & 4 & -2 & 3 \\ 4 & 2 & 5 & 0 & 2 \end{array}\right)$$

3 step solution

Problem 7

Use substitution to solve the system. $$\begin{array}{r}x+3 y=6 \\\3 x-y=2\end{array}$$

5 step solution

Problem 7

Find AB. $$A=\left(\begin{array}{rr} 1 & 2 \\ -4 & 3 \end{array}\right), \quad B=\left(\begin{array}{rrr} -1 & 2 & 3 \\ 7 & 1 & 0 \end{array}\right)$$

4 step solution

Problem 8

In Exercises \(5-8,\) the augmented matrix of a system of equations is given. Express the system in equation notation. $$\left(\begin{array}{rrrr} -1 & 0 & 2 & 6 \\ 0 & 5 & -4 & 1 \\ 8 & -2 & 3 & 4 \end{array}\right)$$

5 step solution

Problem 8

Use substitution to solve the system. $$\begin{aligned}&2 x-3 y=6\\\&5 x+7 y=2\end{aligned}$$

5 step solution

Problem 8

Find AB. $$A=\left(\begin{array}{rrr} 1 & 5 & 9 \\ -2 & 3 & 3 \\ 1 & 7 & 0 \end{array}\right), \quad B=\left(\begin{array}{rrr} 2 & -3 & 0 \\ 4 & 5 & -1 \\ 0 & -1 & 1 \end{array}\right)$$

3 step solution

Problem 9

In Exercises \(9-12\), the reduced row echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. $$\left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & 3 / 2 \\ 0 & 1 & 0 & 0 & 5 \\ 0 & 0 & 1 & 0 & -2 \\ 0 & 0 & 0 & 1 & 0 \end{array}\right)$$

2 step solution

Problem 9

Use substitution to solve the system. $$\begin{aligned}&3 x-2 y=4\\\&2 x+y=-1\end{aligned}$$

6 step solution

Problem 9

Find AB. $$A=\left(\begin{array}{rrr} 1 & 0 & -2 \\ 0 & 3 & -1 \\ 2 & 4 & 0 \end{array}\right), \quad B=\left(\begin{array}{ll} 1 & 1 \\ 1 & 0 \\ 0 & 1 \end{array}\right)$$

4 step solution

Problem 10

In Exercises \(9-12\), the reduced row echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. $$\left(\begin{array}{ccccc} 1 & 0 & 0 & 0 & -1 \\ 0 & 1 & 0 & 0 & 3 \\ 0 & 0 & 1 & 0 & 1 \\ 0 & 0 & 0 & 0 & 1 \end{array}\right)$$

3 step solution

Problem 10

Use substitution to solve the system. $$\begin{aligned}&5 x-3 y=1\\\&5 x-2 y=-7\end{aligned}$$

4 step solution

Problem 10

Find AB. $$A=\left(\begin{array}{rrr} -1 & 7 & 1 \\ -5 & 3 & 2 \\ 0 & 1 & 5 \\ -3 & 6 & 7 \end{array}\right), \quad B=\left(\begin{array}{rrrr} 7 & -2 & 6 & 2 \\ -2 & 8 & 4 & 1 \\ 0 & 7 & 0 & -5 \end{array}\right)$$

3 step solution

Problem 11

In Exercises \(9-12\), the reduced row echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. $$\left(\begin{array}{ccccc} 1 & 0 & 0 & 2 & 3 \\ 0 & 1 & 0 & 3 & 5 \\ 0 & 0 & 1 & 0 & 2 \\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$$

3 step solution

Problem 11

Use substitution to solve the system. $$\begin{aligned}&r+s=0\\\&r-s=5\end{aligned}$$

6 step solution

Problem 11

Find AB. $$A=\left(\begin{array}{rrr} 2 & 0 & -1 \\ 1 & 1 & 2 \\ 0 & 2 & -3 \\ 2 & 3 & 0 \end{array}\right), \quad B=\left(\begin{array}{llll} 1 & 0 & 1 & 1 \\ 1 & 1 & 0 & 1 \\ 1 & 1 & 1 & 0 \end{array}\right)$$

3 step solution

Problem 12

In Exercises \(9-12\), the reduced row echelon form of the augmented matrix of a system of equations is given. Find the solutions of the system. $$\left(\begin{array}{cccc} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 3 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right)$$

2 step solution

Problem 12

Use substitution to solve the system. $$\begin{aligned}&t=7 u-4\\\&t=5 u+6\end{aligned}$$

5 step solution

Problem 13

In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{aligned} -x+3 y+2 z &=0 \\ 2 x-y-z &=3 \\ x+2 y+3 z &=0 \end{aligned}$$

3 step solution

Problem 13

Use substitution to solve the system. \(x+y=c+d \quad\) (where \(c, d\) are constants)\(x-y=2 c-d\)

6 step solution

Problem 13

Show that AB is not equal to BA by computing both products. $$A=\left(\begin{array}{ll} 2 & 3 \\ 1 & 5 \end{array}\right), \quad B=\left(\begin{array}{rr} -1 & 1 \\ 3 & 2 \end{array}\right)$$

5 step solution

Problem 14

In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{aligned} 2 x-3 y+2 z &=8 \\ 3 x+2 y+z &=3 \\ x+2 y+3 z &=1 \end{aligned}$$

4 step solution

Problem 14

Use substitution to solve the system. \(x-2 y=2 c+5 d \quad\) (where \(c, d\) are constants)\(3 x-y=c-2 d\)

4 step solution

Problem 14

Show that AB is not equal to BA by computing both products. $$A=\left(\begin{array}{rr} 5 / 4 & -2 \\ -3 / 4 & 1 \end{array}\right), \quad B=\left(\begin{array}{cc} 9 / 4 & 5 / 4 \\ -2 & 3 \end{array}\right)$$

4 step solution

Problem 15

In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{array}{r} x+y+z=1 \\ x-2 y+2 z=4 \\ 2 x-y+3 z=5 \end{array}$$

3 step solution

Problem 15

Use the elimination method to solve the system. $$\begin{aligned}&2 x-2 y=12\\\&-2 x+3 y=10\end{aligned}$$

5 step solution

Problem 15

Show that AB is not equal to BA by computing both products. $$A=\left(\begin{array}{rrr} 4 & 2 & -1 \\ 0 & 1 & 2 \\ -3 & 0 & 1 \end{array}\right), \quad B=\left(\begin{array}{rrr} 1 & 7 & -5 \\ 2 & -2 & 6 \\ 0 & 0 & 0 \end{array}\right)$$

3 step solution

Problem 16

In Exercises \(13-16,\) use Gaussian elimination to solve the system. $$\begin{aligned} &2 x-z=3\\\ &8 x+y+4 z=-1\\\ &4 x+y+6 z=-7 \end{aligned}$$

5 step solution

Problem 16

Use the elimination method to solve the system. $$\begin{aligned}&4 x-3 y=10\\\&7 x+4 y=-1\end{aligned}$$

7 step solution

Problem 16

Show that AB is not equal to BA by computing both products. $$A=\left(\begin{array}{rrrr} -1 & 3 & 2 & -4 \\ 8 & 0 & 5 & 6 \\ -1 & 0 & 6 & 3 \\ 2 & 3 & -2 & 5 \end{array}\right), \quad B=\left(\begin{array}{rrrr} 4 & -2 & -2 & 0 \\ 1 & -7 & 4 & 1 \\ 0 & 5 & 7 & -2 \\ 1 & 4 & 0 & 5 \end{array}\right)$$

3 step solution

Problem 17

In Exercises \(17-20,\) use the Gauss-Jordan method to solve the system. \begin{array}{rr} x-2 y+4 z= & 6 \\ x+y+13 z= & 6 \\ -2 x+6 y-z= & -10 \end{array}

5 step solution

Problem 17

Use the elimination method to solve the system. $$\begin{array}{rr}x+3 y= & -1 \\\2 x-y= & 5\end{array}$$

6 step solution

Problem 17

Find the inverse of the matrix, if it exists. $$\left(\begin{array}{ll} 1 & 2 \\ 3 & 4 \end{array}\right)$$

4 step solution

Show/ page