Chapter 8

Algebra and Trigonometry Real Mathematics, Real People · 660 exercises

Problem 29

You are given the yearly interest earned from a total of \$18,000 invested in two funds paying the given rates of simple interest. Write and solve a system of equations to find the amount invested at each rate. Yearly Interest \(\$ 400\) Rate 1 \(4 \%\) Rate 2 \(2 \%\)

5 step solution

Problem 29

Solving a Matrix Equation Solve for \(X\) when \(A=\left[\begin{array}{rr}-2 & -1 \\\ 1 & 0 \\ 3 & -4\end{array}\right]\) and \(B=\left[\begin{array}{rr}0 & 3 \\\ 2 & 0 \\ -4 & -1\end{array}\right]\) $$3 X=A+3 B$$

3 step solution

Problem 29

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{l} 3 x+3 y+5 z=1 \\ 3 x+5 y+9 z=0 \\ 5 x+9 y+17 z=0 \end{array}\right.$$

4 step solution

Problem 30

Fill in the blank(s) using elementary row operations to form a row-equivalent matrix. $$\begin{aligned} &\left[\begin{array}{rrrr} 2 & 4 & 8 & 3 \\ 1 & -1 & -3 & 2 \\ 2 & 6 & 4 & 9 \end{array}\right]\\\ &\left[\begin{array}{cccc} 1 & -1 & -3 & 2 \\ 2 & & & \\ 2 & 6 & 4 & 9 \end{array}\right]\\\ &\left[\begin{array}{cccc} 1 & -1 & -3 & 2 \\ 0 & 6 & & \\ 0 & 8 & & \end{array}\right] \end{aligned}$$

3 step solution

Problem 30

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{rrr}1 & 0 & 0 \\\\-1 & -1 & 0 \\\4 & 1 & 5\end{array}\right]$$

4 step solution

Problem 30

You are given the yearly interest earned from a total of \$18,000 invested in two funds paying the given rates of simple interest. Write and solve a system of equations to find the amount invested at each rate. Yearly Interest \$ 840 Rate 1 \(6 \%\) Rate 2 \(3 \%\)

3 step solution

Problem 30

Solving a Matrix Equation Solve for \(X\) when \(A=\left[\begin{array}{rr}-2 & -1 \\\ 1 & 0 \\ 3 & -4\end{array}\right]\) and \(B=\left[\begin{array}{rr}0 & 3 \\\ 2 & 0 \\ -4 & -1\end{array}\right]\) $$2 X=2 A+B$$

2 step solution

Problem 30

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{l} 2 x+y+3 z=1 \\ 2 x+6 y+8 z=3 \\ 6 x+8 y+18 z=5 \end{array}\right.$$

3 step solution

Problem 31

(a) perform the row operations to solve the augmented matrix, (b) write and solve the system of linear equations represented by the augmented matrix, and (c) compare the two solution methods. Which do you prefer? $$\left[\begin{array}{rrrr} -3 & 4 & \vdots & 22 \\ 6 & -4 & \vdots & -28 \end{array}\right]$$ (i) Add \(R_{2}\) to \(R_{1}\) (ii) Add -2 times \(R_{1}\) to \(R_{2}\) (iii) Multiply \(R_{2}\) by \(-\frac{1}{4}\) (iv) Multiply \(R_{1}\) by \(\frac{1}{3}\)

3 step solution

Problem 31

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{llll} 2 & 6 & 6 & 2 \\ 2 & 7 & 3 & 6 \\ 1 & 5 & 0 & 1 \\ 3 & 7 & 0 & 7 \end{array}\right]$$

5 step solution

Problem 31

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{array}{l}5 x+3 y=6 \\ 3 x-y=5\end{array}\right.\)

6 step solution

Problem 31

You are given the yearly interest earned from a total of \$18,000 invested in two funds paying the given rates of simple interest. Write and solve a system of equations to find the amount invested at each rate. Yearly Interest \(\$ 1182\) Rate 1 \(5.6 \%\) Rate 2 \(6.8 \%\)

4 step solution

Problem 31

Solving a Matrix Equation Solve for \(X\) when \(A=\left[\begin{array}{rr}-2 & -1 \\\ 1 & 0 \\ 3 & -4\end{array}\right]\) and \(B=\left[\begin{array}{rr}0 & 3 \\\ 2 & 0 \\ -4 & -1\end{array}\right]\) $$2 A+4 B=-2 X$$

3 step solution

Problem 31

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{aligned} 3 x-3 y+6 z &=6 \\ x+2 y-z &=5 \\ 5 x-8 y+13 z &=7 \end{aligned}\right.$$

4 step solution

Problem 32

(a) perform the row operations to solve the augmented matrix, (b) write and solve the system of linear equations represented by the augmented matrix, and (c) compare the two solution methods. Which do you prefer? $$\left[\begin{array}{rrrrr} 7 & 13 & 1 & \vdots & -4 \\ -3 & -5 & -1 & \vdots & -4 \\ 3 & 6 & 1 & \vdots & -2 \end{array}\right]$$ (i) Add \(R_{2}\) to \(R_{1}\) (ii) Multiply \(R_{1}\) by \(\frac{1}{4}\) (iii) Add \(R_{3}\) to \(R_{2}\) (iv) Add -3 times \(R_{1}\) to \(R_{3}\) (v) Add -2 times \(R_{2}\) to \(R_{1}\)

3 step solution

Problem 32

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{rrrr} 3 & 6 & -5 & 4 \\ -2 & 2 & 6 & 0 \\ 1 & 1 & 2 & 0 \\ 0 & 3 & -1 & -1 \end{array}\right]$$

3 step solution

Problem 32

Finding the Inverse of a \(2 \times 2\) Matrix, use the formula on page 676 to find the inverse of the \(2 \times 2\) matrix (if it exists). $$\left[\begin{array}{rr} -\frac{1}{4} & -\frac{2}{3} \\ \frac{1}{3} & \frac{8}{9} \end{array}\right]$$

3 step solution

Problem 32

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{array}{c}x+5 y=10 \\ 3 x-10 y=-5\end{array}\right.\)

5 step solution

Problem 32

You are given the yearly interest earned from a total of \$18,000 invested in two funds paying the given rates of simple interest. Write and solve a system of equations to find the amount invested at each rate. Yearly Interest \(\$ 684\) Rate 1 \(2.75 \%\) Rate 2 \(4.25 \%\)

5 step solution

Problem 32

Solving a Matrix Equation Solve for \(X\) when \(A=\left[\begin{array}{rr}-2 & -1 \\\ 1 & 0 \\ 3 & -4\end{array}\right]\) and \(B=\left[\begin{array}{rr}0 & 3 \\\ 2 & 0 \\ -4 & -1\end{array}\right]\) $$-3 X-3 A=9 B$$

5 step solution

Problem 32

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{rr} -x+3 y+z= & 4 \\ 4 x-2 y-5 z= & -7 \\ 2 x+4 y-3 z= & 12 \end{array}\right.$$

5 step solution

Problem 33

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{rrrrr} 3 & 2 & 4 & -1 & 5 \\ -2 & 0 & 1 & 3 & 2 \\ 1 & 0 & 0 & 4 & 0 \\ 6 & 0 & 2 & -1 & 0 \\ 3 & 0 & 5 & 1 & 0 \end{array}\right]$$

4 step solution

Problem 33

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{array}{l}\frac{2}{5} x-\frac{3}{2} y=4 \\ \frac{1}{5} x-\frac{3}{4} y=-2\end{array}\right.\)

4 step solution

Problem 33

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{aligned} x^{2}-2 x+y &=8 \\ x-y &=-2 \end{aligned}\right.$$

5 step solution

Problem 33

Finding the Product of Two Matrices Find \(A B,\) if possible. $$A=\left[\begin{array}{rr} 3 & -1 \\ 4 & -5 \\ 2 & 6 \end{array}\right], \quad B=\left[\begin{array}{rr} 6 & 0 \\ 7 & -1 \end{array}\right]$$

3 step solution

Problem 33

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{rr} x-2 y+3 z= & 4 \\ 3 x-y+2 z= & 0 \\ x+3 y-4 z= & -2 \end{array}\right.$$

4 step solution

Problem 34

Find the determinant of the matrix. Expand by cofactors on the row or column that appears to make the computations easiest. $$\left[\begin{array}{lllll} 5 & 2 & 0 & 0 & -2 \\ 0 & 1 & 4 & 3 & 2 \\ 0 & 0 & 2 & 6 & 3 \\ 0 & 0 & 3 & 4 & 1 \\ 0 & 0 & 0 & 0 & 2 \end{array}\right]$$

4 step solution

Problem 34

Finding the Inverse of a \(2 \times 2\) Matrix, use the formula on page 676 to find the inverse of the \(2 \times 2\) matrix (if it exists). $$\left[\begin{array}{rr} 7 & 12 \\ -8 & -5 \end{array}\right]$$

4 step solution

Problem 34

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{array}{l}\frac{2}{3} x+\frac{1}{6} y=\frac{2}{3} \\ 4 x+y=4\end{array}\right.\)

3 step solution

Problem 34

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{aligned} 2 x^{2}-2 x-y &=14 \\ 2 x-y &=-2 \end{aligned}\right.$$

5 step solution

Problem 34

Finding the Product of Two Matrices Find \(A B,\) if possible. $$A=\left[\begin{array}{rr} -1 & 6 \\ -4 & 5 \\ 0 & 3 \end{array}\right], \quad B=\left[\begin{array}{ll} 2 & 3 \\ 0 & 9 \end{array}\right]$$

3 step solution

Problem 34

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{l} x\quad=13 \\ 4 x-2 y+z=1 \\ 2 x-2 y-7 z=-19 \end{array}\right.$$

4 step solution

Problem 35

Determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form. $$\left[\begin{array}{llll} 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 5 \\ 0 & 0 & 0 & 0 \end{array}\right]$$

2 step solution

Problem 35

Use the matrix capabilities of a graphing utility to evaluate the determinant. $$\left|\begin{array}{cccc} 1 & -1 & 8 & 4 \\ 2 & 6 & 0 & -4 \\ 2 & 0 & 2 & 6 \\ 0 & 2 & 8 & 0 \end{array}\right|$$

3 step solution

Problem 35

Finding a Matrix Entry, find the value of the constant \(k\) such that \(B=A^{-1}\). $$A=\left[\begin{array}{rr} 1 & 2 \\ -2 & 0 \end{array}\right], \quad B=\left[\begin{array}{rr} k & -\frac{1}{2} \\ \frac{1}{2} & \frac{1}{4} \end{array}\right]$$

3 step solution

Problem 35

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{array}{l}\frac{3}{4} x+y=\frac{1}{8} \\ \frac{9}{4} x+3 y=\frac{3}{8}\end{array}\right.\)

3 step solution

Problem 35

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{aligned} 2 x^{2}-y &=1 \\ x-y &=2 \end{aligned}\right.$$

5 step solution

Problem 35

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{c} x+2 y+z=1 \\ x-2 y+3 z=-3 \\ 2 x+y+z=-1 \end{array}\right.$$

3 step solution

Problem 36

Determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form. $$\left[\begin{array}{llll} 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 2 \\ 0 & 0 & 1 & 0 \end{array}\right]$$

2 step solution

Problem 36

Use the matrix capabilities of a graphing utility to evaluate the determinant. $$\left|\begin{array}{rrrr} 0 & -3 & 8 & 2 \\ 8 & 1 & -1 & 6 \\ -4 & 6 & 0 & 9 \\ -7 & 0 & 0 & 14 \end{array}\right|$$

3 step solution

Problem 36

Finding a Matrix Entry, find the value of the constant \(k\) such that \(B=A^{-1}\). $$A=\left[\begin{array}{rr} -1 & 1 \\ 2 & 1 \end{array}\right], \quad B=\left[\begin{array}{rr} -\frac{1}{3} & \frac{1}{3} \\ k & \frac{1}{3} \end{array}\right]$$

4 step solution

Problem 36

Solve the system by the method of elimination and check any solutions using a graphing utility. \(\left\\{\begin{aligned} \frac{1}{4} x+\frac{1}{6} y &=1 \\\\-3 x-2 y &=0 \end{aligned}\right.\)

3 step solution

Problem 36

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{array}{r} 2 x^{2}+y=3 \\ x+y=4 \end{array}\right.$$

5 step solution

Problem 36

Solve the system of linear equations and check any solution algebraically. $$\left\\{\begin{array}{r} 3 x-2 y-6 z=-4 \\ -3 x+2 y+6 z=1 \\ x-y-5 z=-3 \end{array}\right.$$

6 step solution

Problem 37

Determine whether the matrix is in row-echelon form. If it is, determine if it is also in reduced row-echelon form. $$\left[\begin{array}{rrrr} 3 & 0 & 3 & 7 \\ 0 & -2 & 0 & 4 \\ 0 & 0 & 1 & 5 \end{array}\right]$$

2 step solution

Problem 37

Determine whether the statement is true or false. Justify your answer. Cramer's Rule cannot be used to solve a system of linear equations when the determinant of the coefficient matrix is zero.

3 step solution

Problem 37

Use the matrix capabilities of a graphing utility to evaluate the determinant. $$\left|\begin{array}{rrrrrr} 3 & -2 & 4 & 3 & 1 & 3 \\ -1 & 0 & 2 & 1 & 0 & 0 \\ 5 & -1 & 0 & 3 & 2 & 1 \\ 4 & 7 & -8 & 0 & 0 & -2 \\ 1 & 2 & 3 & 0 & 2 & 4 \\ 3 & -5 & 1 & 2 & 3 & 1 \end{array}\right|$$

3 step solution

Problem 37

Solve the system by the method of elimination and check any solutions using a graphing utility. $$\left\\{\begin{array}{r} \frac{x+2}{4}+\frac{y-1}{4}=1 \\ x-y=4 \end{array}\right.$$

5 step solution

Problem 37

Solve the system by the method of substitution. Use a graphing utility to verify your results. $$\left\\{\begin{array}{c} x^{3}-y=0 \\ x-y=0 \end{array}\right.$$

5 step solution

Problem 37

Finding the Product of Two Matrices Find \(A B,\) if possible. $$A=\left[\begin{array}{rrr} 6 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & -2 \end{array}\right], \quad B=\left[\begin{array}{rrr} \frac{1}{3} & 0 & 0 \\ 0 & -\frac{1}{4} & 0 \\ 0 & 0 & \frac{1}{6} \end{array}\right]$$

3 step solution

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