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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 11: Q. 65 (page 756)

In Problems 61鈥� 66, show that each matrix has no inverse
$[\begin{array}{r} - 3 & 1 & - 1 \\ 1 & - 4 & - 7 \\ 1 & 2 & 5 \end{array}]$

Short Answer

Expert verified

The given matrix has no inverse.

Step by step solution

01

Step 1. Given information

A matrix is given as,

$[\begin{array}{r} - 3 & 1 & - 1 \\ 1 & - 4 & - 7 \\ 1 & 2 & 5 \end{array}]$

02

Step 2. Finding the inverse.

To find the inverse,

[A 鈭� I_{3}] = [\begin{matrix} - 3 & 1 & - 1 & 1 & 0 & 0 \\ 1 & - 4 & - 7 & 0 & 1 & 0 \\ 1 & 2 & 5 & 0 & 0 & 1 \end{matrix}] R_{1} 鈫� R_{3} [\begin{matrix} - 3 & 1 & - 1 & 1 & 0 & 0 \\ 1 & - 4 & - 7 & 0 & 1 & 0 \\ 1 & 2 & 5 & 0 & 0 & 1 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 1 & - 4 & - 7 & 0 & 1 & 0 \\ - 3 & 1 & - 1 & 1 & 0 & 0 \end{matrix}] R_{2} = r_{2} - r_{1} . [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 1 & - 4 & - 7 & 0 & 1 & 0 \\ - 3 & 1 & - 1 & 1 & 0 & 0 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 1 - 1 & - 4 - 2 & - 7 - 5 & 0 - 0 & 1 - 0 & 0 - 1 \\ - 3 & 1 & - 1 & 1 & 0 & 0 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & - 6 & - 12 & 0 & 1 & - 1 \\ - 3 & 1 & - 1 & 1 & 0 & 0 \end{matrix}] R_{3} = r_{3} + 3 r_{1} [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & - 6 & - 12 & 0 & 1 & - 1 \\ - 3 & 1 & - 1 & 1 & 0 & 0 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & - 6 & - 12 & 0 & 1 & - 1 \\ - 3 + 3 & 1 + 6 & - 1 + 15 & 1 + 0 & 0 + 0 & 0 + 3 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & - 6 & - 12 & 0 & 1 & - 1 \\ 0 & 7 & 14 & 1 & 0 & 3 \end{matrix}] R_{2} = \frac{r_{2}}{- 6} and R_{3} = \frac{r_{3}}{7} [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & - 6 & - 12 & 0 & 1 & - 1 \\ 0 & 7 & 14 & 1 & 0 & 3 \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ \frac{0}{- 6} & \frac{- 6}{- 6} & \frac{- 12}{- 6} & \frac{0}{- 6} & \frac{1}{- 6} & \frac{- 1}{- 6} \\ \frac{0}{7} & \frac{7}{7} & \frac{14}{7} & \frac{1}{7} & \frac{0}{7} & \frac{3}{7} \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & 1 & 2 & 0 & - \frac{1}{6} & \frac{1}{6} \\ 0 & 1 & 2 & \frac{1}{7} & 0 & \frac{3}{7} \end{matrix}] R_{3} = r_{3} - r_{2} [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & 1 & 2 & 0 & - \frac{1}{6} & \frac{1}{6} \\ 0 & 1 & 2 & \frac{1}{7} & 0 & \frac{3}{7} \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & 1 & 2 & 0 & - \frac{1}{6} & \frac{1}{6} \\ 0 - 0 & 1 - 1 & 2 - 2 & \frac{1}{7} - 0 & 0 + \frac{1}{6} & \frac{3}{7} - \frac{1}{6} \end{matrix}] 鈫� [\begin{matrix} 1 & 2 & 5 & 0 & 0 & 1 \\ 0 & 1 & 2 & 0 & - \frac{1}{6} & \frac{1}{6} \\ 0 & 0 & 0 & \frac{1}{7} 0 & \frac{1}{6} & \frac{11}{42} \end{matrix}]

Since the left side does not represent an identity matrix, therefore, inverse does not exist.

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Most popular questions from this chapter

Running a Race : In a 1-mile race, the winner crosses the finish line 10 feet ahead of the second-place runner and 20 feet ahead of the third-place runner. Assuming that each runner maintains a constant speed throughout the race, by how many feet does the second-place runner beat the third-place runner?

In Problems, use the division algorithm to rewrite each improper fraction as the sum of a quotient and proper fraction. Find the partial fraction decomposition of the proper fraction. Finally, express the improper fraction as the sum of a quotient and the partial fraction decomposition.

$\frac{x^{4} + x^{3} - x + 2}{x^{2} - 2 x + 1}$

In Problems 5鈥�24, graph each equation of the system. Then solve the system to find the points of intersection.

$x^{2} + y^{2} = 4; x^{2} + 2 x + y^{2} = 0$

In Problems 35鈥�42, graph each system of inequalities by hand.

$\{\begin{cases} x^{2} + y^{2} 鈮� 9 \\ x + y 鈮� 3 \end{cases}$

Graph each equation of the system. Then solve the system to find the points of intersection.

$\{\begin{cases} y = \sqrt{36 - x^{2}} \\ y = 8 - x \end{cases}$

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