## Algebra and Trigonometry 10th Edition

Published by Cengage Learning

# Chapter 10 - Chapter Test - Page 764: 1

#### Answer

$\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 &0 &1 \end{bmatrix}$

#### Work Step by Step

The general form of a matrix of order $3 \times 3$ is: $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h& i \end{bmatrix}=a(ei-fh) -b(di-fg)+c(dh-eg)$ We will subtract $6 R_1$ from row $2$ and $5 R_1$ from row 3. Then subtract row 2 from row $3$. We have: $\begin{bmatrix} 1 &- 1 & 5 \\ 0 & 8 & -27 \\ 0 &0 & -1 \end{bmatrix}$ Now, we will add row 2 to row 1. We have: $\begin{bmatrix} 1 & 0 & 13/8 \\ 0 & 1 & -27/8 \\ 0 &0 & 1 \end{bmatrix}$ Subtract $\dfrac{13 R_2}{3}$ from row $1$ and add $\dfrac{27 R_3}{8}$ to row 2. Finally, we have: $\begin{bmatrix} 1 &0 & 0 \\ 0 & 1 & 0 \\ 0 &0 &1 \end{bmatrix}$

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