## Linear Algebra: A Modern Introduction

Published by Cengage Learning

# Chapter 2 - Systems of Linear Equations - 2.1 Introduction to Systems of Linear Equations - Exercises 2.1 - Page 63: 11

#### Answer

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

#### Work Step by Step

Given this matrix: $$\begin{bmatrix} 3 & 5 \\ 5 & -2 \\ 2 & 4 \\ \end{bmatrix}$$ First, subtract the first row from the second: $$\begin{bmatrix} 3 & 5 \\ 2 & -7 \\ 2 & 4 \\ \end{bmatrix}$$ Then third from the second, and vice versa, then cancel out the third row again, producing a matrix in row echelon form: $$\begin{bmatrix} 3 & 5 \\ 0 & -11 \\ 0 & 0 \\ \end{bmatrix}$$ Lastly, divide the first and second rows by $3$ and $-11$ to make their leading coefficients $1$, and add $-5$ times the second row to the first row to yield a matrix in reduced row echelon form. $$\begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ \end{bmatrix}$$

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