Linear Algebra: A Modern Introduction

Published by Cengage Learning
ISBN 10: 1285463242
ISBN 13: 978-1-28546-324-7

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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