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 - 2.2 Direct Methods for Solving Linear Systems - Exercises 2.2: 9

Answer

Row Echelon Form: $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ Reduced Row Echelon Form: $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$

Work Step by Step

We're given $$ \begin{bmatrix} 0 & 0 & 1 \\ 0 & 1 & 1 \\ 1 & 1 & 1 \\ \end{bmatrix} $$ First, swap rows 1 and 3, to get the row echelon form: $$ \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ Now, subtract row 3 from rows 1 and 2: $$ \begin{bmatrix} 1 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ And subtract row 2 from row 1: $$ \begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{bmatrix} $$ The matrix is now in reduced row echelon form.
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