College Algebra (6th Edition)

Published by Pearson
ISBN 10: 0-32178-228-3
ISBN 13: 978-0-32178-228-1

Chapter 6 - Matrices and Determinants - Exercise Set 6.1 - Page 600: 33

Answer

Solution set: $\{ ( 1$ , $2$ , $-1 ) \}$

Work Step by Step

Augmented matrix: $\left[\begin{array}{lllll} 2 & 2 & 7 & | & -1\\ 2 & 1 & 2 & | & 2\\ 4 & 6 & 1 & | & 15 \end{array}\right]\left\{\begin{array}{l} .\\ \leftarrow R_{2}-R_{1}\\ \leftarrow R_{3}-2R_{1}. \end{array}\right.$ $\left[\begin{array}{lllll} 2 & 2 & 7 & | & -1\\ 0 & -1 & -5 & | & 3\\ 0 & 2 & -13 & | & 17 \end{array}\right]\left\{\begin{array}{l} .\\ \times(-1)\\ \leftarrow R_{3}+2R_{2}. \end{array}\right.$ $\left[\begin{array}{lllll} 2 & 2 & 7 & | & -1\\ 0 & 1 & 5 & | & -3\\ 0 & 0 & -23 & | & 23 \end{array}\right]\left\{\begin{array}{l} .\\ .\\ \Rightarrow z=-1. \end{array}\right.$ Back substitute into row/equation 2, $y+5(-1)=-3$ $y=-3+5=2$ Back substitute into row/equation 1: $2x+2(2)+7(-1)=-1$ $2x=-1+3$ $2x=2$ $x=1$ Solution set: $\{ ( 1$ , $2$ , $-1 ) \}$
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