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 6.2 - Page 610: 31

Answer

$\{(z+6, z+2, z),\ z\in \mathbb{R}\}$

Work Step by Step

The equations for the intersections: $\left\{\begin{array}{ll} I_{1}: & x-y=4\\ I_{2}: & x-z=6\\ I_{3}: & y-z=2 \end{array}\right.$ Augmented matrix: $\left[\begin{array}{lllll} 1 & -1 & 0 & | & 4\\ 1 & 0 & -1 & | & 6\\ 0 & 1 & -1 & | & 2 \end{array}\right]\left(\begin{array}{l} .\\ \leftarrow(R_{2}-R_{1}).\\ . \end{array}\right)$ $\left[\begin{array}{lllll} 1 & -1 & 0 & | & 4\\ 0 & 1 & -1 & | & 2\\ 0 & 1 & -1 & | & 2 \end{array}\right]\left(\begin{array}{l} .\\ .\\ \leftarrow(R_{3}-R_{2}). \end{array}\right)$ $\left[\begin{array}{lllll} 1 & -1 & 0 & | & 4\\ 0 & 1 & -1 & | & 2\\ 0 & 0 & 0 & | & 0 \end{array}\right]\left(\begin{array}{l} \leftarrow(R_{1}+R_{2}).\\ .\\ . \end{array}\right)$ $\left[\begin{array}{lllll} 1 & 0 & -1 & | & 6\\ 0 & 1 & -1 & | & 2\\ 0 & 0 & 0 & | & 0 \end{array}\right]$ The system is consistent, with infinitely many solutions. Taking z as parameter, express x and y in terms of z. $x-z=6 \Rightarrow x=z+6$ $y-z=2 \Rightarrow y=z+2$ Solution set: $\{(z+6, z+2, z),\ z\in \mathbb{R}\}$.
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