Elementary Linear Algebra 7th Edition

Published by Cengage Learning
ISBN 10: 1-13311-087-8
ISBN 13: 978-1-13311-087-3

Chapter 4 - Vector Spaces - Review Exercises - Page 221: 5


$x=\left( \frac{1}{2},-\frac{9}{2}, -\frac{9}{2}\right)$.

Work Step by Step

Assume that $x=(a,b,c)$, and $2 {x}-{u}+3 {v}+{w}={0}$, then we have \begin{align*} &\Longrightarrow 2 (a,b,c)-(1,-2,1)+3 (0,2,3)+(0,1,1)=(0,0,0)\\ &\Longrightarrow 2 (a,b,c)+(-1,2,-1)+ (0,6,9)+(0,1,1)=(0,0,0)\\ &\Longrightarrow (2a,2b,2c)+(-1,9,9)=(0,0,0)\\ &\Longrightarrow (2a,2b,2c)=(0,0,0)-(-1,9,9)\\ &\Longrightarrow (2a,2b,2c)=(1,-9,-9). \end{align*} Hence, $a=\frac{1}{2}$, $b=-\frac{9}{2}$, $c=-\frac{9}{2}$ and $x=\left( \frac{1}{2},-\frac{9}{2}, -\frac{9}{2}\right)$.
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