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: 6


$x=\left( -\frac{2}{3},\frac{4}{3}, -\frac{1}{3}\right)$.

Work Step by Step

Assume that $x=(a,b,c)$ and $3 x+2 u-v+2 w=0$, then we have \begin{align*} &\Longrightarrow 3 (a,b,c)+2(1,-2,1)- (0,2,3)+2(0,1,1)=(0,0,0)\\ &\Longrightarrow 3 (a,b,c)+(2,-4,2)+ (0,-2,-3)+(0,2,2)=(0,0,0)\\ &\Longrightarrow (3a,3b,3c)+(2,-4,1)=(0,0,0)\\ &\Longrightarrow (3a,3b,3c)=(0,0,0)-(2,-4,1)\\ &\Longrightarrow (3a,3b,3c)=(-2,4,-1). \end{align*} Hence, $a=-\frac{2}{3}$, $b=\frac{4}{3}$, $c=-\frac{1}{3}$ and $x=\left( -\frac{2}{3},\frac{4}{3}, -\frac{1}{3}\right)$.
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