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: 7


$x=\left( \frac{5}{2},-6, 0\right)$.

Work Step by Step

Assume that $x=(a,b,c)$ and $5 {u}-2 {x}=3 {v}+{w}$, then we have \begin{align*} &\Longrightarrow 5 (1,-1,2) -2 (a,b,c)=3(0,2,3)+(0,1,1)\\ &\Longrightarrow (5,-5,10) + (-2a,-2b,-2c)=(0,6,9)+(0,1,1)\\ &\Longrightarrow (5,-5,10) + (-2a,-2b,-2c)=(0,7,10)\\ &\Longrightarrow (-2a,-2b,-2c)=(0,7,10)-(5,-5,10)\\ &\Longrightarrow (-2a,-2b,-2c)=(-5,12,0). \end{align*} Hence, $a=\frac{5}{2}$, $b=-6$, $c=0$ and $x=\left( \frac{5}{2},-6, 0\right)$.
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