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 - 4.1 Vectors in Rn - 4.1 Exercises - Page 153: 24



Work Step by Step

Since we have $u=(1,2,3)$, $v=(2,2,-1)$, $w=(4,0,-41)$ then \begin{align*} 2u+v-w-3z=0 &\Longrightarrow 3z=2u+v-w\\ &\Longrightarrow z=\frac{1}{3}(2u+v-w)\\ &\Longrightarrow z=\frac{1}{3}(2(1,2,3)+(2,2,-1)-(4,0,-41))\\ &\Longrightarrow z=\frac{1}{3}(0,-6,-9)\\ &\Longrightarrow z=(0,-2,-3). \end{align*} Hence, $z=(0,-2,-3).$
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