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: 38

Answer

$w=(\frac{1}{3}, -\frac{8}{3},-\frac{16}{3},3)$.

Work Step by Step

Let $w$ be given by $w=(a,b,c,d)$ \begin{align*} 2u+v-3w &=2(0,0,-8,1)+(1,-8,0,7)-3(a,b,c,d)\\ &=(1-3a,-8-3b,-16-3c,9-3d)\\ &=(0,0,0,0) \end{align*} we get the system $$1-3a=0,\quad -8-3b=0,\quad -16-3c=0,\quad 9-3d=0.$$ By solving the above system we get the solution $$a=\frac{1}{3}, \quad b= -\frac{8}{3}, \quad c=-\frac{16}{3}, \quad d=3.$$ Then, $w=(\frac{1}{3}, -\frac{8}{3},-\frac{16}{3},3)$.
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