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 - 4.5 Basis and Dimension - 4.5 Exercises - Page 188: 47

Answer

$S$ is a basis for $R^3$. $$u=(8,3,8)=2(4,3,2)-(0,3,2)+3(0,0,2).$$

Work Step by Step

The set $S=\{(4,3,2),(0,3,2),(0,0,2)\}$ is a linearly independent set of vectors. Indeed, assume that $$a(4,3,2)+b(0,3,2)+c(0,0,2)=(0,0,0), \quad a,b,c\in R.$$ Then, we have the following system of equations \begin{align*} 4a&=0\\ 3a+3b&=0\\ 2a+2b+2c&=0. \end{align*} Solving the above equations, we find that $a=0,b=0,c=0$ and hence $S$ is linearly independent set of vectors. Now, since $R^3$ is vector space of dimension $3$, then by Theorem 4.12, $S$ is a basis for $R^3$. To write $u=(8,3,8)$ as a linear combination of the vectors in $S$, suppose that $$a(4,3,2)+b(0,3,2)+c(0,0,2)=(8,3,8)$$ Then, we have the following system of equations \begin{align*} 4a&=8\\ 3a+3b&=3\\ 2a+2b+2c&=8. \end{align*} Solving the above equations, we find that $a=2,b=-1,c=3$. Hence $$u=(8,3,8)=2(4,3,2)-(0,3,2)+3(0,0,2).$$
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