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 - 4.4 Spanning Sets and Linear Independence - 4.4 Exercises - Page 178: 39

Answer

$S$ is linearly independent.

Work Step by Step

Assume the combination $$a(4,-3,6,2)+b(1,8,3,1)+c(3,-2,-1,0)=(0,0,0,0).$$ We have the system \begin{align*} 4a+b+3c&=0\\ -3a+8b-2c&=0\\ 6a+3b-c&=0\\ 2a+b&=0 \end{align*} Since the determinant of the coefficient matrix is given by $$\left| \begin{array} {ccc} -4&1&6\\ -3&-2&0 \\ 4&3&0 \end{array} \right|=-6\neq 0$$ one can see that the above system has unique solution, that is, $$a=0, \quad b=0, \quad c=0.$$ Consequently, $S$ is linearly independent.
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