Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 0-32198-238-X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises - Page 62: 16


These vectors are linearly independent.

Work Step by Step

By inspection, we see that the second vector is $\frac{3}{2}$ times the first vector. Hence: $\frac{3}{2}\begin{bmatrix}4\\-2\\6\end{bmatrix}-\begin{bmatrix}6\\-3\\9\end{bmatrix}=\begin{bmatrix}0\\0\\0\end{bmatrix}$. This is a nontrivial solution to the homogeneous vector equation, so the vectors are linearly dependent.
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