Linear Algebra and Its Applications (5th Edition)

Published by Pearson
ISBN 10: 032198238X
ISBN 13: 978-0-32198-238-4

Chapter 1 - Linear Equations in Linear Algebra - 1.7 Exercises - Page 62: 9

Answer

a. There are no values of $h$ such that $\vec{v}_{3}$ is in $Span\{\vec{v}_{1},\vec{v}_{2}\}$. b. The set is linearly dependent for all values of $h$.

Work Step by Step

We can prove both parts with a single line of reasoning. By row reduction, we see that $\begin{bmatrix}1&-3&5\\-3&9&-7\\2&-6&h\end{bmatrix}\sim\begin{bmatrix}1&-3&5\\0&0&8\\0&0&h-10\end{bmatrix}$. That the second column contains no pivot indicates that $\mathbf{A}\vec{x}=\vec{0}$ has infinitely many solutions, and so does the corresponding vector equation. This means that the vectors are linearly dependent for all values of $h$. However, it is also clear that the third column cannot be formed from a sum of the first two columns, owing to the row $(0,0,8)$. Hence, no value of $h$ puts $\vec{v}_{3}$ in $Span\{\vec{v}_{1},\vec{v}_{2}\}$.
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