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 - 1.8 Exercises - Page 69: 4

Answer

$\begin{bmatrix}-5\\-3\\1\end{bmatrix}$ is the unique vector with image $\vec{b}$ under $T$.

Work Step by Step

Finding the vector with a given image under some matrix transformation is equivalent to finding the solution set of a linear system. Hence, we begin with the augmented matrix: $\begin{bmatrix}1&-3&2&6\\0&1&-4&-7\\3&-5&-9&-9\end{bmatrix}$ Add $-3$ times row (1) to row (3): $\begin{bmatrix}1&-3&2&6\\0&1&-4&-7\\0&4&-15&-27\end{bmatrix}$ Add $-4$ times row (2) to row (3): $\begin{bmatrix}1&-3&2&6\\0&1&-4&-7\\0&0&1&1\end{bmatrix}$ Add $4$ times row (3) to row (2) and $-2$ times row (3) to row (1): $\begin{bmatrix}1&-3&0&4\\0&1&0&-3\\0&0&1&1\end{bmatrix}$ Add $3$ times row (2) to row (1): $\begin{bmatrix}1&0&0&-5\\0&1&0&-3\\0&0&1&1\end{bmatrix}$ Since the system contains no free variables, the solution is unique.
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