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.
