## Linear Algebra and Its Applications (5th Edition)

$\begin{bmatrix}3\\1\\2\end{bmatrix}$ is the unique vector with image $\vec{b}$ under $T$.
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: $\begin{bmatrix}1&0&-2&-1\\-2&1&6&7\\3&-2&-5&-3\end{bmatrix}$ Add $2$ times row (1) to row (2) and $-3$ times row (1) to row (3): $\begin{bmatrix}1&0&-2&-1\\0&1&2&5\\0&-2&1&0\end{bmatrix}$ Add $2$ times row (2) to row (3): $\begin{bmatrix}1&0&-2&-1\\0&1&2&5\\0&0&5&10\end{bmatrix}$ Divide row (3) by $5$: $\begin{bmatrix}1&0&-2&-1\\0&1&2&5\\0&0&1&2\end{bmatrix}$ Add $-2$ times row (3) to row (2) and $2$ times row (3) to row (1): $\begin{bmatrix}1&0&0&3\\0&1&0&1\\0&0&1&2\end{bmatrix}$ Because there are no free variables, the solution is unique.