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

$\begin{bmatrix}7-3k\\3-k\\k\\\end{bmatrix}$ for any real number $k$ (not unique)
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: $\begin{bmatrix}1&-2&1&1\\3&-4&5&9\\0&1&1&3\\-3&5&-4&-6\end{bmatrix}$ Add $-3$ times row (1) to row (2) and $3$ times row (1) to row (4): $\begin{bmatrix}1&-2&1&1\\0&2&2&6\\0&1&1&3\\0&-1&-1&-3\end{bmatrix}$ Multiply row (2) by $\frac{1}{2}$; subtract row (2') from row (3) and add row (2') to row (4): $\begin{bmatrix}1&-2&1&1\\0&1&1&3\\0&0&0&0\\0&0&0&0\end{bmatrix}$ Add $2$ times row (2) to row (1): $\begin{bmatrix}1&0&3&7\\0&1&1&3\\0&0&0&0\\0&0&0&0\end{bmatrix}$ Since the third column is not a pivot column, the solution is not unique, and we have: $\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begin{bmatrix}7-3k\\3-k\\k\\\end{bmatrix}$