Answer
$\mathbf{x}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}=x_{3}\begin{bmatrix}-9\\4\\1\\0\end{bmatrix}+x_{4}\begin{bmatrix}8\\-5\\0\\1\end{bmatrix}$
Work Step by Step
To reduce the matrix to rref, we simply add $-3$ times the second row to the first, yielding
$\begin{bmatrix}1&0&9&-8\\0&1&-4&5\end{bmatrix}$.
This is equivalent to the set of equations
$\begin{cases}x_{1}+9x_{3}-8x_{4}=0\\x_{2}-4x_{3}+5x_{4}=0\end{cases}$.
We solve for $x_{1}$ and $x_{2}$ in terms of $x_{3}$ and $x_{4}$,
$\begin{cases}x_{1}=-9x_{3}+8x_{4}\\x_{2}=4x_{3}-5x_{4}\\x_{3}=x_{3}\\x_{4}=x_{4}\end{cases}$,
which we finally rewrite as a single vector equation.
