Answer
$\mathbf{x}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=x_{2}\begin{bmatrix}3\\1\\0\end{bmatrix}+x_{3}\begin{bmatrix}-2\\0\\1\end{bmatrix}$
Work Step by Step
$\begin{bmatrix}3&-9&6\\-1&3&-2\end{bmatrix}$.
We switch places for rows R1 and R2.
$\begin{bmatrix}-1&3&-2\\3&-9&6\end{bmatrix}$.
We add 3*R1 to R2.
$\begin{bmatrix}-1&3&-2\\0&0&0\end{bmatrix}$.
This is equivalent to the set of equations
$\begin{cases}-x_{1}+3x_{2}-2x_{3}=0\\0x_{1}+0x_{2}+0x_{3}=0\end{cases}$.
We solve for each variable in terms of the free variables.
$\begin{cases}x_{1}=3x_{2}-2x_{3}\\x_{2}=x_{2}\\x_{3}=x_{3}\end{cases}$,
which we finally rewrite as a single vector equation.
