Answer
$\mathbf{x}=\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\\x_{4}\end{bmatrix}=x_{2}\begin{bmatrix}-3\\1\\0\\0\end{bmatrix}+x_{4}\begin{bmatrix}4\\0\\0\\1\end{bmatrix}$
Work Step by Step
$\begin{bmatrix}1&3&0&-4\\2&6&0&-8\end{bmatrix}$.
We add $-2*R1$ to R2.
$\begin{bmatrix}1&3&0&-4\\0&0&0&0\end{bmatrix}$.
This is equivalent to the set of equations
$\begin{cases}x_{1}+3x_{2}-4x_{4}=0\\0x_{1}+0x_{2}+0x_{3}+0x_{4}=0\end{cases}$.
We solve for each variable in terms of the free variables.
$\begin{cases}x_{1}=-3x_{2}+4x_{4}\\x_{2}=x_{2}\\x_{3}=x_{3}\\x_{4}=x_{4}\end{cases}$,
which we finally rewrite as a single vector equation.
