Answer
$\mathbf{x}=x_{3}\begin{bmatrix}5\\-2\\1\end{bmatrix}$, for any $x_{3}$ in $\mathbb{R}$
Work Step by Step
We solve the homogeneous system by performing elimination on the coefficient matrix:
$\begin{bmatrix}1&3&1\\-4&-9&2\\0&-3&-6\end{bmatrix}\sim
\begin{bmatrix}1&3&1\\0&3&6\\0&-3&-6\end{bmatrix}\sim
\begin{bmatrix}1&3&1\\0&3&6\\0&0&0\end{bmatrix}\sim
\begin{bmatrix}1&0&-5\\0&3&6\\0&0&0\end{bmatrix}\sim
\begin{bmatrix}1&0&-5\\0&1&2\\0&0&0\end{bmatrix}$
The rref matrix represents the equations
$\begin{cases}x_{1}-5x_{3}=0\\x_{2}+2x_{3}=0\\0=0\end{cases}$
Solving for the basic variables ($x_{1}$ and $x_{2}$) in terms of the free variable ($x_{3}$), we get
$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begin{bmatrix}5x_{3}\\-2x_{3}\\x_{3}\end{bmatrix}=x_{3}\begin{bmatrix}5\\-2\\1\end{bmatrix}$
