Answer
$\mathbf{x}=x_{3}\begin{bmatrix}-4\\3\\1\end{bmatrix}$, for any $x_{3}$ in $\mathbb{R}$
Work Step by Step
We solve the homogeneous system by performing elimination on the coefficient matrix:
$\begin{bmatrix}1&3&-5\\1&4&-8\\-3&-7&9\end{bmatrix}\sim
\begin{bmatrix}1&3&-5\\0&1&-3\\0&2&-6\end{bmatrix}\sim
\begin{bmatrix}1&3&-5\\0&1&-3\\0&0&0\end{bmatrix}\sim
\begin{bmatrix}1&0&4\\0&1&-3\\0&0&0\end{bmatrix}$
This rref matrix represents the equations
$\begin{cases}x_{1}+4x_{3}=0\\x_{2}-3x_{3}=0\\0=0\end{cases}$
Solving for the basic variables ($x_{1}$ and $x_{2}$) in terms of the free variable ($x_{3}$), we get
$\begin{bmatrix}x_{1}\\x_{2}\\x_{3}\end{bmatrix}=\begin{bmatrix}-4x_{3}\\3x_{3}\\x_{3}\end{bmatrix}=x_{3}\begin{bmatrix}-4\\3\\1\end{bmatrix}$
