Answer
The solution is $(3s,r,s,t)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1&0&-3 &0\\
3& 0 & -9 & 0\\
-2&0 &6&0\\
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1&0&-3 &0|0\\
3& 0 & -9 & 0|0\\
-2&0 &6&0|0\\
\end{bmatrix}\approx^1\begin{bmatrix}
1&0&-3 &0|0\\
0& 0 &0 & 0|0\\
0&0 &0&0|0\\
\end{bmatrix} $
This matrix is now in row-echelon form.
$1. A_{12}(-3),A_{13}(2)$
$2.A_{21}(1),A_{23}(-1)$
The solution set is:
$x_1-3x_3=0$
$x_2=r$
$x_3=s$
$x_4=t$
There are three free variables, which we take to be $x_2=r,x_3=s,x_4=t$, where s, r and t can assume any complex value. Applying back substitution yields:
$x_1-3s=0 \rightarrow x_1=3s$
$x_2=r$
$x_3=s$
$x_4=t$
Hence, the solution is $(3s,r,s,t)$
