Answer
The solution is $(-3t,-2t,r,t)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1&-1&0 &1\\
3& -2 & 0 & 5\\
-1 &2 &0&1\\
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1&-1&0 &1|0\\
3& -2 & 0 & 5|0\\
-1 &2 &0&1|0\\
\end{bmatrix} \approx^1\begin{bmatrix}
1&-1&0 &1|0\\
0& 1 & 0 & 2|0\\
0 &1 &0&2|0\\
\end{bmatrix} \approx^2\begin{bmatrix}
1&0&0 &3|0\\
0& 1 & 0 & 2|0\\
0 &0 &0&0|0\\
\end{bmatrix} $
This matrix is now in row-echelon form.
$1. A_{12}(-3),A_{13}(1)$
$2.A_{21}(1),A_{23}(-1)$
The solution set is:
$x_1+3x_4=0$
$x_2+2x^4=0$
$x_3=r$
$x_4=t$
There is one free variable, which we take to be$x_3=r,x_4=t$, where r and t can assume any complex value. Applying back substitution yields:
$x_1+3t=0 \rightarrow x_1=-3t$
$x_2+2t=0 \rightarrow x_2=-2t$
$x_3=r$
$x_4=t$
Hence, the solution is $(-3t,-2t,r,t)$
