Answer
The solution is $(r-5ir+5t+ti,13r,13t)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
2-3i& 1+i&i-1\\
3+2i & -1+i & -1-i\\
5-i &2i &-2
\end{bmatrix}.\begin{bmatrix}
x_1\\
x_2\\
x_3
\end{bmatrix}=\begin{bmatrix}
0\\
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
2-3i& 1+i&i-1|0\\
3+2i & -1+i & -1-i|0\\
5-i &2i &-2 |0
\end{bmatrix} \approx^1\begin{bmatrix}
1&\frac{-1+5i}{13}&\frac{-5-i}{13}|0\\
3+2i & -1+i & -1-i|0\\
5-i &2i &-2 |0
\end{bmatrix} \approx^2\begin{bmatrix}
1&\frac{-1+5i}{13}&\frac{-5-i}{13}|0\\
0& 0 & 0|0\\
0&0 &0 |0
\end{bmatrix} $
This matrix is now in row-echelon form.
$1. M_1(\frac{2+3i}{13})$
$2. A_{2}(-3-2i),A_{13}(-5+i)$
The solution set is:
$x_1+\frac{-1+5i}{13}x_2+\frac{-5-i}{13}x_3=0$
$x_2=13r$
$x_3=13t$
There is one free variable, which we take to be $x_2=13r,x_3=13t$, where r ans t can assume any complex value. Applying back substitution yields:
$x_1+\frac{-1+5i}{13}.13r+\frac{-5-i}{13}.13t=0 \rightarrow x_1=r-5ir+5t+ti$
$x_2=13r$
$x_3=13t$
Hence, the solution is $(r-5ir+5t+ti,13r,13t)$
