Answer
The solution is $(\frac{1+3i}{2}t,t)$
Work Step by Step
$Ax=0 $
$\begin{bmatrix}
1-i & 1-2i\\
-1+i & 2+i
\end{bmatrix}.\begin{bmatrix}
x\\
y
\end{bmatrix}=\begin{bmatrix}
0\\
0
\end{bmatrix} $
$\begin{bmatrix}
1-i & 1-2i |0\\
-1+i & 2+i |0
\end{bmatrix} \approx^1 \begin{bmatrix}
1 & -\frac{1+3i}{2}|0\\
-1+i & 2+i|0
\end{bmatrix} \approx^2 \begin{bmatrix}
1 & -\frac{1+3i}{2}|0\\
0 & 0|0
\end{bmatrix}$
This matrix is now in row-echelon form.
$1. M_1(\frac{1-i}{2})$
$2. A_{12}(1-i)$
The solution set is:
$x_1-\frac{1+3i}{2}x_2=0$
$x_2=t$
There is one free variable, which we take to be $x_2 = t$, where t can assume any complex value. Applying back substitution yields:
$x_2-\frac{1+3i}{2}t=0 \rightarrow x_1=\frac{1+3i}{2}t$
$x_2=t$
Hence, the solution is $(\frac{1+3i}{2}t,t)$
