Answer
The solution of the system is $S=\{\frac{1}{2}t(1-3i),\frac{1}{6}(1+i)t-\frac{1}{3},t\}$
Work Step by Step
Given: $A=\begin{bmatrix}
1 & -3 & 2i\\
-2i & 6 & 2
\end{bmatrix}$
We can reduce A to row-echelon form using the following sequence of elementary row operations:
$\begin{bmatrix}
1 & -3 & 2i | 1\\
-2i & 6 & 2 | -2
\end{bmatrix} \approx^1\begin{bmatrix}
1 & -3 & 2i | 1\\
0 & 6-6i & -2 | -2+2i
\end{bmatrix} \approx^2\begin{bmatrix}
1 & -3 & 2i | 1\\
0 & 1 & -\frac{1}{6}(1+i) | -\frac{1}{3}
\end{bmatrix} $
$1.A_{12}(-32i),A_{13}(-4)$
$3. M_2(\frac{1}{6-6i})$
The equivalent system for this matrix is now:
$x_1-3x_2+2ix_3=-\frac{1}{3}$
$x_2-\frac{1}{6}(1+i)x_3=-\frac{1}{3}$
$x_3=t$
There is one free variable, which we take to be $x_3 = t$, where t can assume any complex value. Applying back substitution yields:
$x_3=t$
$x_2-\frac{1}{6}(1+i)t=-\frac{1}{3} \rightarrow x_2=\frac{1}{6}(1+i)t-\frac{1}{3} $
$x_1-3(\frac{1}{6}(1+i)t-\frac{1}{3})+2it=-\frac{1}{3} \rightarrow x_1=\frac{1}{2}t(1-3i)$
The solution of the system is $S=\{\frac{1}{2}t(1-3i),\frac{1}{6}(1+i)t-\frac{1}{3},t\}$
