Answer
See below
Work Step by Step
$\begin{bmatrix}
1+2i & 1-i & 1|0\\i & 1+i & -i |0 \\2i & 1&3i+1|0
\end{bmatrix} \approx \begin{bmatrix}
i & 1+i & -i |0 \\1+2i & 1-i & 1|0\\2i & 1&3i+1|0
\end{bmatrix} \approx \begin{bmatrix}
1& 1-i & -1 |0 \\1+2i & 1-i & 1|0\\2i & 1&3i+1|0
\end{bmatrix}\approx \begin{bmatrix}
1& 1-i & -1 |0 \\ 0 & -2-2i & 1+2i|0\\0 & -1-2i& 5i+1|0
\end{bmatrix} \approx \begin{bmatrix}
1& 1-i & -1 |0 \\ 0 & -2-2i & 1+2i|0\\0 & 1& 3i|0
\end{bmatrix}\approx \begin{bmatrix}
1& 1-i & -1 |0 \\ 0 & 0& -5+8i|0\\0 & 1& 3i|0
\end{bmatrix} \approx \begin{bmatrix}
1& 1-i & -1 |0 \\ 0 & 1& 3i|0\\ 0& 0& -5+8i|0
\end{bmatrix} \approx \begin{bmatrix}
1& 1-i & -1 |0 \\ 0 & 1& 3i|0\\ 0& 0& 1|0
\end{bmatrix} \approx \begin{bmatrix}
1& 1& 0 |0 \\ 0 & 1& 0|0\\ 0& 0& 1|0
\end{bmatrix}$
$1.P_{12}\\
2.M_{1}(-i)\\
3.A_{12}(-1-2i),A_{13}(-2i)\\
4.A_{23}(-1)\\
5.A_{32}(2+2i)\\
6.P_{23}\\
7.M_3(\frac{1}{8i-5})\\
8.A_{21}(i-1),A_{31}(1),A_{32}(-3i)$
Since $x_1=x_2=x_3$, the original system of equations has two solutions: $(0,0,0)$
