Systems over the complex field

Solve the following systems over the field of complex numbers \(\mathbb C\).

Variant 1
\( \begin{array}{rlrlr} x_1 &+& i x_2 &=\ & i \\ i x_1 &+& (1-i) x_2 &=\ & -1 \end{array} \)
Resolution
\( \left(\begin{array}{cc|c} 1&i&i\\ i&1-i&-1\\ \end{array}\right)\sim \left(\begin{array}{cc|c} 1&i&i\\ 0&2-i&0\\ \end{array}\right), \) now straightforwardly \(\mathbf x=(i,0)^T\)
Variant 2
\( \begin{array}{rlrlrlr} 2 x_1 &+& (2+2i) x_2 &+& 2i x_3 &= & 1 \\ (1-i) x_1 &+& (1+3i) x_2 &+& (i-1) x_3 &= & 0 \\ (1+i) x_1 &+& (1-i) x_2 &+& (1+i) x_3 &= & 1 \end{array} \)
Resolution
\[\left(\begin{array}{ccc|c} 2 & 2+2i & 2i & 1 \\ 1-i & 1+3i & i-1 & 0 \\ 1+i & 1-i & 1+i & 1 \\ \end{array}\right)\sim \left(\begin{array}{ccc|c} 2 & 2+2i & 2i & 1 \\ 2 & 2+2i & 2i & 1 \\ 2 & -2i & 2 & 1-i \\ \end{array}\right)\sim\\ \left(\begin{array}{ccc|c} 2 & -2i & 2 & 1-i \\ 0 & 2+4i & 2i-2 & i \\ 0 & 0 & 0 & 0 \\ \end{array}\right)\sim \left(\begin{array}{ccc|c} 2 & -2i & 2 & 1-i \\ 0 & 10 & 2+6i & 2+i \\ 0 & 0 & 0 & 0 \\ \end{array}\right) \]

By the backward substitution we get \(\mathbf x=(\frac{4-3i}{10},\frac{2+i}{10},0)^T+p(\frac{-4-2i}{10},\frac{-2-6i}{10},1)^T\).

For \(p=1\) and a substitution \(p'=-\frac{p}{5}\) we get \(\mathbf x=(-i/2,-i/2{,}1)^T+p'(2+i,1+3i,-5)^T\).