$\displaystyle \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]

\left[\begin{array} {cc} x \\ y \end{array}\right]=

\left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$

$\displaystyle \left[\begin{array} {cc} 2+i&-3i \\ 3i&3-2i \end{array}\right]^{-1}$ $\displaystyle \left[\begin{array} {cc} 3+i\\3-3i \end{array}\right]$=$\displaystyle \left[\begin{array} {cc} x \\ y \end{array}\right]$

$\displaystyle \frac{1}{17-i}\left[\begin{array} {cc} 3-2i&3i \\ -3i&2+i \end{array}\right]=\left[\begin{array} {cc} x \\ y \end{array}\right]$

After this it gets really messy and I am unable to get a solution.