$A = \begin{pmatrix}i&-i\\-i&i\end{pmatrix}$. Note $\det(A) = i^2-i^2=0$, so $A$ is singular.

$A^2$: compute $(A)^2$. Row1·Col1: $i(i)+(-i)(-i)=-1+(-1)=-2i$... Let's compute: $A^2_{11}=i\cdot i+(-i)(-i)=i^2+i^2=-1-1=-2$. So $A^2 = -2A$... check: $A^2 = -2\begin{pmatrix}i&-i\\-i&i\end{pmatrix}$.

Therefore $A^8 = (-2)^4 A^4 = 16(A^2)^2=16\cdot4A^2=64\cdot(-2A)=-128A$.

Wait: $A^2=-2A \Rightarrow A^4=4A^2=-8A \Rightarrow A^8=-8A^4... $ Correctly: $A^8=(-2)^7A = -128A$.

$-128A\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}8\\64\end{pmatrix}$. Since $\det(A)=0$, check consistency. The equations are linearly dependent but RHS $(8,64)$ is not in column space of $A$. Hence no solution.