Let $$A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$$, $$i = \sqrt{-1}$$. Then, the system of linear equations $$A^8 \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ has:

We have $$A = \begin{bmatrix} i & -i \\ -i & i \end{bmatrix}$$. First, we compute $$A^2 = A \cdot A = \begin{bmatrix} i^2 + i^2 & -i^2 - i^2 \\ -i^2 - i^2 & i^2 + i^2 \end{bmatrix} = \begin{bmatrix} -2 & 2 \\ 2 & -2 \end{bmatrix}$$.

Let $$B = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix}$$, so $$A^2 = -2B$$. We note that $$B^2 = \begin{bmatrix} 2 & -2 \\ -2 & 2 \end{bmatrix} = 2B$$. Therefore $$A^4 = (A^2)^2 = 4B^2 = 8B$$ and $$A^8 = (A^4)^2 = 64B^2 = 128B = \begin{bmatrix} 128 & -128 \\ -128 & 128 \end{bmatrix}$$.

The system $$A^8 \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 8 \\ 64 \end{bmatrix}$$ becomes $$128(x - y) = 8$$ and $$128(-x + y) = 64$$, which gives $$x - y = \frac{1}{16}$$ and $$-x + y = \frac{1}{2}$$.

Adding these two equations: $$0 = \frac{1}{16} + \frac{1}{2} = \frac{9}{16}$$, which is a contradiction. Therefore, the system has no solution.