Let A=[ … ],i=√-1. Then, the system of linear equations A^8[ … ] has :

$$A=[\begin{matrix}i & -i \\ -i & i\end{matrix}]$$A=[i−i​−ii​]

$${A}^{2}=[\begin{matrix}-2 & 2 \\ 2 & -2\end{matrix}]=2[\begin{matrix}-1 & 1 \\ 1 & -1\end{matrix}]$$A2=[−22​2−2​]=2[−11​1−1​]

$${A}^{4}={2}^{2}[\begin{matrix}2 & -2 \\ -2 & 2\end{matrix}]=8[\begin{matrix}1 & -1 \\ -1 & 1\end{matrix}]$$A4=22[2−2​−22​]=8[1−1​−11​]

$${A}^{8}=64[\begin{matrix}2 & -2 \\ -2 & 2\end{matrix}]=128[\begin{matrix}1 & -1 \\ -1 & 1\end{matrix}]$$A8=64[2−2​−22​]=128[1−1​−11​]

$${A}^{8}[\begin{matrix}x \\ y\end{matrix}]=[\begin{matrix}8 \\ 64\end{matrix}]$$A8[xy​]=[864​]

$$\Rightarrow 128[\begin{matrix}1 & -1 \\ -1 & 1\end{matrix}][\begin{matrix}x \\ y\end{matrix}]=[\begin{matrix}8 \\ 64\end{matrix}]$$⇒128[1−1​−11​][xy​]=[864​]

$$\Rightarrow 128[\begin{matrix}x-y \\ -x+y\end{matrix}]=[\begin{matrix}8 \\ 64\end{matrix}]$$⇒128[x−y−x+y​]=[864​]

$\Rightarrow x-y=\frac{1}{16}$⇒x−y=161​ $...(1)$...(1)

and $-x+y=\frac{1}{2}$−x+y=21​ $...(2)$...(2)

$\Rightarrow$⇒ From $(1)$(1) and $(2)$(2) : No solution.