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(C) Given $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} i & -i \ -i & i \end{bmatrix} \begin{

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(C) Given $A = \begin{bmatrix} i & -i \ -i & i \end{bmatrix}$.First,calculate $A^{2}$:$A^{2} = \begin{bmatrix} i & -i \ -i & i \end{bmatrix} \begin{bmatrix} i & -i \ -i & i \end{bmatrix} = \begin{bmatrix} i^{2} + (-i)(-i) & -i^{2} - i^{2} \ -i^{2} - i^{2} & (-i)(-i) + i^{2} \end{bmatrix} = \begin{bmatrix} -1 - 1 & 1 + 1 \ 1 + 1 & -1 - 1 \end{bmatrix} = \begin{bmatrix} -2 & 2 \ 2 & -2 \end{bmatrix} = 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix}$.Next,calculate $A^{4} = (A^{2})^{2}$:$A^{4} = \left( 2 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} ight)^{2} = 4 \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} \begin{bmatrix} -1 & 1 \ 1 & -1 \end{bmatrix} = 4 \begin{bmatrix} 1+1 & -1-1 \ -1-1 & 1+1 \end{bmatrix} = 4 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$.Then,calculate $A^{8} = (A^{4})^{2}$:$A^{8} = \left( 8 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} ight)^{2} = 64 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} = 64 \begin{bmatrix} 1+1 & -1-1 \ -1-1 & 1+1 \end{bmatrix} = 64 \begin{bmatrix} 2 & -2 \ -2 & 2 \end{bmatrix} = 128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix}$.The system is $128 \begin{bmatrix} 1 & -1 \ -1 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix} = \begin{bmatrix} 8 \ 64 \end{bmatrix}$.This implies $128(x - y) = 8 \Rightarrow x - y = \frac{8}{128} = \frac{1}{16}$ and $128(-x + y) = 64 \Rightarrow -x + y = \frac{64}{128} = \frac{1}{2}$.From the first equation,$x - y = \frac{1}{16}$,and from the second,$x - y = -\frac{1}{2}$.Since $\frac{1}{16} 
eq -\frac{1}{2}$,the system has no solution.

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

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