If A = begin{bmatrix} 0 & i -i & 0 end{bmatr | vedclass

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(B) Given $A = \begin{bmatrix} 0 & i \ -i & 0 \end{bmatrix}$.First,calculate $A^2$:$A^2 = A 	imes A = \begin{bmatrix} 0 & i \ -i & 0 \end{bmatrix}

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$\begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$

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(B) Given $A = \begin{bmatrix} 0 & i \ -i & 0 \end{bmatrix}$.First,calculate $A^2$:$A^2 = A 	imes A = \begin{bmatrix} 0 & i \ -i & 0 \end{bmatrix} \begin{bmatrix} 0 & i \ -i & 0 \end{bmatrix} = \begin{bmatrix} (0)(0) + (i)(-i) & (0)(i) + (i)(0) \ (-i)(0) + (0)(-i) & (-i)(i) + (0)(0) \end{bmatrix}$$A^2 = \begin{bmatrix} -i^2 & 0 \ 0 & -i^2 \end{bmatrix} = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix} = I$,where $I$ is the identity matrix.Now,$A^{40} = (A^2)^{20} = I^{20} = I = \begin{bmatrix} 1 & 0 \ 0 & 1 \end{bmatrix}$.

If $A = \begin{bmatrix} 0 & i \\ -i & 0 \end{bmatrix}$,then the value of $A^{40}$ is

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