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(B) Given $P = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$

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$\begin{bmatrix} 2 & -2 \ -1 & 1 \ -1 & 1 \end{bmatrix}$

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(B) Given $P = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$.To find $PQ$,we perform matrix multiplication:$PQ = \begin{bmatrix} i & 0 & -i \ 0 & -i & i \ -i & i & 0 \end{bmatrix} \begin{bmatrix} -i & i \ 0 & 0 \ i & -i \end{bmatrix}$$= \begin{bmatrix} i(-i) + 0(0) + (-i)(i) & i(i) + 0(0) + (-i)(-i) \ 0(-i) + (-i)(0) + i(i) & 0(i) + (-i)(0) + i(-i) \ (-i)(-i) + i(0) + 0(i) & (-i)(i) + i(0) + 0(-i) \end{bmatrix}$Since $i^2 = -1$,we have:$PQ = \begin{bmatrix} -i^2 + 0 - i^2 & i^2 + 0 + i^2 \ 0 + 0 + i^2 & 0 + 0 - i^2 \ i^2 + 0 + 0 & -i^2 + 0 + 0 \end{bmatrix} = \begin{bmatrix} -(-1) - (-1) & -1 + (-1) \ -1 & -(-1) \ -1 & -(-1) \end{bmatrix} = \begin{bmatrix} 1+1 & -1-1 \ -1 & 1 \ -1 & 1 \end{bmatrix} = \begin{bmatrix} 2 & -2 \ -1 & 1 \ -1 & 1 \end{bmatrix}$.

If $P = \begin{bmatrix} i & 0 & -i \\ 0 & -i & i \\ -i & i & 0 \end{bmatrix}$ and $Q = \begin{bmatrix} -i & i \\ 0 & 0 \\ i & -i \end{bmatrix}$,then $PQ$ is equal to

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