Let A = begin{bmatrix} 0 & 0 & -1 0 & -1 & 0 | vedclass

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(B) We are given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 &

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$\begin{bmatrix} 0 & -2 & 0 \ 0 & 0 & -2 \ -2 & 0 & 0 \end{bmatrix}$

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(B) We are given $A = \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix}$.Using the property $(XY)^{-1} = Y^{-1}X^{-1}$, we can simplify the expression:$(A^{-1}B)^{-1} + (AB^{-1})^{-1} = B^{-1}(A^{-1})^{-1} + (B^{-1})^{-1}A^{-1} = B^{-1}A + BA^{-1}$.First, note that $A^2 = I$ and $B^2 = I$, so $A^{-1} = A$ and $B^{-1} = B$.Thus, the expression becomes $BA + BA = 2BA$.Calculating $BA = \begin{bmatrix} 0 & 1 & 0 \ 1 & 0 & 0 \ 0 & 0 & 1 \end{bmatrix} \begin{bmatrix} 0 & 0 & -1 \ 0 & -1 & 0 \ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -1 & 0 \ 0 & 0 & -1 \ -1 & 0 & 0 \end{bmatrix}$.Therefore, $2BA = 2 \begin{bmatrix} 0 & -1 & 0 \ 0 & 0 & -1 \ -1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & -2 & 0 \ 0 & 0 & -2 \ -2 & 0 & 0 \end{bmatrix}$.This matches option $B$.

Let $A = \begin{bmatrix} 0 & 0 & -1 \\ 0 & -1 & 0 \\ -1 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$. Then find the value of $(A^{-1}B)^{-1} + (AB^{-1})^{-1}$.

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