If A = begin{bmatrix} -2 & 2 -3 & 2 end{bmat | vedclass

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(A) We know that $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} = AB$. Given $A = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix}$ and $B = \begin

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$\begin{bmatrix} 2 & 2 \ 2 & 3 \end{bmatrix}$

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(A) We know that $(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} = AB$. Given $A = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$. Now,calculate the product $AB$: $AB = \begin{bmatrix} -2 & 2 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ $AB = \begin{bmatrix} (-2)(0) + (2)(1) & (-2)(-1) + (2)(0) \ (-3)(0) + (2)(1) & (-3)(-1) + (2)(0) \end{bmatrix}$ $AB = \begin{bmatrix} 0 + 2 & 2 + 0 \ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 2 & 2 \ 2 & 3 \end{bmatrix}$.

If $A = \begin{bmatrix} -2 & 2 \\ -3 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$,then $(B^{-1} A^{-1})^{-1}$ is equal to

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