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

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(A) We use the property of matrix inversion: $(XY)^{-1} = Y^{-1} X^{-1}$.Applying this to the given expression:$(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (

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

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(A) We use the property of matrix inversion: $(XY)^{-1} = Y^{-1} X^{-1}$.Applying this to the given expression:$(B^{-1} A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1}$Since $(M^{-1})^{-1} = M$,we have:$(B^{-1} A^{-1})^{-1} = A B$Now,calculate the product $AB$:$AB = \begin{bmatrix} 2 & 3 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$$AB = \begin{bmatrix} (2)(0) + (3)(1) & (2)(-1) + (3)(0) \ (-3)(0) + (2)(1) & (-3)(-1) + (2)(0) \end{bmatrix}$$AB = \begin{bmatrix} 0 + 3 & -2 + 0 \ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 3 & -2 \ 2 & 3 \end{bmatrix}$

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

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