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

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(A) Using the reversal law of inverses,we know that $(B^{-1}A^{-1})^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB$. Now,we calculate the product $AB$: $AB =

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

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(A) Using the reversal law of inverses,we know that $(B^{-1}A^{-1})^{-1} = (A^{-1})^{-1}(B^{-1})^{-1} = AB$. Now,we calculate the product $AB$: $AB = \begin{bmatrix} 2 & 2 \ -3 & 2 \end{bmatrix} \begin{bmatrix} 0 & -1 \ 1 & 0 \end{bmatrix}$ $AB = \begin{bmatrix} (2 	imes 0) + (2 	imes 1) & (2 	imes -1) + (2 	imes 0) \ (-3 	imes 0) + (2 	imes 1) & (-3 	imes -1) + (2 	imes 0) \end{bmatrix}$ $AB = \begin{bmatrix} 0 + 2 & -2 + 0 \ 0 + 2 & 3 + 0 \end{bmatrix} = \begin{bmatrix} 2 & -2 \ 2 & 3 \end{bmatrix}$.

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} = $

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