If A = begin{bmatrix} 2 & 3 1 & 2 end{bmatri | vedclass

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(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.First,we find the product $AB$:$AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \

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$\begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$

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(A) We know that $(AB)^{-1} = B^{-1} A^{-1}$.First,we find the product $AB$:$AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2)(1) + (3)(3) & (2)(0) + (3)(1) \ (1)(1) + (2)(3) & (1)(0) + (2)(1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \ 7 & 2 \end{bmatrix}$.Now,we find $(AB)^{-1} = \frac{1}{|AB|} 	ext{adj}(AB)$.$|AB| = (11)(2) - (3)(7) = 22 - 21 = 1$.$	ext{adj}(AB) = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$.Therefore,$B^{-1} A^{-1} = (AB)^{-1} = \frac{1}{1} \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$.

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

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