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

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(A) First,calculate the product $AB$: $AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix} = \begin{bmatrix

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

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(A) First,calculate the product $AB$: $AB = \begin{bmatrix} 2 & 3 \ 1 & 2 \end{bmatrix} \begin{bmatrix} 1 & 0 \ 3 & 1 \end{bmatrix} = \begin{bmatrix} (2 	imes 1 + 3 	imes 3) & (2 	imes 0 + 3 	imes 1) \ (1 	imes 1 + 2 	imes 3) & (1 	imes 0 + 2 	imes 1) \end{bmatrix} = \begin{bmatrix} 11 & 3 \ 7 & 2 \end{bmatrix}$ Next,find the determinant $|AB|$: $|AB| = (11 	imes 2) - (3 	imes 7) = 22 - 21 = 1$ Now,find the adjoint of $AB$: For a matrix $\begin{bmatrix} a & b \ c & d \end{bmatrix}$,the adjoint is $\begin{bmatrix} d & -b \ -c & a \end{bmatrix}$. Thus,$	ext{adj}(AB) = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$ Finally,$(AB)^{-1} = \frac{1}{|AB|} 	ext{adj}(AB) = \frac{1}{1} \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix} = \begin{bmatrix} 2 & -3 \ -7 & 11 \end{bmatrix}$

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

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