If A = begin{bmatrix} 2 & 3 5 & -2 end{bmatr | vedclass

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(B) We use the property $(AB)^{-1} = B^{-1}A^{-1}$.First, find $A^{-1}$. Given $A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$, $|A| = (2)(-2) - (

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$\frac{1}{95} \begin{bmatrix} 12 & -1 \ -1 & 8 \end{bmatrix}$

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(B) We use the property $(AB)^{-1} = B^{-1}A^{-1}$.First, find $A^{-1}$. Given $A = \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$, $|A| = (2)(-2) - (3)(5) = -4 - 15 = -19$.$A^{-1} = \frac{1}{|A|} 	ext{adj}(A) = \frac{1}{-19} \begin{bmatrix} -2 & -3 \ -5 & 2 \end{bmatrix} = \frac{1}{19} \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$.Now, $(AB)^{-1} = B^{-1}A^{-1} = \begin{bmatrix} 1/5 & 2/5 \ 2/5 & -1/5 \end{bmatrix} 	imes \frac{1}{19} \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix}$.$= \frac{1}{5 	imes 19} \begin{bmatrix} 1 & 2 \ 2 & -1 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 5 & -2 \end{bmatrix} = \frac{1}{95} \begin{bmatrix} (1)(2)+(2)(5) & (1)(3)+(2)(-2) \ (2)(2)+(-1)(5) & (2)(3)+(-1)(-2) \end{bmatrix}$.$= \frac{1}{95} \begin{bmatrix} 12 & -1 \ -1 & 8 \end{bmatrix}$.

If $A = \begin{bmatrix} 2 & 3 \\ 5 & -2 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix} 1/5 & 2/5 \\ 2/5 & -1/5 \end{bmatrix}$, then $(AB)^{-1} = $

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