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

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(A) Step $1$: Calculate the product $AB$. $AB = \begin{bmatrix} 1 & 2 & 1 \ 3 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 1 & 2 \ 1 & 2 \end{bmat

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$\begin{bmatrix} \frac{-17}{5} & \frac{9}{5} \ 2 & -1 \end{bmatrix}$

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(A) Step $1$: Calculate the product $AB$. $AB = \begin{bmatrix} 1 & 2 & 1 \ 3 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 3 \ 1 & 2 \ 1 & 2 \end{bmatrix} = \begin{bmatrix} (1)(2) + (2)(1) + (1)(1) & (1)(3) + (2)(2) + (1)(2) \ (3)(2) + (1)(1) + (3)(1) & (3)(3) + (1)(2) + (3)(2) \end{bmatrix} = \begin{bmatrix} 5 & 9 \ 10 & 17 \end{bmatrix}$. Step $2$: Find the determinant of $AB$. $|AB| = (5)(17) - (9)(10) = 85 - 90 = -5$. Step $3$: Find the inverse $(AB)^{-1} = \frac{1}{|AB|} 	ext{adj}(AB)$. $(AB)^{-1} = \frac{1}{-5} \begin{bmatrix} 17 & -9 \ -10 & 5 \end{bmatrix} = \begin{bmatrix} \frac{-17}{5} & \frac{9}{5} \ 2 & -1 \end{bmatrix}$.

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

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