If A = begin{bmatrix} 1 & 2 & 1 2 & 1 & 0 en | vedclass

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

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$\left(\frac{1}{5}ight) \begin{bmatrix} 5 & -5 \ -4 & 5 \end{bmatrix}$

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(B) First,calculate the product $AB$: $AB = \begin{bmatrix} 1 & 2 & 1 \ 2 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & 2 \ 2 & 1 \ 0 & 1 \end{bmatrix} = \begin{bmatrix} (1)(1)+(2)(2)+(1)(0) & (1)(2)+(2)(1)+(1)(1) \ (2)(1)+(1)(2)+(0)(0) & (2)(2)+(1)(1)+(0)(1) \end{bmatrix} = \begin{bmatrix} 5 & 5 \ 4 & 5 \end{bmatrix}$ Now,find the determinant $|AB|$: $|AB| = (5)(5) - (5)(4) = 25 - 20 = 5$ Next,find the adjoint of $AB$: $adj(AB) = \begin{bmatrix} 5 & -5 \ -4 & 5 \end{bmatrix}$ Finally,calculate the inverse $(AB)^{-1} = \frac{1}{|AB|} adj(AB)$: $(AB)^{-1} = \frac{1}{5} \begin{bmatrix} 5 & -5 \ -4 & 5 \end{bmatrix}$ Thus,the correct option is $B$.

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

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