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

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(B) First,we calculate the product $AB$: $AB = \begin{bmatrix} 1 & -2 & 1 \ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \ 3 & 2 \ 1 & 1 \end{bmat

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$\begin{bmatrix} -3 & 10 \ -2 & 7 \end{bmatrix}$

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(B) First,we calculate the product $AB$: $AB = \begin{bmatrix} 1 & -2 & 1 \ 2 & 1 & 3 \end{bmatrix} \begin{bmatrix} 2 & 1 \ 3 & 2 \ 1 & 1 \end{bmatrix}$ $AB = \begin{bmatrix} (1)(2) + (-2)(3) + (1)(1) & (1)(1) + (-2)(2) + (1)(1) \ (2)(2) + (1)(3) + (3)(1) & (2)(1) + (1)(2) + (3)(1) \end{bmatrix}$ $AB = \begin{bmatrix} 2 - 6 + 1 & 1 - 4 + 1 \ 4 + 3 + 3 & 2 + 2 + 3 \end{bmatrix} = \begin{bmatrix} -3 & -2 \ 10 & 7 \end{bmatrix}$ Now,we find the transpose $(AB)^T$ by swapping rows and columns: $(AB)^T = \begin{bmatrix} -3 & 10 \ -2 & 7 \end{bmatrix}$ Thus,the correct option is $B$.

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

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