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

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(A) To find the product $AB$, we multiply the rows of matrix $A$ by the columns of matrix $B$: $AB = \begin{bmatrix} 2 & 1 & 3 \ 4 & 1 & 0 \end{bmatr

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$\begin{bmatrix} 17 & 0 \ 4 & -2 \end{bmatrix}$

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(A) To find the product $AB$, we multiply the rows of matrix $A$ by the columns of matrix $B$: $AB = \begin{bmatrix} 2 & 1 & 3 \ 4 & 1 & 0 \end{bmatrix} \begin{bmatrix} 1 & -1 \ 0 & 2 \ 5 & 0 \end{bmatrix}$ $AB = \begin{bmatrix} (2 	imes 1 + 1 	imes 0 + 3 	imes 5) & (2 	imes -1 + 1 	imes 2 + 3 	imes 0) \ (4 	imes 1 + 1 	imes 0 + 0 	imes 5) & (4 	imes -1 + 1 	imes 2 + 0 	imes 0) \end{bmatrix}$ $AB = \begin{bmatrix} (2 + 0 + 15) & (-2 + 2 + 0) \ (4 + 0 + 0) & (-4 + 2 + 0) \end{bmatrix}$ $AB = \begin{bmatrix} 17 & 0 \ 4 & -2 \end{bmatrix}$

If the matrices $A = \begin{bmatrix} 2 & 1 & 3 \\ 4 & 1 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & -1 \\ 0 & 2 \\ 5 & 0 \end{bmatrix}$, then $AB$ will be

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