If A = begin{bmatrix} 1 & 3 & 0 -1 & 2 & 1 | vedclass

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(B) To find the product $AB$,we multiply matrix $A$ and matrix $B$ row by column: $AB = \begin{bmatrix} 1 & 3 & 0 \ -1 & 2 & 1 \ 0 & 0 & 2 \end{bmat

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$\begin{bmatrix} 5 & 9 & 13 \ -1 & 2 & 4 \ -2 & 2 & 4 \end{bmatrix}$

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(B) To find the product $AB$,we multiply matrix $A$ and matrix $B$ row by column: $AB = \begin{bmatrix} 1 & 3 & 0 \ -1 & 2 & 1 \ 0 & 0 & 2 \end{bmatrix} \begin{bmatrix} 2 & 3 & 4 \ 1 & 2 & 3 \ -1 & 1 & 2 \end{bmatrix}$ $= \begin{bmatrix} (1)(2)+(3)(1)+(0)(-1) & (1)(3)+(3)(2)+(0)(1) & (1)(4)+(3)(3)+(0)(2) \ (-1)(2)+(2)(1)+(1)(-1) & (-1)(3)+(2)(2)+(1)(1) & (-1)(4)+(2)(3)+(1)(2) \ (0)(2)+(0)(1)+(2)(-1) & (0)(3)+(0)(2)+(2)(1) & (0)(4)+(0)(3)+(2)(2) \end{bmatrix}$ $= \begin{bmatrix} 2+3+0 & 3+6+0 & 4+9+0 \ -2+2-1 & -3+4+1 & -4+6+2 \ 0+0-2 & 0+0+2 & 0+0+4 \end{bmatrix}$ $= \begin{bmatrix} 5 & 9 & 13 \ -1 & 2 & 4 \ -2 & 2 & 4 \end{bmatrix}$ Thus,the correct option is $B$.

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

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