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

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(A) Given matrices $A = \begin{bmatrix} 1 & 2 & -1 \ 3 & 0 & 2 \ 4 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 0 & 1 &

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$\begin{bmatrix} 5 & 1 & -3 \ 3 & 2 & 6 \ 14 & 5 & 0 \end{bmatrix}$

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(A) Given matrices $A = \begin{bmatrix} 1 & 2 & -1 \ 3 & 0 & 2 \ 4 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 0 \ 0 & 1 & 3 \end{bmatrix}$.To find $AB$,we perform matrix multiplication:$AB = \begin{bmatrix} (1)(1)+(2)(2)+(-1)(0) & (1)(0)+(2)(1)+(-1)(1) & (1)(0)+(2)(0)+(-1)(3) \ (3)(1)+(0)(2)+(2)(0) & (3)(0)+(0)(1)+(2)(1) & (3)(0)+(0)(0)+(2)(3) \ (4)(1)+(5)(2)+(0)(0) & (4)(0)+(5)(1)+(0)(1) & (4)(0)+(5)(0)+(0)(3) \end{bmatrix}$Calculating each element:Row $1$: $(1+4+0) = 5$,$(0+2-1) = 1$,$(0+0-3) = -3$Row $2$: $(3+0+0) = 3$,$(0+0+2) = 2$,$(0+0+6) = 6$Row $3$: $(4+10+0) = 14$,$(0+5+0) = 5$,$(0+0+0) = 0$Thus,$AB = \begin{bmatrix} 5 & 1 & -3 \ 3 & 2 & 6 \ 14 & 5 & 0 \end{bmatrix}$.

If $A = \begin{bmatrix} 1 & 2 & -1 \\ 3 & 0 & 2 \\ 4 & 5 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 2 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$,then $AB$ is:

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