If A = [1, 2, 3] and B = begin{bmatrix} -5 & | vedclass

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(C) Given $A = [1\, 2\, 3]$ (a $1 	imes 3$ matrix) and $B = \begin{bmatrix} -5 & 4 & 0 \ 0 & 2 & -1 \ 1 & -3 & 2 \end{bmatrix}$ (a $3 	imes 3$ mat

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

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(C) Given $A = [1\, 2\, 3]$ (a $1 	imes 3$ matrix) and $B = \begin{bmatrix} -5 & 4 & 0 \ 0 & 2 & -1 \ 1 & -3 & 2 \end{bmatrix}$ (a $3 	imes 3$ matrix). To find $AB$,we multiply the row of $A$ by each column of $B$: $AB = [1 	imes (-5) + 2 	imes 0 + 3 	imes 1, \quad 1 	imes 4 + 2 	imes 2 + 3 	imes (-3), \quad 1 	imes 0 + 2 	imes (-1) + 3 	imes 2]$ $AB = [-5 + 0 + 3, \quad 4 + 4 - 9, \quad 0 - 2 + 6]$ $AB = [-2, \quad -1, \quad 4]$ Thus,$AB = [-2\, -1\, 4]$.

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

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