begin{bmatrix} 7 & 1 & 2 9 & 2 & 1 end{bmatr | vedclass

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(A) First,calculate the product of the matrix and the column vector: $\begin{bmatrix} 7 & 1 & 2 \ 9 & 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \ 4 \ 5

Vedclass · Accepted Answer

$\begin{bmatrix} 43 \ 44 \end{bmatrix}$

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(A) First,calculate the product of the matrix and the column vector: $\begin{bmatrix} 7 & 1 & 2 \ 9 & 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \ 4 \ 5 \end{bmatrix} = \begin{bmatrix} (7 	imes 3) + (1 	imes 4) + (2 	imes 5) \ (9 	imes 3) + (2 	imes 4) + (1 	imes 5) \end{bmatrix} = \begin{bmatrix} 21 + 4 + 10 \ 27 + 8 + 5 \end{bmatrix} = \begin{bmatrix} 35 \ 40 \end{bmatrix}$. Next,calculate the scalar multiplication: $2 \begin{bmatrix} 4 \ 2 \end{bmatrix} = \begin{bmatrix} 8 \ 4 \end{bmatrix}$. Finally,add the two resulting matrices: $\begin{bmatrix} 35 \ 40 \end{bmatrix} + \begin{bmatrix} 8 \ 4 \end{bmatrix} = \begin{bmatrix} 35 + 8 \ 40 + 4 \end{bmatrix} = \begin{bmatrix} 43 \ 44 \end{bmatrix}$. Thus,the correct option is $A$.

$\begin{bmatrix} 7 & 1 & 2 \\ 9 & 2 & 1 \end{bmatrix} \begin{bmatrix} 3 \\ 4 \\ 5 \end{bmatrix} + 2 \begin{bmatrix} 4 \\ 2 \end{bmatrix}$ is equal to

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