If A = [2] and B = begin{bmatrix} 3 4 end{bm | vedclass

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(D) Given $A = [2]$ and $B = \begin{bmatrix} 3 \ 4 \end{bmatrix}$.First,calculate the product $BA$:$BA = \begin{bmatrix} 3 \ 4 \end{bmatrix} [2] = \

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$\begin{bmatrix} 3 & 4 \ 6 & 8 \end{bmatrix}$

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(D) Given $A = [2]$ and $B = \begin{bmatrix} 3 \ 4 \end{bmatrix}$.First,calculate the product $BA$:$BA = \begin{bmatrix} 3 \ 4 \end{bmatrix} [2] = \begin{bmatrix} 3 	imes 2 \ 4 	imes 2 \end{bmatrix} = \begin{bmatrix} 6 \ 8 \end{bmatrix}$.Now,find the transpose $(BA)'$:$(BA)' = \begin{bmatrix} 6 & 8 \end{bmatrix}$.Note: Based on the provided options,there seems to be a typo in the original question's matrix $A$. If $A = [2]$,the result is $[6 \quad 8]$. If the question intended $B = \begin{bmatrix} 3 \ 4 \end{bmatrix}$ and $A = [1 \quad 2]$,then $BA = \begin{bmatrix} 3 \ 4 \end{bmatrix} [1 \quad 2] = \begin{bmatrix} 3 & 6 \ 4 & 8 \end{bmatrix}$.Then $(BA)' = \begin{bmatrix} 3 & 4 \ 6 & 8 \end{bmatrix}$,which matches option $D$.

If $A = [2]$ and $B = \begin{bmatrix} 3 \\ 4 \end{bmatrix}$,then $(BA)' = $ . . . . . . .

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