For matrices A and B,A^{prime} = begin{bmatri | vedclass

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(A) Given $A^{\prime} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$,which is a $3 	imes 1$ matrix. Therefore,$A$ is a $1 	imes 3$ matrix. Given $B^{\

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Square matrix

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(A) Given $A^{\prime} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$,which is a $3 	imes 1$ matrix. Therefore,$A$ is a $1 	imes 3$ matrix. Given $B^{\prime} = \begin{bmatrix} 4 & 3 & 2 \end{bmatrix}$,which is a $1 	imes 3$ matrix. Therefore,$B$ is a $3 	imes 1$ matrix. We need to find $(BA)^{\prime}$. Using the property of transpose,$(BA)^{\prime} = A^{\prime} B^{\prime}$. However,$A^{\prime}$ is $3 	imes 1$ and $B^{\prime}$ is $1 	imes 3$. Thus,$A^{\prime} B^{\prime} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix} \begin{bmatrix} 4 & 3 & 2 \end{bmatrix} = \begin{bmatrix} 4 & 3 & 2 \ 8 & 6 & 4 \ 12 & 9 & 6 \end{bmatrix}$. This is a $3 	imes 3$ matrix,which is a square matrix. Therefore,the correct option is $A$.

For matrices $A$ and $B$,$A^{\prime} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$ and $B^{\prime} = \begin{bmatrix} 4 & 3 & 2 \end{bmatrix}$,then $(BA)^{\prime}$ is . . . . . . .

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