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(C) Given $A = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{bmatrix} = 2I$,where $I$ is the identity matrix of order $3 	imes 3$.We know

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$16$

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(C) Given $A = \begin{bmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{bmatrix} = 2I$,where $I$ is the identity matrix of order $3 	imes 3$.We know that the determinant of the product of two matrices is the product of their determinants,i.e.,$|AB| = |A| \cdot |B|$.First,calculate $|A|$:$|A| = \begin{vmatrix} 2 & 0 & 0 \ 0 & 2 & 0 \ 0 & 0 & 2 \end{vmatrix} = 2(2 	imes 2 - 0 	imes 0) = 2(4) = 8$.Next,calculate $|B|$:Since $B$ is an upper triangular matrix,its determinant is the product of its diagonal elements:$|B| = \begin{vmatrix} 1 & 2 & 3 \ 0 & 1 & 3 \ 0 & 0 & 2 \end{vmatrix} = 1 	imes 1 	imes 2 = 2$.Therefore,$|AB| = |A| 	imes |B| = 8 	imes 2 = 16$.

If $A = \begin{bmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 3 \\ 0 & 0 & 2 \end{bmatrix}$,then $|AB|$ is equal to

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