If $A =\begin{bmatrix}2&0&3\\-1&1&3\\0&-4&0\end{bmatrix}$, then the value of $\text{det (2A)}$ is

288

The correct answer is Option (3) → 288

$A = \begin{bmatrix} 2 & 0 & 3 \\ -1 & 1 & 3 \\ 0 & -4 & 0 \end{bmatrix}$

$\det(2A) = 2^3 \cdot \det(A) = 8 \cdot \det(A)$

\[ \det(A) = \begin{vmatrix} 2 & 0 & 3 \\ -1 & 1 & 3 \\ 0 & -4 & 0 \end{vmatrix} = 2 \cdot \begin{vmatrix} 1 & 3 \\ -4 & 0 \end{vmatrix} - 0 \cdot \begin{vmatrix} -1 & 3 \\ 0 & 0 \end{vmatrix} + 3 \cdot \begin{vmatrix} -1 & 1 \\ 0 & -4 \end{vmatrix} \]

\[ = 2(1 \cdot 0 - 3 \cdot (-4)) + 0 + 3((-1) \cdot (-4) - 1 \cdot 0) = 2(12) + 0 + 3(4) = 24 + 12 = 36 \]

$\det(2A) = 8 \cdot 36 = 288$