If $A = \begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$, then the value of det (adj (2A)) is:

For any scalar $k$ and $n \times n$ matrix $M$:

$\det(kM) = k^n \times \det(M)$

For any $n \times n$ matrix $M$:

$\det(\text{adj}(M)) = [\det(M)]^{n-1}$

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$A = \begin{bmatrix} 2 & -1 & -2 \\ 0 & 2 & -1 \\ 3 & -5 & 0 \end{bmatrix}$

Expanding along Row 1:

$\det(A) = 2 \times \begin{vmatrix} 2 & -1 \\ -5 & 0 \end{vmatrix} - (-1) \times \begin{vmatrix} 0 & -1 \\ 3 & 0 \end{vmatrix} + (-2) \times \begin{vmatrix} 0 & 2 \\ 3 & -5 \end{vmatrix}$

$\det(A) = 2[(2)(0) - (-1)(-5)] + 1[(0)(0) - (-1)(3)] - 2[(0)(-5) - (2)(3)]$

$\det(A) = 2(0 - 5) + 1(0 + 3) - 2(0 - 6)$

$\det(A) = 2(-5) + 1(3) - 2(-6)$

$\det(A) = -10 + 3 + 12$

$\det(A) = 5$

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Since $A$ is a $3 \times 3$ matrix:

$\det(2A) = 2^3 \times \det(A)$

$\det(2A) = 8 \times 5$

$\det(2A) = 40$

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Since $2A$ is a $3 \times 3$ matrix:

$\det(\text{adj}(2A)) = [\det(2A)]^{3-1}$

$\det(\text{adj}(2A)) = [\det(2A)]^2$

$\det(\text{adj}(2A)) = (40)^2$

$\det(\text{adj}(2A)) = 1600$