If A= […] then the value of det (adj (2A)) is:

The correct answer is Option (4) → 1600 

For any scalar $k$ and $n × n$ matrix $M$:

$\text{det (kM)} = k^n × \text{det (M)}$

For any $n × n$ matrix $M$:

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

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

Expanding along Row 1:

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

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

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

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

$\text{det(A) 10+ 3 + 12}$

$\text{det(A) = 5}$

Since A is a $3 × 3$ matrix:

$\text{det (24)} = 2^3× \text{det(A)}$

$\text{det (24)} = 8 × 5$

$\text{det (24) = 40}$

Since 2A is a $3 × 3$ matrix:

$\text{det (adj(24))= [det (24)]}^{3-1}$

$\text{det (adj(24))= [det (24)]}^2$

$\text{det (adj(24)) = (40)}^2$

$\text{det (adj(24)) = 1600}$