If $A =\begin{bmatrix}a&1&-1\\0&b&4\\4&4&c\end{bmatrix}$ and $abc = 12, b = 4a$, then the value of $\text{|A (adjA)|}$ is:

$28^3$

The correct answer is Option (1) → $28^3$

Given matrix: $A = \begin{bmatrix} a & 1 & -1 \\ 0 & b & 4 \\ 4 & 4 & c \end{bmatrix}$

Given: $abc = 12$, $b = 4a$

Property: $A \cdot \text{adj}(A) = |A| I \Rightarrow |A \cdot \text{adj}(A)| = |A|^3$ for a 3×3 matrix

Compute $|A|$:

$|A| = a \begin{vmatrix} b & 4 \\ 4 & c \end{vmatrix} - 1 \begin{vmatrix} 0 & 4 \\ 4 & c \end{vmatrix} + (-1) \begin{vmatrix} 0 & b \\ 4 & 4 \end{vmatrix}$

So, $|A| = a(bc-16) -1(-16) + (-1)(-4b) = a(bc-16) + 16 +4b = a(bc-16) +16 +4b$

Substitute $b=4a$ and $c = \frac{12}{ab} = \frac{12}{a*4a} = \frac{12}{4a^2} = \frac{3}{a^2}$

Then $bc = 4a * (3/a^2) = 12/a$

$|A| = a(12/a -16) +16 - 4*4a = (12 -16a) +16 +16a = 28 $

Then $|A \cdot \text{adj}(A)| = |A|^3 = (28 )^3$