Question

Sol. [Key Concept: $det(B) = 66$ fixes $\alpha = 1$ so $|A| = 21$; then $det(adj\ A) = {|A|}^{(n{- 1)}^{2}}{= 21}^{2} = 441$.]

$$\ A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}\&|\text{ }A| = 5\alpha + 6$$

Now adj $A = \begin{bmatrix} 15 & 3 & - 6 \\ - 10 & 5\alpha & 4 \\ 8 & - 4\alpha & 3\alpha - 2 \end{bmatrix}$

$$\therefore B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & - 5\alpha & 0 \\ 0 & 4\alpha & - 2\alpha \end{bmatrix} + adj(A)$$

$B = \begin{bmatrix} 16 & 3 & - 6 \\ - 10 & 0 & 4 \\ 8 & 0 & \alpha - 2 \end{bmatrix}\ \ \&\ |B| = 6(5\alpha + 6) = 66$ (given in question)

$$\Rightarrow \alpha = 1$$

$$\Rightarrow |A| = 21$$

$$\therefore det(adjA) = |A|^{(n - 1)^{2}}$$

Here $n = 3$

$$= 21^{2}$$

$$= 441 $$