Let $$A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$$ and $$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$$. If $$\det(B) = 66$$, then $$\det(\text{adj}(A))$$ equals :

$$\det(\text{adj}(A)) = (\det(A))^{n-1}$$

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

$$B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \begin{bmatrix} 15 & 3 & -6 \\ -10 & 5\alpha & 4 \\ 8 & -4\alpha & 3\alpha-2 \end{bmatrix}$$

$$B = \begin{bmatrix} 16 & 3 & -6 \\ -10 & 0 & 4 \\ 8 & 0 & \alpha-2 \end{bmatrix}$$

$$\det(B) = -3 \left[ -10\alpha + 20 - 32 \right] = -3(-10\alpha - 12) = 30\alpha + 36$$

$$30\alpha + 36 = 66 \implies 30\alpha = 30 \implies \alpha = 1$$

$$\det(A) = 15(1) + 6 = 21$$

$$\det(\text{adj}(A)) = 21^2 = 441$$