Matrix Determinant and Adjugate Calculation

This solution provides the steps to calculate the determinant of the adjugate of matrix A, given information about matrix B and its determinant.

1. Calculate Determinant of A

First, find the determinant of matrix $A = \begin{bmatrix} \alpha & 1 & 2 \\ 2 & 3 & 0 \\ 0 & 4 & 5 \end{bmatrix}$.

Using cofactor expansion along the first row:

$ \det(A) = \alpha \begin{vmatrix} 3 & 0 \\ 4 & 5 \end{vmatrix} - 1 \begin{vmatrix} 2 & 0 \\ 0 & 5 \end{vmatrix} + 2 \begin{vmatrix} 2 & 3 \\ 0 & 4 \end{vmatrix} $

$ \det(A) = \alpha(15 - 0) - 1(10 - 0) + 2(8 - 0) $

$ \det(A) = 15\alpha - 10 + 16 $

$ \det(A) = 15\alpha + 6 $

2. Determinant of Adjugate Formula

The determinant of the adjugate matrix is related to the determinant of the original matrix by the formula:

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

For a $3 \times 3$ matrix ($n=3$), this simplifies to:

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

3. Construct and Evaluate Determinant of B

Matrix $B$ is defined as $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$.

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

Adding the matrices to find B:

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

Calculate the determinant of B:

$ \det(B) = 16 \begin{vmatrix} 0 & 4 \\ 0 & \alpha - 2 \end{vmatrix} - 3 \begin{vmatrix} -10 & 4 \\ 8 & \alpha - 2 \end{vmatrix} + (-6) \begin{vmatrix} -10 & 0 \\ 8 & 0 \end{vmatrix} $

$ \det(B) = 16(0) - 3((-10)(\alpha - 2) - 32) - 6(0) $

$ \det(B) = -3(-10\alpha + 20 - 32) = -3(-10\alpha - 12) $

$ \det(B) = 30\alpha + 36 $

4. Solve for Alpha

Given that $\det(B) = 66$, we set the expression equal to 66:

$ 30\alpha + 36 = 66 $

Solving for $\alpha$:

$ 30\alpha = 30 $

$ \alpha = 1 $

5. Final Calculation of Det(adj(A))

Substitute the value $\alpha = 1$ back into the expression for $\det(A)$:

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

Now, apply the formula for the determinant of the adjugate matrix:

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

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