Given:
\[ \det(\text{adj}(2A - A^T) \cdot \text{adj}(A - 2A^T)) = 2^8. \]
Recall the Property of Determinants:
For any square matrix \(B\), we have:
\[ \det(\text{adj}(B)) = (\det(B))^{n-1} \quad \text{for an \(n \times n\) matrix.} \]
Since \(A\) is a \(3 \times 3\) matrix, we consider:
\[ \det(A - 2A^T) = \pm 4, \quad (\det(A - 2A^T))^2 = 16. \]
Matrix Calculations:
Consider:
\[ A - 2A^T = \begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ \alpha & 1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2\alpha & -1 & -2 \end{bmatrix}. \]
Equating Determinants:
Given \(\alpha = 1\), the determinant becomes: \[ \det(A) = -4, \quad (\det(A))^2 = 16. \]
The portion of the line \( 4x + 5y = 20 \) in the first quadrant is trisected by the lines \( L_1 \) and \( L_2 \) passing through the origin. The tangent of an angle between the lines \( L_1 \) and \( L_2 \) is: