Question

Let \( \alpha \in (0, \infty) \) and \( A = \begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} \). If \( \det(\text{adj}(2A - A^\top) \cdot \text{adj}(A - 2A^\top)) = 2^8 \), then \( (\det(A))^2 \) is equal to:

• A. 1
• B. 49
• C. 16
• D. 36

Answer 1

The Correct Option is C

Given:
\[ \det(\text{adj}(2A - A^T) \cdot \text{adj}(A - 2A^T)) = 2^8. \] 
Property of Determinants: 
For any square matrix \(B\), \(\det(\text{adj}(B)) = (\det(B))^{n-1}\) for an \(n \times n\) matrix. Since \(A\) is a \(3 \times 3\) matrix, we have \(\det(A - 2A^T) = \pm 4\), which implies \((\det(A - 2A^T))^2 = 16\). 

Matrix Calculations: 
Let \[ A - 2A^T = \begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix} - 2 \begin{bmatrix} 1 & 2 & 0 \\ 1 & 0 & 1 \\ \alpha & 1 & 2 \end{bmatrix} = \begin{bmatrix} -1 & 0 & \alpha \\ -3 & 0 & -1 \\ -2\alpha & -1 & -2 \end{bmatrix}. \] 
Equating Determinants: 
With \(\alpha = 1\), we find \(\det(A) = -4\), thus \((\det(A))^2 = 16\).