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

Answer: (C)

16

Answer 2

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: $\det(A) = -4, \quad (\det(A))^2 = 16.$