Question

Let A be a \( 2 \times 2 \) matrix with real entries such that \( A^T = \alpha A + I \), where \( \alpha \in \mathbb{R} \setminus \{-1, 1\} \). If \( \text{det}(A^2 - A) = 4 \), then the sum of all possible values of \( \alpha \) is equal to:

• A. 0
• B. \( \frac{5}{2} \)
• C. 2
• D. \( \frac{3}{2}

Hint:

In problems involving matrix determinants and operations, it's important to break down the matrix expressions step by step and apply algebraic operations correctly.

Answer 1

The Correct Option is B

We are given that: \[ A^T = \alpha A + I \quad \text{and} \quad \text{det}(A^2 - A) = 4 \] We start by simplifying the expression for \( A^2 - A \). First, express \( A^T \) as: \[ A^T = \alpha A + I \] Thus, we have: \[ A = \alpha(A + I) + I \] \[ A = \alpha A + (\alpha + 1)I \] Now, calculate the determinant \( |A - I| \). From the above, we know that: \[ |A - I| = \frac{1}{(1 - \alpha^2)} \quad \text{(Equation 3)} \] Next, from the equation \( \text{det}(A^2 - A) \), we have: \[ A^2 - A = \left|A - I\right| \] Substituting the value, we find that the determinant of \( A^2 - A \) is 4: \[ \text{det}(A^2 - A) = 4 \] After solving the quadratic equation, we find that the sum of possible values of \( \alpha \) is \( \frac{5}{2} \).