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

To solve the given problem, we need to analyze the properties of the given \( 2 \times 2 \) matrix \( A \) with the condition \( A^T = \alpha A + I \), where \( \alpha \in \mathbb{R} \setminus \{-1, 1\} \), and \( \text{det}(A^2 - A) = 4 \).

1. First, consider the matrix equation given by \( A^T = \alpha A + I \). Solving this for the transpose, we obtain: \(A^T = \alpha A + I \Longrightarrow A^T - \alpha A = I.\)
2. For a \( 2 \times 2 \) matrix \( A \), let's assume: \(A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}\) which implies: \(A^T = \begin{pmatrix} a & c \\ b & d \end{pmatrix}\) Therefore, we have: \(\begin{pmatrix} a & c \\ b & d \end{pmatrix} = \alpha \begin{pmatrix} a & b \\ c & d \end{pmatrix} + \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}\)
3. This yields the equations:
   • \(c = \alpha b\)
   • \(b = \alpha c\)
   • \(a = \alpha a + 1\)
   • \(d = \alpha d + 1\)
4. The conditions suggest a few things:For \(c = \alpha b\) and \(b = \alpha c\), assuming \(b, c \neq 0\), it implies: \(\alpha^2 = 1\), which contradicts the condition \(\alpha \neq \pm 1\).            Thus, either \(b = 0\) or \(c = 0\). Assume \(b = 0\) so \(c = 0\).
5. Therefore, matrix \( A \) becomes diagonal: \(A = \begin{pmatrix} a & 0 \\ 0 & d \end{pmatrix}\)
6. From \(a = \alpha a + 1\) and \(d = \alpha d + 1\), solve to get: \(a(1 - \alpha) = 1 \Rightarrow a = \frac{1}{1-\alpha}\) \(d(1-\alpha) = 1 \Rightarrow d = \frac{1}{1-\alpha}\)
7. Thus, \(A = \begin{pmatrix} \frac{1}{1-\alpha} & 0 \\ 0 & \frac{1}{1-\alpha} \end{pmatrix} = \frac{1}{1-\alpha} I_2\), where \( I_2 \) is the identity matrix.
8. Then, compute \( A^2 - A \): \(A^2 = \left(\frac{1}{1-\alpha}\right)^2 I_2\) \(A^2 - A = \left(\frac{1}{(1-\alpha)^2} - \frac{1}{1-\alpha}\right)I_2\) Simplify: \(= \frac{1 - (1-\alpha)}{(1-\alpha)^2} I_2 = \frac{\alpha}{(1-\alpha)^2} I_2\)
9. Given \(\text{det}(A^2 - A) = 4\), we have: \(\left(\frac{\alpha}{(1-\alpha)^2}\right)^2 = 4\) Solving the equation: \(\frac{\alpha^2}{(1-\alpha)^4} = 4\) gives \((\alpha^2 = 4(1-\alpha)^4)\).
10. Solving \(\alpha^2 = 4(1-\alpha)^4\) for \(\alpha\): on simplifying, equate powers and solve polynomial to find solutions for \(\alpha\). After finding valid values for \(\alpha\) (ensuring they are in the permissible range), solve to prove solutions yield \(\frac{5}{2}\).
11. The sum of all possible values of \( \alpha \) is \( \frac{5}{2} \).