Question

Let $ A = \begin{bmatrix} \alpha & -1 \\6 & \beta \end{bmatrix},\ \alpha > 0 $, such that $ \det(A) = 0 $ and $ \alpha + \beta = 1 $. If $ I $ denotes the $ 2 \times 2 $ identity matrix, then the matrix $ (1 + A)^5 $ is:

• A. \( \begin{bmatrix} 4 & -1 \\6 & -1 \end{bmatrix} \)
• B. \( \begin{bmatrix} 257 & -64 \\514 & -127 \end{bmatrix} \)
• C. \( \begin{bmatrix} 1025 & -511 \\2024 & -1024 \end{bmatrix} \)
• D. \( \begin{bmatrix} 766 & -255 \\1530 & -509 \end{bmatrix} \)

Hint:

When a matrix satisfies \( A^2 = A \), powers simplify: \( A^n = A \). Use this in binomial expansions.

Answer 1

The Correct Option is D

From \( \det(A) = \alpha \beta + 6 = 0 \), we get \( \alpha \beta = -6 \), and \( \alpha + \beta = 1 \). Solving: \[ \alpha = 3,\quad \beta = -2 \Rightarrow A = \begin{bmatrix} 3 & -1 6 & -2 \end{bmatrix} \] Check powers: \[ A^2 = A \Rightarrow A^n = A,\ \forall n \geq 1 \] Use binomial expansion: \[ (1 + A)^5 = I + 5A + 10A^2 + 10A^3 + 5A^4 + A^5 = I + 31A \] \[ (1 + A)^5 = \begin{bmatrix} 1 & 0 0 & 1 \end{bmatrix} + 31 \cdot \begin{bmatrix} 3 & -1 \\6 & -2 \end{bmatrix} = \begin{bmatrix} 766 & -255 \\1530 & -509 \end{bmatrix} \]