Question

• A. \( \begin{pmatrix} 25 & 0 \\ 0 & 25 \end{pmatrix} \)
• B. \( \begin{pmatrix} 125 & 0 \\ 0 & 125 \end{pmatrix} \)
• C. \( \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix} \)
• D. \( \begin{pmatrix} 3125 & 0 \\ 0 & 3125 \end{pmatrix} \)

Hint:

For matrix exponentiation problems in GATE, diagonalizing the matrix using eigenvalues and eigenvectors simplifies exponentiation calculations significantly.

Answer 1

The Correct Option is C

To compute \( A^8 \), we first diagonalize \( A \).

1. Compute the eigenvalues \( \lambda \) of \( A \) from \( \det(A - \lambda I) = 0 \):

\[ \begin{vmatrix} 1 - \lambda & 2 \\ 2 & -1 - \lambda \end{vmatrix} = 0 \] Expanding: \[ (1 - \lambda)(-1 - \lambda) - (2 \times 2) = 0 \] \[ -1 - \lambda + \lambda + \lambda^2 - 4 = 0 \] \[ \lambda^2 - 5 = 0 \] \[ \lambda = \pm \sqrt{5} \]

2. The matrix \( A \) is diagonalizable as \( A = P D P^{-1} \), where:

\[ D = \begin{pmatrix} \sqrt{5} & 0 \\ 0 & -\sqrt{5} \end{pmatrix} \] Then: \[ A^8 = P D^8 P^{-1} \] Since \( D^8 = \begin{pmatrix} (\sqrt{5})^8 & 0 \\ 0 & (-\sqrt{5})^8 \end{pmatrix} = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix} \), We get: \[ A^8 = \begin{pmatrix} 625 & 0 \\ 0 & 625 \end{pmatrix} \]

Thus, the correct answer is (C).