Question

Given the matrix \( A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \), find \( A^8 \).

• A. \( 625I \)
• B. \( 625A \)
• C. \( I \)
• D. \( 25I \)

Hint:

To find high powers of matrices, use eigenvalue decomposition when possible. This simplifies the computation, especially for diagonalizable matrices.

Answer 1

The Correct Option is A

Step 1: Finding the eigenvalues and eigenvectors of matrix \( A \).

The matrix \( A \) is:

\[ A = \begin{bmatrix} 1 & 2 \\ 2 & -1 \end{bmatrix} \]

The characteristic equation of \( A \) is given by:

\[ \det(A - \lambda I) = 0 \] \[ \begin{vmatrix} 1 - \lambda & 2 \\ 2 & -1 - \lambda \end{vmatrix} = 0 \] \[ (1 - \lambda)(-1 - \lambda) - 4 = 0 \] \[ \lambda^2 - 2 = 0 \quad \Rightarrow \quad \lambda = \pm \sqrt{2} \]

Step 2: Using the properties of eigenvalues.

For a matrix \( A \) with eigenvalues \( \lambda_1 \) and \( \lambda_2 \), the powers of \( A \) can be expressed in terms of its eigenvalues as:

\[ A^n = P \begin{bmatrix} \lambda_1^n & 0 \\ 0 & \lambda_2^n \end{bmatrix} P^{-1} \]

Since \( \lambda_1 = \sqrt{2} \) and \( \lambda_2 = -\sqrt{2} \), we have:

\[ A^8 = P \begin{bmatrix} (\sqrt{2})^8 & 0 \\ 0 & (-\sqrt{2})^8 \end{bmatrix} P^{-1} = P \begin{bmatrix} 256 & 0 \\ 0 & 256 \end{bmatrix} P^{-1} \] \[ A^8 = 256I \]

Thus, the correct answer is \( 256I \).