Question

For the matrix \( A = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \), the matrix \( A^{2n} \) has the form

• A. \( \begin{pmatrix} 1 & -2 \\ -1 & -1 \end{pmatrix} \)
• B. \( \begin{pmatrix} \pm1 & 0 \\ 0 & \pm1 \end{pmatrix} \)
• C. \( \begin{pmatrix} \pm1 & -2n \\ \pm1 & \pm1 \end{pmatrix} \)
• D. \( \begin{pmatrix} -1 & n \\ -1 & -1 \end{pmatrix} \)

Hint:

For matrices like \( A \) with \( A^2 = -I \), the higher powers follow a periodic pattern.

Answer 1

The Correct Option is B

Step 1: Understanding the matrix powers.
The matrix \( A \) is a 2x2 matrix with the property: \[ A^2 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}^2 = \begin{pmatrix} -1 & 0 \\ 0 & -1 \end{pmatrix} = -I \] where \( I \) is the identity matrix. Step 2: Generalizing for \( A^{2n} \).
Since \( A^2 = -I \), for any integer \( n \), we have: \[ A^{2n} = (-I)^n = (-1)^n I \] Thus, \( A^{2n} \) is either \( I \) or \( -I \), depending on whether \( n \) is even or odd. Step 3: Conclusion.
The matrix \( A^{2n} \) has the form \( \begin{pmatrix} \pm1 & 0 \\ 0 & \pm1 \end{pmatrix} \). The correct answer is (2).