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).

Answer 2

Rather than computing \( A^2 \) and generalizing directly, let's check the characteristic behaviour of \( A \) and test each option for consistency with a rotation-matrix interpretation.

1. \( \begin{pmatrix} 1 & -2 \\ -1 & -1 \end{pmatrix} \): This matrix is not even symmetric in the way powers of \( A \) must be (since \( A \) represents a 90-degree rotation, all its even powers must be scalar multiples of the identity matrix), so this form is inconsistent with the structure of \( A \).
2. \( \begin{pmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{pmatrix} \): The matrix \( A \) represents a rotation by 90 degrees (since it sends \( (1,0) \to (0,-1) \) and \( (0,1) \to (1,0) \)). Rotating by 90 degrees twice gives a 180-degree rotation, i.e. \( A^2 = -I \). Rotating by 180 degrees \( n \) times gives either the identity \( I \) (for even \( n \)) or \( -I \) (for odd \( n \)), both of which are diagonal matrices with \( \pm 1 \) entries, matching this form exactly.
3. \( \begin{pmatrix} \pm 1 & -2n \\ \pm 1 & \pm 1 \end{pmatrix} \): This has off-diagonal entries growing with \( n \), but rotation-type matrices raised to even powers stay bounded (in fact become exactly \( \pm I \)), so growing entries don't fit.
4. \( \begin{pmatrix} -1 & n \\ -1 & -1 \end{pmatrix} \): Also has an entry that grows with \( n \) and is not diagonal, inconsistent with \( A^{2n} \) always collapsing to \( \pm I \).

Since \( A \) is a 90-degree rotation matrix, all its even powers must reduce to \( \pm \) the identity matrix.

Therefore, the correct answer is \( \begin{pmatrix} \pm 1 & 0 \\ 0 & \pm 1 \end{pmatrix} \).