Question

Let \( A = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} \), where \( \theta \in \mathbb{R} \). Then \( A^n \), for any positive integer \( n \), is equal to:

• A. \( \begin{bmatrix} \cos n\theta & -\sin n\theta \\ \sin n\theta & \cos n\theta \end{bmatrix} \)
• B. \( \begin{bmatrix} \cos\theta^n & -\sin\theta^n \\ \sin\theta^n & \cos\theta^n \end{bmatrix} \)
• C. \( n \cdot A \)
• D. None of these

Hint:

The \( n \)-th power of a 2D rotation matrix is equivalent to a rotation by \( n\theta \), not \( \theta^n \) or scalar multiplication.

Answer 1

The Correct Option is A

Step 1: Recognize the structure of the matrix.
The matrix \( A = \begin{bmatrix} \cos\theta & -\sin\theta
\sin\theta & \cos\theta \end{bmatrix} \) is a standard rotation matrix.
It represents a counterclockwise rotation by angle \( \theta \) in the 2D plane. Step 2: Apply matrix powers.
The matrix \( A^n \) corresponds to rotating by \( \theta \) \( n \)-times, i.e., a total rotation of \( n\theta \).
Hence, \[ A^n = \begin{bmatrix} \cos(n\theta) & -\sin(n\theta) \\ \sin(n\theta) & \cos(n\theta) \end{bmatrix} \] Step 3: Eliminate incorrect options.
- Option (2): Incorrect because \( \cos(\theta^n) \neq \cos(n\theta) \) in general.
- Option (3): Incorrect as matrix powers are not the same as scalar multiplication.
- Option (4): Incorrect since Option (1) is valid. \[ \boxed{A^n = \begin{bmatrix} \cos(n\theta) & -\sin(n\theta)
\sin(n\theta) & \cos(n\theta) \end{bmatrix}} \]