If A=`[[cos theta,sin theta],[-sin theta,cos theta]], ` then `lim _(n rarr infty )A^(n)/n ` is (where `theta in R`)

Options

A. a zero matrix
B. an identity matrix
C. `[[0,1],[-1,0]]`
D. `[[0,1],[0,-1]]`

Explanation

To solve the problem, we need to find the limit of the matrix \( A^n/n \) as \( n \) approaches infinity, where \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \).

### Step-by-Step Solution:

1. **Understanding the Matrix \( A \)**:
The matrix \( A \) is a rotation matrix. It represents a rotation by an angle \( \theta \) in the counter-clockwise direction.

2. **Finding \( A^2 \)**:
To find \( A^2 \), we multiply \( A \) by itself:
\[ A^2 = A \cdot A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \cdot \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]
Performing the multiplication:
\[ A^2 = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2 \sin \theta \cos \theta \\ -2 \sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{bmatrix} = \begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ -\sin(2\theta) & \cos(2\theta) \end{bmatrix} \]

3. **Finding \( A^n \)**:
By induction or using the properties of rotation matrices, we can derive that:
\[ A^n = \begin{bmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{bmatrix} \]

4. **Calculating \( \frac{A^n}{n} \)**:
Now, we need to divide each element of \( A^n \) by \( n \):
\[ \frac{A^n}{n} = \begin{bmatrix} \frac{\cos(n\theta)}{n} & \frac{\sin(n\theta)}{n} \\ -\frac{\sin(n\theta)}{n} & \frac{\cos(n\theta)}{n} \end{bmatrix} \]

5. **Taking the Limit as \( n \to \infty \)**:
We take the limit of each element in the matrix:
\[ \lim_{n \to \infty} \frac{A^n}{n} = \begin{bmatrix} \lim_{n \to \infty} \frac{\cos(n\theta)}{n} & \lim_{n \to \infty} \frac{\sin(n\theta)}{n} \\ \lim_{n \to \infty} -\frac{\sin(n\theta)}{n} & \lim_{n \to \infty} \frac{\cos(n\theta)}{n} \end{bmatrix} \]
Since \( \cos(n\theta) \) and \( \sin(n\theta) \) are bounded between -1 and 1, we have:
\[ \lim_{n \to \infty} \frac{\cos(n\theta)}{n} = 0 \quad \text{and} \quad \lim_{n \to \infty} \frac{\sin(n\theta)}{n} = 0 \]
Thus, the limit of the matrix becomes:
\[ \lim_{n \to \infty} \frac{A^n}{n} = \begin{bmatrix} 0 & 0 \\ 0 & 0 \end{bmatrix} \]

6. **Conclusion**:
Therefore, the final answer is:
\[ \lim_{n \to \infty} \frac{A^n}{n} = \mathbf{0} \]