`If A=[[costheta,sintheta],[-sintheta,costheta]],then Lim_(x_>oo)1/nA^n` is

Options

A. (A) an identity matrix
B. (B) [0 10 -1 0 ]
C. (C) a null matrix
D. (D) none of these

Explanation

To solve the problem, we need to find the limit of the expression \( \lim_{n \to \infty} \frac{1}{n} A^n \) where \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \).

### Step 1: Calculate \( A^2 \)

First, we calculate \( A^2 \) by multiplying \( A \) by itself:

\[ A^2 = A \cdot A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \cdot \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

Calculating the elements:

- First row, first column: \( \cos^2 \theta + \sin^2 \theta = 1 \)
- First row, second column: \( \cos \theta \sin \theta + \sin \theta \cos \theta = 2 \cos \theta \sin \theta \)
- Second row, first column: \( -\sin \theta \cos \theta + \cos \theta \sin \theta = 0 \)
- Second row, second column: \( -\sin^2 \theta + \cos^2 \theta = \cos(2\theta) \)

Thus, we have:

\[ A^2 = \begin{pmatrix} \cos(2\theta) & \sin(2\theta) \\ -\sin(2\theta) & \cos(2\theta) \end{pmatrix} \]

### Step 2: Generalize \( A^n \)

By observing the pattern from \( A \) and \( A^2 \), we can generalize \( A^n \):

\[ A^n = \begin{pmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{pmatrix} \]

### Step 3: Calculate \( \frac{1}{n} A^n \)

Now, we compute \( \frac{1}{n} A^n \):

\[ \frac{1}{n} A^n = \frac{1}{n} \begin{pmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{pmatrix} = \begin{pmatrix} \frac{\cos(n\theta)}{n} & \frac{\sin(n\theta)}{n} \\ -\frac{\sin(n\theta)}{n} & \frac{\cos(n\theta)}{n} \end{pmatrix} \]

### Step 4: Take the limit as \( n \to \infty \)

Now we take the limit of each element as \( n \to \infty \):

\[ \lim_{n \to \infty} \frac{1}{n} A^n = \begin{pmatrix} \lim_{n \to \infty} \frac{\cos(n\theta)}{n} & \lim_{n \to \infty} \frac{\sin(n\theta)}{n} \\ \lim_{n \to \infty} -\frac{\sin(n\theta)}{n} & \lim_{n \to \infty} \frac{\cos(n\theta)}{n} \end{pmatrix} \]

Since both \( \cos(n\theta) \) 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 becomes:

\[ \lim_{n \to \infty} \frac{1}{n} A^n = \begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix} \]

### Final Result

The final result is:

\[ \lim_{n \to \infty} \frac{1}{n} A^n = \text{Zero Matrix} \]