If `A = [{:("cos" theta, -"sin"theta),("sin"theta, "cos"theta):}]`, then the matrix `A^(-50) " when " theta = (pi)/(12)`, is equal to

Options

A. `[(1/2, (sqrt(3))/2),(-(sqrt(3))/2,1/2)]`
B. `[((sqrt(3))/2,-1/2),(1/2,(sqrt(3))/2)]`
C. `[(1/2,(sqrt(3))/2),(1/2,1/2)]`
D. `[((sqrt(3))/2,1/2),(-1/2,(sqrt(3))/2)]`

Explanation

To solve the problem, we need to find the matrix \( A^{-50} \) when \( A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \) and \( \theta = \frac{\pi}{12} \).

### Step-by-step Solution:

1. **Define the Matrix A**:
\[ A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

2. **Find the Inverse of A**:
The inverse of a rotation matrix \( A \) is given by:
\[ A^{-1} = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]
Since \( A \) is an orthogonal matrix, we have \( A^{-1} = A^T \).

3. **Calculate \( A^{-50} \)**:
We know that:
\[ A^{-n} = A^{-1} \cdot A^{-1} \cdots A^{-1} \quad (n \text{ times}) \]
Therefore:
\[ A^{-50} = (A^{-1})^{50} \]

4. **Express \( A^{-n} \)**:
From the properties of rotation matrices, we have:
\[ A^{-n} = \begin{pmatrix} \cos(n\theta) & \sin(n\theta) \\ -\sin(n\theta) & \cos(n\theta) \end{pmatrix} \]
Thus:
\[ A^{-50} = \begin{pmatrix} \cos(50\theta) & \sin(50\theta) \\ -\sin(50\theta) & \cos(50\theta) \end{pmatrix} \]

5. **Substitute \( \theta = \frac{\pi}{12} \)**:
We need to calculate \( 50\theta \):
\[ 50\theta = 50 \cdot \frac{\pi}{12} = \frac{50\pi}{12} = \frac{25\pi}{6} \]

6. **Find \( \cos\left(\frac{25\pi}{6}\right) \) and \( \sin\left(\frac{25\pi}{6}\right) \)**:
To simplify \( \frac{25\pi}{6} \):
\[ \frac{25\pi}{6} = 4\pi + \frac{\pi}{6} \quad \text{(since } 4\pi = \frac{24\pi}{6}\text{)} \]
Thus:
\[ \cos\left(\frac{25\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
\[ \sin\left(\frac{25\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

7. **Write the Final Matrix**:
Now substituting back into our expression for \( A^{-50} \):
\[ A^{-50} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]

### Final Answer:
\[ A^{-50} = \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]