If `A = [(costheta,-sintheta),(sintheta,costheta)]`, then the matrix `A^(-50)`, when `theta = (pi)/(12)`, is equal to

Options

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

Answer

To solve the problem, we need to find the matrix \( A^{-50} \) where

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

and \( \theta = \frac{\pi}{12} \).

### Step 1: Verify that A is an Orthogonal Matrix

First, we need to check if \( A \) is an orthogonal matrix. We can do this by calculating \( A A^T \) (where \( A^T \) is the transpose of \( A \)):

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

Now, we compute \( A A^T \):

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

Calculating the elements:

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

Thus,

\[ A A^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = I \]

This confirms that \( A \) is an orthogonal matrix, and therefore \( A^{-1} = A^T \).

### Step 2: Find \( A^{-1} \)

From the property of orthogonal matrices, we have:

\[ A^{-1} = A^T = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

### Step 3: General Formula for \( A^{-n} \)

Using the properties of the matrix, we can derive that:

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

### Step 4: Calculate \( A^{-50} \)

Now, we need to compute \( A^{-50} \):

\[ A^{-50} = \begin{pmatrix} \cos(50\theta) & \sin(50\theta) \\ -\sin(50\theta) & \cos(50\theta) \end{pmatrix} \]

Substituting \( \theta = \frac{\pi}{12} \):

\[ A^{-50} = \begin{pmatrix} \cos\left(50 \cdot \frac{\pi}{12}\right) & \sin\left(50 \cdot \frac{\pi}{12}\right) \\ -\sin\left(50 \cdot \frac{\pi}{12}\right) & \cos\left(50 \cdot \frac{\pi}{12}\right) \end{pmatrix} \]

Calculating \( 50 \cdot \frac{\pi}{12} = \frac{50\pi}{12} = \frac{25\pi}{6} \).

### Step 5: Simplify \( \cos\left(\frac{25\pi}{6}\right) \) and \( \sin\left(\frac{25\pi}{6}\right) \)

To simplify these trigonometric functions, we can reduce \( \frac{25\pi}{6} \):

\[ \frac{25\pi}{6} = 4\pi + \frac{\pi}{6} \quad (\text{since } 4\pi = 24\pi/6) \]

Thus,

\[ \cos\left(\frac{25\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \]
\[ \sin\left(\frac{25\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \]

### Step 6: Write the Final Matrix

Substituting these values back into the matrix:

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

### Final Answer

Thus, the matrix \( A^{-50} \) is:

\[ \begin{pmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{pmatrix} \]