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: Find \( A^{-1} \)

The inverse of the matrix \( A \) can be calculated as follows:

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

### Step 2: Find \( A^{-50} \)

Since \( A^{-1} \) is given, we can find \( A^{-50} \) using the property of matrix exponentiation:

\[ A^{-50} = (A^{-1})^{50} \]

Using the formula for the powers of rotation matrices, we have:

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

Thus,

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

### Step 3: Calculate \( 50\theta \)

Now we need to calculate \( 50\theta \):

\[ 50\theta = 50 \cdot \frac{\pi}{12} = \frac{50\pi}{12} = \frac{25\pi}{6} \]

### Step 4: Find \( \cos(50\theta) \) and \( \sin(50\theta) \)

To find \( \cos\left(\frac{25\pi}{6}\right) \) and \( \sin\left(\frac{25\pi}{6}\right) \), we can first reduce \( \frac{25\pi}{6} \) to an equivalent angle within \( [0, 2\pi] \):

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

Thus,

\[ \frac{25\pi}{6} \equiv \frac{\pi}{6} \quad (\text{mod } 2\pi) \]

Now we can find:

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

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

### Step 5: Substitute back into \( A^{-50} \)

Now 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 \( A^{-50} \) when \( \theta = \frac{\pi}{12} \) is:

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