If `A=[[cos theta , sin theta],[sin theta,-costheta]], B = [[1,0],[-1,1]], C=ABA^(T),` then
`A^(T) C^(n) A, n in I^(+)` equals to

Options

A. `[[-n,1],[1,0]]`
B. `[[1,-n],[0,1]]`
C. `[[0,1],[1,-n]]`
D. `[[1,0],[-n,1]]`

Explanation

To solve the problem step by step, we will follow the reasoning presented in the video transcript, ensuring clarity and detail at each stage.

### Step 1: Define the Matrices
We have the matrices:
\[ A = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}, \quad B = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \]

### Step 2: Calculate \( A^T \)
The transpose of matrix \( A \) is:
\[ A^T = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix}^T = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \]
Since \( A \) is symmetric, \( A^T = A \).

### Step 3: Calculate \( C = A B A^T \)
Now we compute \( C \):
\[ C = A B A^T \]
First, calculate \( AB \):
\[ AB = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} = \begin{bmatrix} \cos \theta - \sin \theta & \sin \theta \\ \sin \theta + \cos \theta & -\cos \theta \end{bmatrix} \]

Next, we multiply this result by \( A^T \):
\[ C = \begin{bmatrix} \cos \theta - \sin \theta & \sin \theta \\ \sin \theta + \cos \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \]

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

Thus, we have:
\[ C = \begin{bmatrix} 1 - \sin \theta \cos \theta & -\sin^2 \theta \\ \cos^2 \theta & 1 + \sin \theta \cos \theta \end{bmatrix} \]

### Step 4: Calculate \( A^T C^n A \)
We need to find \( A^T C^n A \). Since \( A^T = A \), we can write:
\[ A^T C^n A = A C^n A \]

Using the property \( A^T C = B A^T \):
\[ A C = B A \]
Thus, we can express \( A C^n A \) recursively:
\[ A C^n A = A (C^{n-1} C) A = A C^{n-1} (B A) \]
Continuing this process, we find:
\[ A C^n A = B^n A \]

### Step 5: Calculate \( B^n A \)
Now we compute \( B^n \):
\[ B^n = \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix}^n = \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \]
Thus,
\[ A^T C^n A = B^n A = \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \]

Calculating this product:
1. First row:
\[ \begin{bmatrix} 1 \cdot \cos \theta + 0 \cdot \sin \theta & 1 \cdot \sin \theta + 0 \cdot (-\cos \theta) \end{bmatrix} = \begin{bmatrix} \cos \theta & \sin \theta \end{bmatrix} \]
2. Second row:
\[ \begin{bmatrix} -n \cdot \cos \theta + 1 \cdot \sin \theta & -n \cdot \sin \theta + 1 \cdot (-\cos \theta) \end{bmatrix} = \begin{bmatrix} \sin \theta - n \cos \theta & -n \sin \theta - \cos \theta \end{bmatrix} \]

Thus, the final result is:
\[ A^T C^n A = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta - n \cos \theta & -n \sin \theta - \cos \theta \end{bmatrix} \]

### Final Answer
The final expression for \( A^T C^n A \) is:
\[ \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \]