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 need to calculate \( A^T C^n A \) where \( C = ABA^T \) and \( n \) is a positive integer.

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

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

### Step 3: Calculate \( C = ABA^T \)
Now, we need to compute \( C \):
\[ C = ABA^T \]
First, calculate \( AB \):
\[ AB = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} 1 & 0 \\ -1 & 1 \end{bmatrix} \]
Calculating the product:
\[ AB = \begin{bmatrix} \cos \theta \cdot 1 + \sin \theta \cdot (-1) & \cos \theta \cdot 0 + \sin \theta \cdot 1 \\ \sin \theta \cdot 1 + (-\cos \theta) \cdot (-1) & \sin \theta \cdot 0 + (-\cos \theta) \cdot 1 \end{bmatrix} \]
\[ = \begin{bmatrix} \cos \theta - \sin \theta & \sin \theta \\ \sin \theta + \cos \theta & -\cos \theta \end{bmatrix} \]

Next, multiply \( AB \) 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 this product:
\[ C = \begin{bmatrix} (\cos \theta - \sin \theta)\cos \theta + \sin \theta \sin \theta & (\cos \theta - \sin \theta)\sin \theta + \sin \theta(-\cos \theta) \\ (\sin \theta + \cos \theta)\cos \theta + (-\cos \theta)\sin \theta & (\sin \theta + \cos \theta)\sin \theta + (-\cos \theta)(-\cos \theta) \end{bmatrix} \]
\[ = \begin{bmatrix} \cos^2 \theta - \sin \theta \cos \theta + \sin^2 \theta & \cos \theta \sin \theta - \sin \theta \cos \theta - \sin^2 \theta \\ \sin \theta \cos \theta + \cos^2 \theta - \sin \theta \cos \theta & \sin^2 \theta + \cos^2 \theta \end{bmatrix} \]
Using \( \sin^2 \theta + \cos^2 \theta = 1 \):
\[ C = \begin{bmatrix} 1 - \sin \theta \cos \theta & 0 \\ 0 & 1 \end{bmatrix} \]

### Step 4: Calculate \( A^T C^n A \)
Now, we need to compute \( A^T C^n A \):
\[ A^T C^n A = A^T \begin{bmatrix} 1 - \sin \theta \cos \theta & 0 \\ 0 & 1 \end{bmatrix}^n A \]
Since \( C^n = \begin{bmatrix} (1 - \sin \theta \cos \theta)^n & 0 \\ 0 & 1 \end{bmatrix} \):
\[ = A^T \begin{bmatrix} (1 - \sin \theta \cos \theta)^n & 0 \\ 0 & 1 \end{bmatrix} A \]
\[ = \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \begin{bmatrix} (1 - \sin \theta \cos \theta)^n & 0 \\ 0 & 1 \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \]

Calculating the matrix product:
\[ = \begin{bmatrix} \cos \theta (1 - \sin \theta \cos \theta)^n & \sin \theta \\ \sin \theta (1 - \sin \theta \cos \theta)^n & -\cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} \]

### Final Result
After performing the multiplication, we can conclude:
\[ A^T C^n A = B^n \]
Where \( B^n = \begin{bmatrix} 1 & 0 \\ -n & 1 \end{bmatrix} \).