if `A = [[cos theta, sin theta],[-sin theta, cos theta]]` where `theta = (2pi)/(7)` then `sum_(r=1)^(6) A^(r)` =

Options

A. `[[1,0],[0,1]]`
B. `[[3,0],[0,3]]`
C. `[[-1,0],[0,-1]]`
D. None of these

Explanation

To solve the problem, we need to calculate the sum \( \sum_{r=1}^{6} A^r \) where \( A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \) and \( \theta = \frac{2\pi}{7} \).

### Step-by-Step Solution:

1. **Understanding the Matrix A**:
The matrix \( A \) represents a rotation matrix. The angle of rotation is \( \theta = \frac{2\pi}{7} \).

2. **Finding Powers of A**:
We will first compute \( A^2 \):
\[ A^2 = A \cdot A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]
Using matrix multiplication:
\[ A^2 = \begin{bmatrix} \cos^2 \theta - \sin^2 \theta & 2\sin \theta \cos \theta \\ -2\sin \theta \cos \theta & \cos^2 \theta - \sin^2 \theta \end{bmatrix} \]
Recognizing that \( \cos^2 \theta - \sin^2 \theta = \cos(2\theta) \) and \( 2\sin \theta \cos \theta = \sin(2\theta) \):
\[ A^2 = \begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ -\sin(2\theta) & \cos(2\theta) \end{bmatrix} \]

3. **Continuing to Higher Powers**:
By continuing this process, we find:
\[ A^3 = A^2 \cdot A = \begin{bmatrix} \cos(2\theta) & \sin(2\theta) \\ -\sin(2\theta) & \cos(2\theta) \end{bmatrix} \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \]
This results in:
\[ A^3 = \begin{bmatrix} \cos(3\theta) & \sin(3\theta) \\ -\sin(3\theta) & \cos(3\theta) \end{bmatrix} \]
Continuing this way, we find:
\[ A^r = \begin{bmatrix} \cos(r\theta) & \sin(r\theta) \\ -\sin(r\theta) & \cos(r\theta) \end{bmatrix} \]

4. **Summing the Powers**:
Now we need to compute:
\[ \sum_{r=1}^{6} A^r = A + A^2 + A^3 + A^4 + A^5 + A^6 \]
This can be expressed as:
\[ \sum_{r=1}^{6} A^r = \begin{bmatrix} \sum_{r=1}^{6} \cos(r\theta) & \sum_{r=1}^{6} \sin(r\theta) \\ -\sum_{r=1}^{6} \sin(r\theta) & \sum_{r=1}^{6} \cos(r\theta) \end{bmatrix} \]

5. **Using Properties of Roots of Unity**:
The sums \( \sum_{r=0}^{6} \cos(r\theta) \) and \( \sum_{r=0}^{6} \sin(r\theta) \) correspond to the real and imaginary parts of the 7th roots of unity, which sum to zero:
\[ \sum_{r=0}^{6} \cos\left(\frac{2\pi r}{7}\right) = 0 \quad \text{and} \quad \sum_{r=0}^{6} \sin\left(\frac{2\pi r}{7}\right) = 0 \]
Therefore:
\[ \sum_{r=1}^{6} \cos(r\theta) = -1 \quad \text{and} \quad \sum_{r=1}^{6} \sin(r\theta) = 0 \]

6. **Final Result**:
Substituting these sums back into our matrix:
\[ \sum_{r=1}^{6} A^r = \begin{bmatrix} -1 & 0 \\ 0 & -1 \end{bmatrix} \]

### Conclusion:
Thus, the final result is:
\[ \sum_{r=1}^{6} A^r = -I \]
where \( I \) is the identity matrix.