If `A_(alpha), = [{:(cos alpha , sin alpha),(-sin alpha , cos alpha):}]` , then

Options

A. `(A_(alpha))^(n) = [{:(-cos n alpha , cos n alpha),(-sin n alpha , sin n alpha):}]`
B. `A_(alpha) A_(beta) = A_(alpha + beta)`
C. `(A_(alpha))^(n) = [{:(-cos n alpha , sin n alpha),(-sin n alpha , cos n alpha):}]`
D. `A_(alpha) A_(beta) = - A_(alpha + beta)`

Answer

To solve the problem involving the matrix \( A_\alpha = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \), we will find \( A_\alpha^n \) and the product \( A_\alpha A_\beta \).

### Step 1: Calculate \( A_\alpha^2 \)

To find \( A_\alpha^2 \), we multiply \( A_\alpha \) by itself:

\[ A_\alpha^2 = A_\alpha \cdot A_\alpha = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \]

Calculating the elements:

1. First row, first column:
\[ \cos \alpha \cdot \cos \alpha + \sin \alpha \cdot (-\sin \alpha) = \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha \]

2. First row, second column:
\[ \cos \alpha \cdot \sin \alpha + \sin \alpha \cdot \cos \alpha = 2 \cos \alpha \sin \alpha = \sin 2\alpha \]

3. Second row, first column:
\[ -\sin \alpha \cdot \cos \alpha + \cos \alpha \cdot (-\sin \alpha) = -2 \sin \alpha \cos \alpha = -\sin 2\alpha \]

4. Second row, second column:
\[ -\sin \alpha \cdot \sin \alpha + \cos \alpha \cdot \cos \alpha = \cos^2 \alpha - \sin^2 \alpha = \cos 2\alpha \]

Thus, we have:
\[ A_\alpha^2 = \begin{pmatrix} \cos 2\alpha & \sin 2\alpha \\ -\sin 2\alpha & \cos 2\alpha \end{pmatrix} \]

### Step 2: Generalize \( A_\alpha^n \)

From the pattern observed, we can generalize:
\[ A_\alpha^n = \begin{pmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{pmatrix} \]

### Step 3: Calculate \( A_\alpha A_\beta \)

Next, we calculate the product \( A_\alpha A_\beta \):
\[ A_\alpha A_\beta = \begin{pmatrix} \cos \alpha & \sin \alpha \\ -\sin \alpha & \cos \alpha \end{pmatrix} \begin{pmatrix} \cos \beta & \sin \beta \\ -\sin \beta & \cos \beta \end{pmatrix} \]

Calculating the elements:

1. First row, first column:
\[ \cos \alpha \cdot \cos \beta + \sin \alpha \cdot (-\sin \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta) \]

2. First row, second column:
\[ \cos \alpha \cdot \sin \beta + \sin \alpha \cdot \cos \beta = \cos \alpha \sin \beta + \sin \alpha \cos \beta = \sin(\alpha + \beta) \]

3. Second row, first column:
\[ -\sin \alpha \cdot \cos \beta + \cos \alpha \cdot (-\sin \beta) = -\sin \alpha \cos \beta - \cos \alpha \sin \beta = -\sin(\alpha + \beta) \]

4. Second row, second column:
\[ -\sin \alpha \cdot \sin \beta + \cos \alpha \cdot \cos \beta = \cos \alpha \cos \beta - \sin \alpha \sin \beta = \cos(\alpha + \beta) \]

Thus, we have:
\[ A_\alpha A_\beta = \begin{pmatrix} \cos(\alpha + \beta) & \sin(\alpha + \beta) \\ -\sin(\alpha + \beta) & \cos(\alpha + \beta) \end{pmatrix} \]

### Conclusion

The results we derived are:
1. \( A_\alpha^n = \begin{pmatrix} \cos n\alpha & \sin n\alpha \\ -\sin n\alpha & \cos n\alpha \end{pmatrix} \)
2. \( A_\alpha A_\beta = A_{\alpha + \beta} \)