Let `A_(alpha)=[(cosalpha, -sinalpha,0),(sinalpha, cosalpha, 0),(0,0,1)]`, then :

Options

A. `A_(alpha+beta)=A_(alpha)A_(beta)`
B. `A_(alpha)^(-1)=A_(-alpha)`
C. `A_(alpha)^(-1)=-A_(alpha)`
D. `A_(alpha)^(2)=-I`

Answer

To solve the problem, we need to analyze the given matrix \( A_{\alpha} \) and derive the necessary results step by step.

### Step 1: Define the Matrix
The matrix \( A_{\alpha} \) is given as:
\[ A_{\alpha} = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 2: Find \( A_{-\alpha} \)
To find \( A_{-\alpha} \), we replace \( \alpha \) with \( -\alpha \) in the matrix:
\[ A_{-\alpha} = \begin{pmatrix} \cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Using the properties of cosine and sine, we have:
\[ \cos(-\alpha) = \cos \alpha \quad \text{and} \quad \sin(-\alpha) = -\sin \alpha \]
Thus,
\[ A_{-\alpha} = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Find the Cofactor Matrix
The cofactor matrix \( C \) of \( A_{\alpha} \) is computed as follows:
1. **C11**: Minor of \( A_{\alpha} \) by removing the first row and first column:
\[ C_{11} = \cos \alpha \]
2. **C12**: Minor of \( A_{\alpha} \) by removing the first row and second column:
\[ C_{12} = -\sin \alpha \]
3. **C13**: Minor of \( A_{\alpha} \) by removing the first row and third column:
\[ C_{13} = 0 \]
4. **C21**: Minor of \( A_{\alpha} \) by removing the second row and first column:
\[ C_{21} = \sin \alpha \]
5. **C22**: Minor of \( A_{\alpha} \) by removing the second row and second column:
\[ C_{22} = \cos \alpha \]
6. **C23**: Minor of \( A_{\alpha} \) by removing the second row and third column:
\[ C_{23} = 0 \]
7. **C31**: Minor of \( A_{\alpha} \) by removing the third row and first column:
\[ C_{31} = 0 \]
8. **C32**: Minor of \( A_{\alpha} \) by removing the third row and second column:
\[ C_{32} = 0 \]
9. **C33**: Minor of \( A_{\alpha} \) by removing the third row and third column:
\[ C_{33} = 1 \]

Thus, the cofactor matrix \( C \) is:
\[ C = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 4: Find the Determinant of \( A_{\alpha} \)
The determinant of \( A_{\alpha} \) is calculated as:
\[ \text{det}(A_{\alpha}) = \cos \alpha \cdot \cos \alpha \cdot 1 - (-\sin \alpha)(\sin \alpha) = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 5: Find the Inverse of \( A_{\alpha} \)
Since the determinant is 1, the inverse of \( A_{\alpha} \) can be found using the adjoint:
\[ A_{\alpha}^{-1} = \frac{1}{\text{det}(A_{\alpha})} \cdot C^T = C^T \]
The transpose of the cofactor matrix \( C \) is:
\[ C^T = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
Thus,
\[ A_{\alpha}^{-1} = A_{-\alpha} \]

### Step 6: Verify the Results
1. **Result 1**: \( A_{-\alpha} = A_{\alpha}^{-1} \)
2. **Result 2**: \( A_{\alpha} A_{\beta} = A_{\alpha + \beta} \)

### Conclusion
The correct options based on the analysis are:
- \( A_{\alpha} A_{\beta} = A_{\alpha + \beta} \)
- \( A_{-\alpha} = A_{\alpha}^{-1} \)