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 matrix \( A_\alpha \) given by:

\[ A_\alpha = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

We will explore the properties of this matrix, specifically focusing on the statements provided in the options.

### Step 1: Verify \( A_{\alpha + \beta} = A_\alpha \cdot A_\beta \)

To verify this, we need to compute \( A_{\alpha + \beta} \) and \( A_\alpha \cdot A_\beta \).

**1.1 Compute \( A_{\alpha + \beta} \)**

\[ A_{\alpha + \beta} = \begin{pmatrix} \cos(\alpha + \beta) & -\sin(\alpha + \beta) & 0 \\ \sin(\alpha + \beta) & \cos(\alpha + \beta) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Using the angle addition formulas:
- \( \cos(\alpha + \beta) = \cos \alpha \cos \beta - \sin \alpha \sin \beta \)
- \( \sin(\alpha + \beta) = \sin \alpha \cos \beta + \cos \alpha \sin \beta \)

Thus, we can write:

\[ A_{\alpha + \beta} = \begin{pmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -(\sin \alpha \cos \beta + \cos \alpha \sin \beta) & 0 \\ \sin \alpha \cos \beta + \cos \alpha \sin \beta & \cos \alpha \cos \beta - \sin \alpha \sin \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

**1.2 Compute \( A_\alpha \cdot A_\beta \)**

Now, we compute \( A_\alpha \cdot A_\beta \):

\[ A_\beta = \begin{pmatrix} \cos \beta & -\sin \beta & 0 \\ \sin \beta & \cos \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Now, multiplying \( A_\alpha \) and \( A_\beta \):

\[ A_\alpha \cdot A_\beta = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \cdot \begin{pmatrix} \cos \beta & -\sin \beta & 0 \\ \sin \beta & \cos \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Calculating the product:

\[ = \begin{pmatrix} \cos \alpha \cos \beta - \sin \alpha \sin \beta & -(\sin \alpha \cos \beta + \cos \alpha \sin \beta) & 0 \\ \sin \alpha \cos \beta + \cos \alpha \sin \beta & \cos \alpha \cos \beta - \sin \alpha \sin \beta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

**Conclusion for Step 1:**

Since \( A_{\alpha + \beta} = A_\alpha \cdot A_\beta \), the first statement is correct.

### Step 2: Verify \( A_\alpha^{-1} = A_{-\alpha} \)

**2.1 Compute \( A_\alpha^{-1} \)**

To find the inverse, we can use the property that for a rotation matrix, the inverse is the transpose:

\[ A_\alpha^{-1} = A_{-\alpha} = \begin{pmatrix} \cos(-\alpha) & -\sin(-\alpha) & 0 \\ \sin(-\alpha) & \cos(-\alpha) & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Using the fact that \( \cos(-\alpha) = \cos \alpha \) and \( \sin(-\alpha) = -\sin \alpha \):

\[ A_{-\alpha} = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

**Conclusion for Step 2:**

Thus, \( A_\alpha^{-1} = A_{-\alpha} \) is also correct.

### Final Conclusion

Both statements are correct:
1. \( A_{\alpha + \beta} = A_\alpha \cdot A_\beta \)
2. \( A_\alpha^{-1} = A_{-\alpha} \)