If `F(alpha)=[(cos alpha, -sin alpha, 0),(sin alpha, cos alpha, 0),(0,0,1)]` and `G(beta)=[(cos beta, 0, sin beta),(0,1,0),(-sin beta, 0, cos beta)]`, then `[F(alpha)G(beta)]^(-1)` is equal to

Options

A. `F(-alpha)G(-beta)`
B. `F(alpha^(-1))G(beta^(-1))`
C. `G(-beta)F(-alpha)`
D. `G(beta^(-1))F(alpha^(-1))`

Answer

To find the inverse of the product of the matrices \( F(\alpha) \) and \( G(\beta) \), we will follow these steps:

### Step 1: Define the matrices
We have:
\[ F(\alpha) = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]
and
\[ G(\beta) = \begin{pmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{pmatrix} \]

### Step 2: Multiply the matrices \( F(\alpha) \) and \( G(\beta) \)
To find \( F(\alpha)G(\beta) \), we perform matrix multiplication:
\[ F(\alpha)G(\beta) = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \cos \beta & 0 & \sin \beta \\ 0 & 1 & 0 \\ -\sin \beta & 0 & \cos \beta \end{pmatrix} \]

Calculating the product:
- The first row:
- First element: \( \cos \alpha \cdot \cos \beta + (-\sin \alpha) \cdot 0 + 0 \cdot (-\sin \beta) = \cos \alpha \cos \beta \)
- Second element: \( \cos \alpha \cdot 0 + (-\sin \alpha) \cdot 1 + 0 \cdot 0 = -\sin \alpha \)
- Third element: \( \cos \alpha \cdot \sin \beta + (-\sin \alpha) \cdot 0 + 0 \cdot \cos \beta = \cos \alpha \sin \beta \)

- The second row:
- First element: \( \sin \alpha \cdot \cos \beta + \cos \alpha \cdot 0 + 0 \cdot (-\sin \beta) = \sin \alpha \cos \beta \)
- Second element: \( \sin \alpha \cdot 0 + \cos \alpha \cdot 1 + 0 \cdot 0 = \cos \alpha \)
- Third element: \( \sin \alpha \cdot \sin \beta + \cos \alpha \cdot 0 + 0 \cdot \cos \beta = \sin \alpha \sin \beta \)

- The third row:
- First element: \( 0 \cdot \cos \beta + 0 \cdot 0 + 1 \cdot (-\sin \beta) = -\sin \beta \)
- Second element: \( 0 \cdot 0 + 0 \cdot 1 + 1 \cdot 0 = 0 \)
- Third element: \( 0 \cdot \sin \beta + 0 \cdot 0 + 1 \cdot \cos \beta = \cos \beta \)

Thus, we have:
\[ F(\alpha)G(\beta) = \begin{pmatrix} \cos \alpha \cos \beta & -\sin \alpha & \cos \alpha \sin \beta \\ \sin \alpha \cos \beta & \cos \alpha & \sin \alpha \sin \beta \\ -\sin \beta & 0 & \cos \beta \end{pmatrix} \]

### Step 3: Find the inverse of the product
To find the inverse of a \( 3 \times 3 \) matrix, we can use the formula:
\[ A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) \]
However, for rotation matrices, we know that the inverse of a rotation matrix is its transpose.

Thus, we can take the transpose of \( F(\alpha)G(\beta) \):
\[ [F(\alpha)G(\beta)]^{-1} = (F(\alpha)G(\beta))^T \]

Calculating the transpose:
\[ [F(\alpha)G(\beta)]^T = \begin{pmatrix} \cos \alpha \cos \beta & \sin \alpha \cos \beta & -\sin \beta \\ -\sin \alpha & \cos \alpha & 0 \\ \cos \alpha \sin \beta & \sin \alpha \sin \beta & \cos \beta \end{pmatrix} \]

### Final Result
Thus, the inverse of \( F(\alpha)G(\beta) \) is:
\[ [F(\alpha)G(\beta)]^{-1} = \begin{pmatrix} \cos \alpha \cos \beta & \sin \alpha \cos \beta & -\sin \beta \\ -\sin \alpha & \cos \alpha & 0 \\ \cos \alpha \sin \beta & \sin \alpha \sin \beta & \cos \beta \end{pmatrix} \]