Let `F(alpha)=[(cos alpha, -sin alpha, 0),(sin alpha, cos alpha, 0),(0,0,1)]` where alpha in `R`.
Then `[F(alpha)]^(-1)` is equal to

Options

A. `F(-alpha)`
B. `F(alpha^(-1))`
C. `F(2alpha)`
D. None of these

Answer

To find the inverse of the matrix \( F(\alpha) = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \), we will follow these steps:

### Step 1: Calculate the Determinant of \( F(\alpha) \)

The determinant of a \( 3 \times 3 \) matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by the formula:
\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( F(\alpha) \):
- \( a = \cos \alpha \)
- \( b = -\sin \alpha \)
- \( c = 0 \)
- \( d = \sin \alpha \)
- \( e = \cos \alpha \)
- \( f = 0 \)
- \( g = 0 \)
- \( h = 0 \)
- \( i = 1 \)

Calculating the determinant:
\[ \text{det}(F(\alpha)) = \cos \alpha ( \cos \alpha \cdot 1 - 0 \cdot 0 ) - (-\sin \alpha)(\sin \alpha \cdot 1 - 0 \cdot 0) + 0 \]
\[ = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 2: Verify the Determinant is Non-zero

Since \( \text{det}(F(\alpha)) = 1 \neq 0 \), the inverse \( F(\alpha)^{-1} \) exists.

### Step 3: Find the Adjoint of \( F(\alpha) \)

The adjoint of a matrix is obtained by taking the transpose of the cofactor matrix. The cofactor matrix \( C \) for \( F(\alpha) \) is calculated as follows:

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det} \begin{pmatrix} \cos \alpha & 0 \\ 0 & 1 \end{pmatrix} = \cos \alpha \]

2. **Cofactor \( C_{12} \)**:
\[ C_{12} = -\text{det} \begin{pmatrix} \sin \alpha & 0 \\ 0 & 1 \end{pmatrix} = -\sin \alpha \]

3. **Cofactor \( C_{13} \)**:
\[ C_{13} = \text{det} \begin{pmatrix} \sin \alpha & \cos \alpha \\ 0 & 0 \end{pmatrix} = 0 \]

4. **Cofactor \( C_{21} \)**:
\[ C_{21} = -\text{det} \begin{pmatrix} -\sin \alpha & 0 \\ 0 & 1 \end{pmatrix} = \sin \alpha \]

5. **Cofactor \( C_{22} \)**:
\[ C_{22} = \text{det} \begin{pmatrix} \cos \alpha & 0 \\ 0 & 1 \end{pmatrix} = \cos \alpha \]

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det} \begin{pmatrix} \cos \alpha & -\sin \alpha \\ 0 & 0 \end{pmatrix} = 0 \]

7. **Cofactor \( C_{31} \)**:
\[ C_{31} = \text{det} \begin{pmatrix} -\sin \alpha & 0 \\ \cos \alpha & 0 \end{pmatrix} = 0 \]

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det} \begin{pmatrix} \cos \alpha & 0 \\ \sin \alpha & 0 \end{pmatrix} = 0 \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det} \begin{pmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{pmatrix} = \cos^2 \alpha + \sin^2 \alpha = 1 \]

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 Adjoint Matrix

The adjoint \( \text{adj}(F(\alpha)) \) is the transpose of the cofactor matrix:
\[ \text{adj}(F(\alpha)) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 5: Calculate the Inverse

Using the formula for the inverse:
\[ F(\alpha)^{-1} = \frac{1}{\text{det}(F(\alpha))} \cdot \text{adj}(F(\alpha)) \]
Since \( \text{det}(F(\alpha)) = 1 \):
\[ F(\alpha)^{-1} = \text{adj}(F(\alpha)) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Final Result

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