If `A=[(cos alpha, -sin alpha, 0),(sin alpha, cos alpha,0),(0,0,1)]` then `A^(-1)` is

Options

A. A
B. `-A`
C. `adj(A)`
D. `-adj(A)`

Explanation

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

### Step 1: Calculate the Determinant of \( A \)

The determinant of a 3x3 matrix \( A = \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \) is given by:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

For our matrix \( A \):

- \( a = \cos \alpha \)
- \( b = -\sin \alpha \)
- \( c = 0 \)
- \( d = \sin \alpha \)
- \( e = \cos \alpha \)
- \( f = 0 \)
- \( g = 0 \)
- \( h = 0 \)
- \( i = 1 \)

Calculating the determinant:

\[ \text{det}(A) = \cos \alpha (\cos \alpha \cdot 1 - 0 \cdot 0) - (-\sin \alpha)(\sin \alpha \cdot 1 - 0 \cdot 0) + 0 \]
\[ = \cos^2 \alpha + \sin^2 \alpha = 1 \]

### Step 2: Calculate the Adjoint of \( A \)

The adjoint of a matrix is the transpose of the cofactor matrix. We calculate the cofactors of each element of \( A \):

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

The cofactor matrix is:

\[ \text{Cof}(A) = \begin{pmatrix} \cos \alpha & -\sin \alpha & 0 \\ \sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Taking the transpose gives us the adjoint:

\[ \text{adj}(A) = \begin{pmatrix} \cos \alpha & \sin \alpha & 0 \\ -\sin \alpha & \cos \alpha & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

### Step 3: Calculate the Inverse of \( A \)

Using the formula for the inverse of a matrix:

\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]

Since \( \text{det}(A) = 1 \):

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

### Final Answer

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