If `A=[(cos theta, - sin theta, 0),(sin theta, cos theta, 0),(0,0,1)]` then `A^(-1)`=………..

Answer

To find the inverse of the matrix \( A \) given by

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

we can use the formula for the inverse of a matrix, which is given by

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

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

To find the determinant of \( A \), we can use the formula for the determinant of a \( 3 \times 3 \) matrix:

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

For our matrix \( A \):

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

Calculating this gives:

\[ \text{det}(A) = \cos^2 \theta + \sin^2 \theta = 1 \]

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

The adjoint of a \( 3 \times 3 \) matrix is the transpose of the cofactor matrix. We will compute the cofactors for each element of \( A \).

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

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

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

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

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

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

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

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

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

Now, we can write the cofactor matrix:

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

Taking the transpose gives us the adjoint:

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

### Step 3: Calculate \( A^{-1} \)

Using the formula for the inverse:

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

Thus, we have:

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