Then inverse of the matrix `[[1, 0, 0], [0, costheta, sintheta], [0, sintheta, -costheta]]` is

Options

A. `[[-1, 0, 0], [0, -costheta, -sintheta], [0, -sintheta, costheta]]`
B. `[[costheta, sintheta, 0], [-sintheta, costheta, 0], [0, 0, 1]]`
C. `[[1, 0, 0], [0, costheta, sintheta], [0, sintheta, -costheta]]`
D. `[[1, 0, 0], [0, -costheta, -sintheta], [1, sintheta, costheta]]`

Answer

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

### Step 1: Calculate the Determinant of Matrix A

The determinant of a 3x3 matrix can be calculated using the formula:

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

For our matrix \( A \):

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

Calculating the determinant:

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

### Step 2: Check if the Determinant is Non-Zero

Since \(\text{det}(A) = -1\) (which is not equal to 0), the inverse of the matrix exists.

### Step 3: Calculate the Adjoint of Matrix A

The adjoint of a matrix is the transpose of the cofactor matrix. We will calculate the cofactor matrix \( C \).

1. **Cofactor \( C_{11} \)**:
\[ C_{11} = \text{det} \begin{bmatrix} \cos \theta & \sin \theta \\ \sin \theta & -\cos \theta \end{bmatrix} = (-\cos^2 \theta - \sin^2 \theta) = -1 \]

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

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

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

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

6. **Cofactor \( C_{23} \)**:
\[ C_{23} = -\text{det} \begin{bmatrix} 1 & 0 \\ 0 & \sin \theta \end{bmatrix} = -\sin \theta \]

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

8. **Cofactor \( C_{32} \)**:
\[ C_{32} = -\text{det} \begin{bmatrix} 1 & 0 \\ 0 & \sin \theta \end{bmatrix} = -\sin \theta \]

9. **Cofactor \( C_{33} \)**:
\[ C_{33} = \text{det} \begin{bmatrix} 1 & 0 \\ 0 & \cos \theta \end{bmatrix} = \cos \theta \]

Now, we can construct the cofactor matrix \( C \):

\[ C = \begin{bmatrix} -1 & 0 & 0 \\ 0 & -\cos \theta & -\sin \theta \\ 0 & -\sin \theta & \cos \theta \end{bmatrix} \]

### Step 4: Calculate the Adjoint of A

The adjoint \( \text{adj}(A) \) is the transpose of the cofactor matrix \( C \):

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

### Step 5: Calculate the Inverse of Matrix A

Using the formula for the inverse of a matrix:

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

Substituting the values we found:

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

This simplifies to:

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

### Final Result

The inverse of the matrix \( A \) is:

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