If `f(theta)=[(cos theta, -sin theta,0),(sin theta, cos theta, 0),(0,0,1)]` then `{f(theta)^(-1)}` is equal to

Options

A. `f(-theta)`
B. `f(theta)^(-1)`
C. `f(2theta)`
D. None of these

Explanation

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

### Step 1: Calculate the Determinant of \( f(\theta) \)

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 \( f(\theta) \):

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

Calculating this gives:

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

### Step 2: Find the Adjoint of \( f(\theta) \)

The adjoint of a matrix is the transpose of its cofactor matrix. For \( f(\theta) \):

1. The cofactor of the element at position (1,1) is \( \text{det}\begin{pmatrix} \cos \theta & 0 \\ 0 & 1 \end{pmatrix} = \cos \theta \).
2. The cofactor of the element at position (1,2) is \( -\text{det}\begin{pmatrix} \sin \theta & 0 \\ 0 & 1 \end{pmatrix} = -\sin \theta \).
3. The cofactor of the element at position (1,3) is \( 0 \).
4. The cofactor of the element at position (2,1) is \( -\text{det}\begin{pmatrix} -\sin \theta & 0 \\ 0 & 1 \end{pmatrix} = \sin \theta \).
5. The cofactor of the element at position (2,2) is \( \text{det}\begin{pmatrix} \cos \theta & 0 \\ 0 & 1 \end{pmatrix} = \cos \theta \).
6. The cofactor of the element at position (2,3) is \( 0 \).
7. The cofactor of the element at position (3,1) is \( 0 \).
8. The cofactor of the element at position (3,2) is \( 0 \).
9. The cofactor of the element at position (3,3) is \( 1 \).

Thus, the cofactor matrix is:

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

Taking the transpose gives us the adjoint:

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

### Step 3: Calculate the Inverse

Using the formula for the inverse of a matrix:

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

Since \( \text{det}(f(\theta)) = 1 \):

\[ f(\theta)^{-1} = 1 \cdot \begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{pmatrix} \]

Thus, we have:

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

### Final Answer

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