If `A=[{:(cosx,sinx,0),(-sinx,cosx,0),(0,0,1):}]` =f(x), then `A^(-1)` is equal to

Options

A. `f(-x)`
B. `f(x)`
C. `-f(x)`
D. `-f(-x)`

Answer

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

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

we will follow these steps:

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

The determinant of a \( 3 \times 3 \) matrix

\[ \begin{pmatrix} a & b & c \\ d & e & f \\ g & h & i \end{pmatrix} \]

is calculated using the formula:

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

For our matrix \( A \):

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

Calculating the determinant:

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

\[ = \cos^2 x + \sin^2 x \]

Using the Pythagorean identity \( \sin^2 x + \cos^2 x = 1 \):

\[ \text{det}(A) = 1 \]

### Step 2: Verify that the Inverse Exists

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

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

The adjoint of a matrix is the transpose of its cofactor matrix. We will calculate the cofactor matrix first.

#### Cofactor Matrix Calculation:

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

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

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

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

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

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

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

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

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

The cofactor matrix is:

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

### Step 4: Transpose the Cofactor Matrix to Get the Adjoint

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

### Step 5: Calculate the Inverse of \( A \)

Using the formula \( A^{-1} = \frac{\text{Adj}(A)}{\text{det}(A)} \):

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

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

### Step 6: Relate \( A^{-1} \) to \( f(-x) \)

From the original matrix \( A \):

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

We can see that \( A^{-1} \) can be expressed as:

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

Thus, we have:

\[ A^{-1} = f(-x) \]

### Conclusion

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

\[ A^{-1} = f(-x) \]