Find `A^(-1)` using adjoint method, where `A = [(cos theta,sin theta),(-sin theta,cos theta)]`

Answer

To find the inverse of the matrix \( A \) using the adjoint method, we follow these steps:

### Step 1: Write down the matrix \( A \)
Given:
\[ A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \]

### Step 2: Find the determinant of \( A \)
The determinant of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:
\[ \text{det}(A) = ad - bc \]
For our matrix \( A \):
- \( a = \cos \theta \)
- \( b = \sin \theta \)
- \( c = -\sin \theta \)
- \( d = \cos \theta \)

Calculating the determinant:
\[ \text{det}(A) = \cos \theta \cdot \cos \theta - (\sin \theta \cdot -\sin \theta) = \cos^2 \theta + \sin^2 \theta \]
Using the Pythagorean identity:
\[ \cos^2 \theta + \sin^2 \theta = 1 \]
Thus,
\[ \text{det}(A) = 1 \]

### Step 3: Find the adjoint of \( A \)
The adjoint of a \( 2 \times 2 \) matrix \( A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by:
\[ \text{adj}(A) = \begin{pmatrix} d & -b \\ -c & a \end{pmatrix} \]
For our matrix \( A \):
- \( d = \cos \theta \)
- \( -b = -\sin \theta \)
- \( -c = \sin \theta \)
- \( a = \cos \theta \)

Thus, the adjoint of \( A \) is:
\[ \text{adj}(A) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

### Step 4: Calculate the inverse of \( A \)
The inverse of \( A \) is given by:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]
Since we found that \( \text{det}(A) = 1 \):
\[ A^{-1} = \text{adj}(A) = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

### Final Answer
Thus, the inverse of the matrix \( A \) is:
\[ A^{-1} = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

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