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

Answer

To find the inverse of the matrix \( A = \begin{pmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{pmatrix} \), we will follow the steps outlined below.

### Step 1: Calculate the Determinant of A
The determinant of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For our matrix \( A \):
- \( a = \cos \theta \)
- \( b = -\sin \theta \)
- \( c = -\sin \theta \)
- \( d = -\cos \theta \)

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

### Step 2: Calculate the Adjoint of A
The adjoint of a 2x2 matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) 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 3: Calculate the Inverse of A
The inverse of a matrix is given by:
\[ A^{-1} = \frac{\text{adj}(A)}{\text{det}(A)} \]

Substituting the values we found:
\[ A^{-1} = \frac{1}{-1} \begin{pmatrix} -\cos \theta & \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} = \begin{pmatrix} \cos \theta & -\sin \theta \\ -\sin \theta & -\cos \theta \end{pmatrix} \]

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