`if A= [{:(costheta,sintheta),(-sintheta,costheta):}],` then `A A^T`=?`

Options

A. `[{:(1,0),(0,1):}]`
B. `[{:(0,1),(1,0):}]`
C. `[{:(1,0),(0,-1):}]`
D. None of these

Explanation

To solve the problem, we need to calculate \( A A^T \) where \( A = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \).

### Step-by-Step Solution:

1. **Find the Transpose of Matrix A**:
\[ A^T = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix}^T = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

**Hint**: To find the transpose of a matrix, swap its rows and columns.

2. **Multiply A by A^T**:
\[ A A^T = \begin{pmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \]

**Hint**: When multiplying two matrices, the element in the resulting matrix at position (i, j) is the dot product of the i-th row of the first matrix and the j-th column of the second matrix.

3. **Calculate Each Element of the Resulting Matrix**:
- First row, first column:
\[ \cos \theta \cdot \cos \theta + \sin \theta \cdot \sin \theta = \cos^2 \theta + \sin^2 \theta = 1 \]
- First row, second column:
\[ \cos \theta \cdot (-\sin \theta) + \sin \theta \cdot \cos \theta = -\cos \theta \sin \theta + \sin \theta \cos \theta = 0 \]
- Second row, first column:
\[ -\sin \theta \cdot \cos \theta + \cos \theta \cdot \sin \theta = -\sin \theta \cos \theta + \cos \theta \sin \theta = 0 \]
- Second row, second column:
\[ -\sin \theta \cdot (-\sin \theta) + \cos \theta \cdot \cos \theta = \sin^2 \theta + \cos^2 \theta = 1 \]

4. **Combine the Results**:
\[ A A^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

5. **Conclusion**:
The product \( A A^T \) is the identity matrix of order 2:
\[ A A^T = I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Final Answer:
\[ A A^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]