If `A=[(cosx,-sinx),(sinx,cosx)]`, then `A``A^(T) `is

Options

A. Zero matrix
B. `I_(2)`
C. `[(1,1),(1,1)]`
D. None of these

Answer

To solve the problem, we need to find the product of the matrix \( A \) and its transpose \( A^T \). Let's go through the steps systematically.

### Step 1: Write down the matrix \( A \)

The matrix \( A \) is given as:
\[ A = \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} \]

### Step 2: Find the transpose of matrix \( A \)

The transpose of a matrix is obtained by interchanging its rows and columns. Thus, the transpose \( A^T \) is:
\[ A^T = \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \]

### Step 3: Multiply \( A \) and \( A^T \)

Now we need to calculate the product \( A A^T \):
\[ A A^T = \begin{pmatrix} \cos x & -\sin x \\ \sin x & \cos x \end{pmatrix} \begin{pmatrix} \cos x & \sin x \\ -\sin x & \cos x \end{pmatrix} \]

To multiply these matrices, we use the formula for matrix multiplication, which involves taking the dot product of rows of the first matrix with columns of the second matrix.

#### Calculation of each element:

1. **First row, first column**:
\[ (\cos x)(\cos x) + (-\sin x)(-\sin x) = \cos^2 x + \sin^2 x = 1 \]

2. **First row, second column**:
\[ (\cos x)(\sin x) + (-\sin x)(\cos x) = \cos x \sin x - \sin x \cos x = 0 \]

3. **Second row, first column**:
\[ (\sin x)(\cos x) + (\cos x)(-\sin x) = \sin x \cos x - \sin x \cos x = 0 \]

4. **Second row, second column**:
\[ (\sin x)(\sin x) + (\cos x)(\cos x) = \sin^2 x + \cos^2 x = 1 \]

### Step 4: Write the resulting matrix

Putting all the calculated elements together, we get:
\[ A A^T = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

This is the identity matrix \( I_2 \).

### Final Answer

Thus, the result of \( A A^T \) is:
\[ A A^T = I_2 \]