Simplify : `sinA[(sinA,-cosA),(cosA, sinA)]+cosA[(cosA,sinA),(-sinA, cosA)]`.

Answer

To simplify the expression \( \sin A \begin{pmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{pmatrix} + \cos A \begin{pmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{pmatrix} \), we will follow these steps:

### Step 1: Distribute \( \sin A \) and \( \cos A \) into their respective matrices.

\[ \sin A \begin{pmatrix} \sin A & -\cos A \\ \cos A & \sin A \end{pmatrix} = \begin{pmatrix} \sin^2 A & -\sin A \cos A \\ \sin A \cos A & \sin^2 A \end{pmatrix} \]

\[ \cos A \begin{pmatrix} \cos A & \sin A \\ -\sin A & \cos A \end{pmatrix} = \begin{pmatrix} \cos^2 A & \cos A \sin A \\ -\cos A \sin A & \cos^2 A \end{pmatrix} \]

### Step 2: Add the two resulting matrices together.

Now, we add the two matrices:

\[ \begin{pmatrix} \sin^2 A & -\sin A \cos A \\ \sin A \cos A & \sin^2 A \end{pmatrix} + \begin{pmatrix} \cos^2 A & \cos A \sin A \\ -\cos A \sin A & \cos^2 A \end{pmatrix} \]

This results in:

\[ \begin{pmatrix} \sin^2 A + \cos^2 A & -\sin A \cos A + \cos A \sin A \\ \sin A \cos A - \cos A \sin A & \sin^2 A + \cos^2 A \end{pmatrix} \]

### Step 3: Simplify the entries of the resulting matrix.

The entries simplify as follows:

1. The first entry: \( \sin^2 A + \cos^2 A = 1 \) (using the Pythagorean identity).
2. The second entry: \( -\sin A \cos A + \cos A \sin A = 0 \).
3. The third entry: \( \sin A \cos A - \cos A \sin A = 0 \).
4. The fourth entry: \( \sin^2 A + \cos^2 A = 1 \) (again using the Pythagorean identity).

Thus, the resulting matrix is:

\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]

### Final Answer:

The simplified expression is:

\[ \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} \]