Question

If \( A(t) = \begin{pmatrix} \cos t & \sin t \\ -\sin t & \cos t \end{pmatrix} \), then the product of \( A(t) \) and \( A(-t) \) is

• A. Identity matrix
• B. \( A^2(t) \)
• C. Null matrix
• D. \( A^2(-t) \)

Hint:

Rotation matrices satisfy \( A(t)A(-t) = I \), since they represent inverse rotations.

Answer 1

The Correct Option is A

Step 1: Write the given matrix.
\[ A(t) = \begin{pmatrix} \cos t & \sin t -\sin t & \cos t \end{pmatrix}. \]

Step 2: Find \( A(-t) \).
Using trigonometric identities:
\[ \cos(-t) = \cos t, \quad \sin(-t) = -\sin t. \]
So,
\[ A(-t) = \begin{pmatrix} \cos t & -\sin t \sin t & \cos t \end{pmatrix}. \]

Step 3: Multiply \( A(t)A(-t) \).
\[ A(t)A(-t) = \begin{pmatrix} \cos t & \sin t -\sin t & \cos t \end{pmatrix} \begin{pmatrix} \cos t & -\sin t \sin t & \cos t \end{pmatrix}. \]

Step 4: Perform matrix multiplication.
First row:
\[ (\cos^2 t + \sin^2 t,\ -\cos t\sin t + \sin t\cos t). \]
Second row:
\[ (-\sin t\cos t + \cos t\sin t,\ \sin^2 t + \cos^2 t). \]

Step 5: Simplify using identities.
\[ \cos^2 t + \sin^2 t = 1, \] \[ -\cos t\sin t + \sin t\cos t = 0. \]
Thus,
\[ A(t)A(-t) = \begin{pmatrix} 1 & 0 0 & 1 \end{pmatrix}. \]

Step 6: Interpret the result.
The result is the identity matrix.

Step 7: Final conclusion.
Thus, the product is the identity matrix.
Final Answer:
\[ \boxed{\text{Identity matrix}}. \]