Question

Simplify: \(\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix}\).

Hint:

Multiply out each matrix by its scalar and add entrywise, using sin^2+cos^2=1.

Answer 1

Step 1: Recognising the rotation-matrix structure:
Both matrices are related to the standard rotation matrix \(R(\theta)=\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\); the second is \(R(\theta)\) rotated by \(90^\circ\) in structure.

Step 2: Computing entrywise using the Pythagorean identity:
Every diagonal entry becomes \(\cos^2\theta+\sin^2\theta=1\) and every off-diagonal entry cancels to \(0\) since the cross terms have opposite signs.

Final Answer:
\[ \boxed{I=\begin{bmatrix}1&0\\0&1\end{bmatrix}} \]