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: Multiplying each matrix by its scalar:
\(\cos\theta\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}=\begin{bmatrix}\cos^{2}\theta&\sin\theta\cos\theta\\-\sin\theta\cos\theta&\cos^{2}\theta\end{bmatrix}\), and \(\sin\theta\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}=\begin{bmatrix}\sin^{2}\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^{2}\theta\end{bmatrix}\).

Step 2: Adding entrywise:
Top-left: \(\cos^2\theta+\sin^2\theta=1\). Top-right: \(\sin\theta\cos\theta-\sin\theta\cos\theta=0\). Bottom-left: \(-\sin\theta\cos\theta+\sin\theta\cos\theta=0\). Bottom-right: \(\cos^2\theta+\sin^2\theta=1\).

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