Q.If $A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$, show that $A^2=\begin{bmatrix}\cos2\alpha&\sin2\alpha\\-\sin2\alpha&\cos2\alpha\end{bmatrix}$.

Given $A=\begin{bmatrix}\cos\alpha&\sin\alpha\\-\sin\alpha&\cos\alpha\end{bmatrix}$.

Compute $A^2=A\cdot A$:

$(A^2)_{11}=\cos\alpha\cos\alpha+\sin\alpha(-\sin\alpha)=\cos^2\alpha-\sin^2\alpha=\cos2\alpha$

$(A^2)_{12}=\cos\alpha\sin\alpha+\sin\alpha\cos\alpha=2\sin\alpha\cos\alpha=\sin2\alpha$

$(A^2)_{21}=-\sin\alpha\cos\alpha+\cos\alpha(-\sin\alpha)=-2\sin\alpha\cos\alpha=-\sin2\alpha$

$(A^2)_{22}=-\sin\alpha\sin\alpha+\cos\alpha\cos\alpha=\cos^2\alpha-\sin^2\alpha=\cos2\alpha$ …

Compute A·A directly using the compound-angle identities $\cos^2\alpha-\sin^2\alpha=\cos2\alpha …

• Forgetting to apply the double-angle identities and leaving the answer in the unsimplified form $\cos^2\alpha-\sin^2\alpha$ instead of $\cos2\alpha$.
• Sign error on the (2,1) entry — both terms are negative ($-\sin\alpha\cos\alpha - \cos\alpha\sin\alpha$), giving $-\sin2\alpha$, easy to mis-simplify to $+\sin2\alpha$. …