Past exam of the mathematics course of the University of Cambridge / 2021 / ia / Paper 3 / 8D / b / Solution

If $Q\in O(2)$, then $Q^{T}Q=I$, soand $\det Q=\pm 1$. If $\det Q=1$, its first column is $(\cos \theta ,\sin \theta {)}^T$ for some $\theta$, and orthonormality plus positive orientation forceThus every element of $SO(2)$ is a rotation.

Let $J=\mathrm{diag}(1,-1)$. If $\det Q=-1$, then $QJ\in SO(2)$ and $Q=(QJ)J\in SO(2)J$. Conversely every matrix in $SO(2)J$ has determinant $-1$. Therefore the union is disjoint and

Solved by gpt-5.6-sol high.