Past exam of the mathematics course of the University of Oxford / 2021 / Prelims / Mathematics IV resit / 2 / b / ii / Solution

A possible center $\mathbf{c}$ must be a fixed point, so $(I-R_{\theta })\mathbf{c}=\mathbf{b}$. Let $c=\cos \theta$ and $s=\sin \theta$. The determinant iswhen $\theta$ is not an integer multiple of $2\pi$. The unique center is thereforeThe fixed point equation implies $\mathbf{b}=\mathbf{c}-R_{\theta }\mathbf{c}$, whenceRelative to $\mathbf{c}$, this is precisely a planar rotation through $\theta$. This also explains the excluded case: when $\theta \in 2\pi \mathbb{Z}$, $T$ is a translation, with no fixed point unless $\mathbf{b}=0$.