Past exam of the mathematical methods I course of the University of Oxford / 2022 / 5 / ii / Solution

Directly multiplying the columns' scalar products gives $C^{T}C=I$, and direct evaluation gives $\det C=1$. Thus the operation is a proper rotation. Its trace is zero. A three-dimensional proper rotation has eigenvalues $1,e^{i\phi },e^{-i\phi }$, hence $\mathrm{tr}C=1+2\cos \phi$. ThereforeThis determines the usual unsigned angle in $[0,\pi ]$ without changing coordinates. The axis can be checked, if desired, from $C(0,1,-1{)}^T=(0,1,-1{)}^T$. The determinant test must precede the trace formula: an improper rotation instead has trace $-1+2\cos \phi$.