Call the two proposed <matrices> $A$ and $B$. Direct multiplication gives $A^{2n}=I$, $B^2=\epsilon^nI=A^n$, and $B^{-1}AB=A^{-1}$. Both are invertible since $\epsilon\ne0$. They therefore satisfy every defining relation, and the <universal property of a group presentation> gives the desired <group representation>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2015 / ii / Paper 1 / 15F / b / i / Solution

 Solution (source code)