Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 3 / 2G / Solution

Introduce $p_{-1}=1$, $q_{-1}=0$, $p_0=a_0$ and $q_0=1$. Composition of the Möbius transformations $z\mapsto a_j+1/z$ represents a finite continued fraction. Their matrices giveIndeed, multiplying the last factor gives the linear recurrences $p_n=a_n{p}_{n-1}+p_{n-2}$ and $q_n=a_n{q}_{n-1}+q_{n-2}$, which reproduce the continued fraction convergents of the continued fraction. Each factor has determinant $-1$. Taking determinants therefore yieldsThis identity also proves that the numerator and denominator in the first column are coprime, so no unnoticed cancellation changes the displayed matrix.