= 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> give
$$
\boxed{\begin{pmatrix}p_n&p_{n-1}\\q_n&q_{n-1}\end{pmatrix}=\begin{pmatrix}a_0&1\\1&0\end{pmatrix}\begin{pmatrix}a_1&1\\1&0\end{pmatrix}\cdots\begin{pmatrix}a_n&1\\1&0\end{pmatrix}.}
$$
Indeed, multiplying the last factor gives the <linear recurrences> $p_n=a_np_{n-1}+p_{n-2}$ and $q_n=a_nq_{n-1}+q_{n-2}$, which reproduce the <continued fraction convergents> of the <continued fraction>. Each factor has <determinant> $-1$. Taking <determinants> therefore yields
$$
\boxed{p_nq_{n-1}-q_np_{n-1}=(-1)^{n+1}.}
$$
This identity also proves that the numerator and denominator in the first column are <coprime>, so no unnoticed cancellation changes the displayed <matrix>.
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