Solution
= Solution
Expanding $\det A^{(n)}$ along its last row gives the last diagonal contribution $a_nd_{n-1}$. The only other nonzero entry is $c_{n-1}$; expanding its minor along the last column contributes
$$
-b_{n-1}c_{n-1}d_{n-2}.
$$
Thus the <tridiagonal determinant recurrence> is
$$
\boxed{d_n=a_nd_{n-1}-b_{n-1}c_{n-1}d_{n-2}},
$$
so
$$
\boxed{X_n=a_n,\qquad Y_n=-b_{n-1}c_{n-1}}.
$$