<s>[INST] closed formula for determinant Consider the following matrix
$$
\begin{equation}A_{r-1} := 
\begin{bmatrix}
\frac{1}{x_{1}} & -p & \dots  & 0  &\dots &0    \\
-q & \frac{1}{x_{2}} & -p &0  & \dots  & 0    \\
0 & -q & \frac{1}{x_{3}} &-p  & ~...  & 0    \\
0 & 0 &-q &\frac{1}{x_{4}} &-p   & 0    \\
0  &\vdots    & \ddots & -q & \frac{1}{x_{r-2}}  & -p \\
0  &0 &0 &\dots &-q  &\frac{1}{x_{r-1}}
\end{bmatrix}
\end{equation}
$$
where $x_{i},p,q \in \mathbb{R}$, $x_{i} \neq 0$ for all $i = 1,2,...,r-1$ and $r \in \mathbb{N}$, $r \geq 3$. I want to find a closed formula for $\det(A)$. For $r=3$ we have
$$
\begin{equation}
\det(A) = 
\begin{vmatrix}
\frac{1}{x_1} & -p   \\
-q & \frac{1}{x_2} \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2)}{x_1
x_2}\end{equation} .
$$
For $r = 4$ we have
$$
\begin{equation}
\det(A) = 
\begin{vmatrix}
\frac{1}{x_1} & -p & 0   \\
-q & \frac{1}{x_2} & -p \\
0  &-q &\frac{1}{x_3}  \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2 + x_2 x_3)}{x_1 x_2 x_3}
\end{equation} .
$$
Up to this point I think the formula is given by
$$
\det(A) = \frac{1 - pq(x_{r-1}x_{r-2} + x_{r-2}x_{r-3})}{x_{1}x_{2}...x_{r-1}}.
$$
But this is not correct. For $r = 5$ I get
$$
\begin{equation}
\det(A) = 
\begin{vmatrix}
\frac{1}{x_1} & -p & 0 &0   \\
-q & \frac{1}{x_2} & -p &0 \\
0  &-q &\frac{1}{x_3} &-p  \\
0 & 0 &-q &\frac{1}{x_4 } \\
\end{vmatrix} = \frac{1 -pq(x_1 x_2 + x_2 x_3 + x_3 x_4) + p^2 q^2x_1 x_2 x_3x_4}{x_1 x_2 x_3 x_4}.
\end{equation} .
$$
Has someone has an idea how to find a closed formula? Thanks for any hints!
 [/INST] Since the matrix is tridiagonal, its determinant satisfies the following recurrence:
$$
\det A_n = \frac{1}{x_n} \det A_{n-1} -pq \det A_{n-2}
$$
Not a closed formula, but probably just as good.
</s>