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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!

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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.

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