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Consider the following matrix

$$ \begin{equation}A_{n} := \begin{bmatrix} a_{1} & -p & \dots & 0 &\dots &0 \\ -q & a_{2} & -p &0 & \dots & 0 \\ 0 & -q & a_{3} &-p & ~... & 0 \\ 0 & 0 &-q &a_{4} &-p & 0 \\ 0 &\vdots & \ddots & -q & a_{n-1} & -p \\ 0 &0 &0 &\dots &-q &a_{n} \end{bmatrix} \end{equation} $$

where $a_{j},p,q \in \mathbb{R}$, $a_{j} \neq 0$ for all $j = 1,2,...,n$ and $n \in \mathbb{N}$. We have

$$ \det A_{n} = \frac{1}{x_{n}} \det A_{n-1} - pq \det A_{n-2}. $$

Now consider the vector $b := (q,0,..,0,p)^{t} \in \mathbb{R}^{r-1}$ and define $[A_{n}]_{i}$ to be the matrix obtained from $A_{n}$ except that I replace the i-th column of $A_{n}$ with $b$ for $i=1,...,n$. Define for simplicity

$$ \begin{equation}D_{k,n} := \begin{bmatrix} a_{k} & -p & \dots & 0 &\dots &0 \\ -q & a_{k+1} & -p &0 & \dots & 0 \\ 0 & -q & a_{k+2} &-p & ~... & 0 \\ 0 & 0 &-q &a_{k+3} &-p & 0 \\ 0 &\vdots & \ddots & -q & a_{n-1} & -p \\ 0 &0 &0 &\dots &-q &a_{n} \end{bmatrix} \end{equation} $$ where $k,n \in \mathbb{N}$, $k \leq n$.

Now for example for $i = 1$ I would get after expanding by the first column

\begin{align} \det([A_{n}]_{1}) &= \begin{vmatrix} q & -p \\ 0 & a_2 & -p \\ 0 &-q & a_3 & -p \\ & & \ddots & \ddots & -p \\ & & & -q & a_{n-2} & -p \\ & & & & -q & a_{n-1} & -p \\ p& & & & & -q & a_n \end{vmatrix} &= q \cdot \det(D_{2,n}) + p^{n} \end{align}

For $i = 2$ I have also compute the determinant after expanding by the second column. I get

\begin{align} \det([A_{n}]_{2}) &= \begin{vmatrix} a_1 & q \\ -q & 0 & -p \\ 0 & 0 & a_3 & -p \\ & & \ddots & \ddots & -p \\ & & & -q & a_{n-2} & -p \\ & & & & -q & a_{n-1} & -p \\ & p& & & & -q & a_n \end{vmatrix} &= q^{2} \cdot \det(D_{3,n}) + a_1p^{n-1} \end{align}

Up to this point I thought I have the following formula

$$ \det([A_{n}]_{i}) = q^{i} \cdot \det(D_{i+1,n}) + a_1 ... a_{i-1}p^{n-i+1} $$

But then I compute $\det([A_{n}]_{3})$ and after expanding by the third column I get

$$ \det([A_{n}]_{3}) = q^{3} \cdot \det(D_{4,n}) + a_1 a_2 p^{n-2} - p^{n-1}q $$

So my first thought is not correct. My question is, wether there is a closed formula for $\det([A_{n}]_{i})$. Or maybe a recurrence equation? I Hope someone can help me. Thanks!

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  • $\begingroup$ This answer to the almost identical question that you asked earlier gives you a recurrence for the determinant and a link to more general results for a tridiagonal matrix. $\endgroup$
    – amd
    Oct 12, 2018 at 20:46
  • $\begingroup$ I don't understand what you mean. Where can I find the link? $\endgroup$
    – wayne
    Oct 13, 2018 at 9:11
  • $\begingroup$ The word “recurrence” in the first sentence is a link. $\endgroup$
    – amd
    Oct 13, 2018 at 20:44
  • $\begingroup$ there is no link $\endgroup$
    – wayne
    Oct 14, 2018 at 11:36
  • $\begingroup$ Sure, there is. It links to en.m.wikipedia.org/wiki/Tridiagonal_matrix#Determinant. $\endgroup$
    – amd
    Oct 16, 2018 at 0:44

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