For a few days now I've been trying to find a closed form expression for the determinant of the following $n\times n$ tridiagonal matrix

$$\begin{pmatrix}c_1+b_1+a_1 & b_1 & 0 & \ddots & 0 \\ c_2 & c_2+b_2+a_2 & b_2 & \ddots & 0 \\ 0 & c_3 & c_3+b_3+a_3 & \ddots & \vdots \\ \vdots & \ddots & \ddots & \ddots & b_{n-1}\\ 0 & ... & ... & c_{n} & c_{n}+b_n +a_n\end{pmatrix}$$

For the sequences $c_n$, $b_n$, and $a_n$. I've figured out closed form expression for special cases. Namely, when $a_n=0$, the determinant is $$\Big(\prod_{i=1}^nb_i\Big)\sum_{l=0}^n\prod_{k=1}^l\frac{c_{k}}{b_k}$$ When $l=0$ in the product series, that returns a $1$. Additionally, if $c_1=0$, then the determinant is simply $$\prod_{i=1}^nb_i.$$

I would really like to find an analogous formula in the case where $a_n \neq 0$. For your benefit I will list the first few determinants for small $n$ $$n=1:\quad\quad c_1+b_1+a_1$$ $$n=2:\quad\quad a_1a_2+b_1a_2+a_1b_2+b_1b_2+c_1a_2+c_1b_2+a_1c_2+c_1c_2$$ $$n=3:\quad\quad a_1a_2a_3+b_1a_2a_3+a_1b_2a_3+b_1b_2a_3+a_1a_2b_3+b_1a_2b_3+a_1b_2b_3+b_1b_2b_3+c_1a_2a_3+c_1b_2a_3+c_1a_2b_3+c_1b_2b_3+a_1c_2a_3+a_1c_2b_3+c_1c_2a_3+c_1c_2b_3+a_1a_2c_3+b_1a_2c_3+c_1a_2c_3+a_1c_2c_3+c_1c_2c_3$$

When you look at this, you may suspect that it is just the sum of every $n$th order product of $a$'s $b$'s and $c$'s with no subscript repeated, however this is not the case. For instance, $b_1c_2$ does not appear in the $n=2$ formula. Similarly there are $6$ terms which do not appear in the $n=3$ formula.

I would really appreciate anyones input on this!

  • $\begingroup$ Perhaps something simplifies if you use the recursive formula for the determinant of a tridiagonal matrix: en.wikipedia.org/wiki/Tridiagonal_matrix#Determinant $\endgroup$
    – EuxhenH
    Oct 3, 2020 at 20:49
  • 1
    $\begingroup$ I may have potentially found a nice simplification $\endgroup$ Oct 3, 2020 at 20:59
  • $\begingroup$ @Buraian did you ever find the nice simplification? $\endgroup$
    – fewfew4
    Oct 23, 2020 at 19:18
  • $\begingroup$ I did but it was too big to be written as an answer :-) $\endgroup$ Oct 23, 2020 at 19:25

2 Answers 2


Your matrix is a general tridiagonal matrix, with $d_i:=a_i+b_i+c_i$ along the diagonal. If we denote the determinant of the $n\times n$-matrix by $f_n$, then we have the recurrence relation $$f_n=d_nf_{n-1}-b_{n-1}c_{n-1}f_{n-2}.$$ Not much more can be said for general sequences $b_n$, $c_n$ and $d_n$. For more information see Wikipedia.


I believe I have an explicit solution!

Using the case that I had already figured out (when $a_k=0$), we can Taylor expand around this solution. For finite $n$, this will be a finite expansion.

First I define the quantity $\theta_{km}$, with $1\leq k,m\leq n$, which satisfies the following recursive relations

$$\theta_{km}=(c_m+b_m+a_m)\theta_{k,m-1}-b_{m-1}c_m\theta_{k,m-2},\quad \theta_{kk}=c_k+b_k+a_k,\quad \theta_{k,k-1}=1$$ $$\theta_{km}=(c_k+b_k+a_k)\theta_{k+1,m}-b_{k}c_{k+1}s\theta_{k+2,m},\quad \theta_{mm}=c_m+b_m+a_m,\quad \theta_{m+1,m}=1$$ and $\theta_{km}=0$ when $k> m+1$ and $m< k-1$.

Note that this quantity combines the $\theta_n$ and $\phi_n$ which is defined in this Wikipedia article. And $\theta_{1n}$ is the determinant of the matrix.

When $a_k=0$, this quantity has an explicit solution:


Using the recursive relations, it can be shown that this quantity satisfies


Thus the general solution for nonzero $a_k$ is

$$\theta_{1n}+\sum_{k=1}^n\theta_{1k-1}a_k\theta_{k+1n}+\cdots+\sum_{k_1\cdots k_p=1}^n\theta_{1k_1-1}a_{k_1}\theta_{k_1+1,k_2-1}\cdots a_{k_p}\theta_{k_p+1,n}+\cdots+a_1\cdots a_n$$

Where all of the $\theta$'s in the above expression are for the case where $a_k=0$.

To tidy up the formula a bit more, one can note that $(a\theta)_{nm}=a_n\theta_{n+1,m-1}$ is a nilpotent upper triangular matrix. So this formula can actually be cast as


That's about as explicit as I can do for now.


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