# How to find deteminant of the tridiagonal matrix

Given the matrix:

$$A = \begin{pmatrix} x & 1 & 0 & 0 & \cdots & 0 & 0 \\ n - 1 & x & 2 & 0 & \cdots & 0 & 0 \\ 0 & n - 2 & x & 3 & \cdots & 0 & 0 \\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & 0 & \cdots & x & n - 1 \\ 0 & 0 & 0 & 0 & \cdots & 1 & x \end{pmatrix}$$

How to find the determinant of $A$. I considered examples for $n = 3,4,5,6$ and obtained some expressions. But I can't understand what the expression for the general case.

I guess that the formula is like the following: $$\det(A) = x^n - x^{n-2}\sum_{i = 1}^{n - 1}( i \cdot (n - i) ) + ...$$

• There is a linear recurrence of order $2$ for the determinant of a general tridiagonal matrix. – Bernard Nov 18 '15 at 1:06
• @Bernard $A_n = x A_{n - 1} + (n - 1) A_{n - 2}$ ? – J.Exactor Nov 18 '15 at 1:10
• $-$, not $+$, as far as I remember. – Bernard Nov 18 '15 at 1:14
• @Bernard $A_k = x A_{k - 1} - (k - 1) (n - k + 1) A_{k - 2}$ ? – J.Exactor Nov 18 '15 at 1:19
• I would say so, yes. – Bernard Nov 18 '15 at 1:30

Hint. The matrix in question is in the form of $xI+K$, where $K$ is a Kac matrix whose spectrum has a closed-form formula.