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My question is similar to How to compute the determinant of a tridiagonal matrix with constant diagonals? However, all of the options seem to include computations that are not permitted for integers. I am trying to find the determinant for a matrix in integers, more specifically, this matrix:

$X = \begin{bmatrix} 2 & -1 & 0 & 0 & \cdots & 0 & 0 \\ -1 & 2 & -1 & 0 & \ddots & 0 & 0 \\ 0 & -1 & 2 & -1 & \ddots & 0 & 0 \\ \vdots & \ddots & \ddots & \ddots & \ddots & \vdots & \vdots \\ 0 & 0 & 0 & \cdots & -1 & 2 & -1 \\ 0 & 0 & 0 & \cdots & 0 & -1 & 2 \\ \end{bmatrix} \in \mathbb{Z}^{n,n}$

Computing the determinant for smaller $n$ with the Rule of Sarrus gave me the suspicion that it is always $n + 1$, however, how do I show this for larger $n$ as well?

Max
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  • You can't use Sarrus on a $n\times n$ matrix with $n\geq 4$. – Davide Trono Jan 13 '21 at 22:47
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    I know, I did it for $n < 4$ which gave me my suspicion. Now I am unsure how to confirm my suspicion or get a new one that also holds for larger $n$ – Max Jan 13 '21 at 22:50

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While I don't quite understand why you only want to use integers, you can prove this by induction. Let $d_n$ be the determinant of your matrix of size $n \times n$. Then $d_1 = 2$, $d_2 = 3$. So assume that $d_k = k+1$ for all $k < n$. By Laplace expansion and the induction hypothesis, we get $$d_n = 2d_{n-1} - d_{n-2} = 2n - (n-1) = n+1$$ and we are done.

Klaus
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  • @Max You don't get $d_{n-1}$ for the second term. You get $1$ times a different determinant, which happens to be $(-1)$ times $d_{n-2}$ after another Laplace expansion. – Klaus Jan 19 '21 at 13:39
  • I realised that quickly after posting and deleted my comment. Thanks again! – Max Jan 19 '21 at 15:20