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I tried to find the inverse matrix of

$$ \begin{pmatrix} 2 & -1 & 0 & \cdots& 0 \\ -1 & 2 & -1 & \cdots& 0\\ 0 & -1 & 2 & \ddots& \vdots\\ \vdots & \vdots & \ddots & \ddots &-1\\ 0 & 0 & \cdots & -1& 2\\ \end{pmatrix} $$

unsuccesfuly.

  • For $n=2$ the inverse matrix is $$\frac 1 3 \begin{pmatrix} 2 & 1 \\ 1 & 2 \\ \end{pmatrix} $$

  • For $n=3$ the inverse matrix is

$$ \frac 1 4\begin{pmatrix} 3 & 2 & 1 \\ 2 & 4 & 2 \\ 1 & 2 & 3 \\ \end{pmatrix} $$

  • In general I figured out:

$$ \frac 1 {n+1} \begin{pmatrix} n & n-1 & n-2 & \cdots& 1 \\ n-1 & ? & ? & ?& 2\\ n-2 & ? & ? & ? & \vdots\\ \vdots &? & ? & ? &n-1\\ 1 & 2 & \cdots & n-1& n\\ \end{pmatrix} $$

I just don't know how to write it generally even though I see those numbers in the middle of my matrices for $n = 2$ and $n = 3$. I tried it even for the larger $n$.

Leif
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1 Answers1

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Your examples suggest that the $(i,j)$-th entry in the upper triangular part (i.e. when $i\le j$) is $\frac{1}{n+1}i(n+1-j)$. It should be straightforward to verify this formula.

user1551
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