# linear algebra: inverse of a matrix

The inverse of the matrix

$A=\left( \matrix{1 & \frac{1}{2} & \frac{1}{3} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} \\ \frac{1}{3} & \frac{1}{4} & \frac{1}{5} }\right)$

is

$A^{-1}=\left( \matrix{ 9 & -36 & 30 \\ -36 & 192 & -180 \\ 30 & -180 & 180 } \right)$.

Then, perhaps the matrix

$B=\left( \matrix{1 & \frac{1}{2} & ... & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & ... & \frac{1}{n+1} \\ ... & ... & & ...\\ \frac{1}{n} & \frac{1}{n+1} & ... & \frac{1}{2n-1} }\right)$

is invertible and $B^{-1}$ has integer entries. How can I prove it?

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This is a Hilbert matrix and the elements of its inverse can be represented as the product of binomial coefficients. This might help you get started. –  mathematician1975 Jul 1 '12 at 10:32
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## 1 Answer

Here you will find your answer and many other things about Hilbert matrices : http://www.jstor.org/stable/2975779

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Especially for those who don't access to the article through JSTOR, will you please indicate how the article gives an answer to this question? –  Jonas Meyer Jul 1 '12 at 10:48
Google spits out quite a few additional links for that paper: 1 2 and also a related MO thread. –  Martin Sleziak Jul 1 '12 at 14:16
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