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This question already has an answer here:

Let $n$ be a positive integer. Show that the matrix

$$\begin{pmatrix} 1 & 1/2 & 1/3 & \cdots & 1/n \\ 1/2 & 1/3 & 1/4 & \cdots & 1/(n+1) \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1/n & 1/(n+1) & 1/(n+2) & \cdots & 1/(2n-1) \end{pmatrix}$$

is invertible and all the entries of its inverse are integers. This is an exercise in Hoffman and Kunze's linear algebra book. Any hints will be appreciated!

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marked as duplicate by Theoretical Economist, Paul Frost, Namaste linear-algebra Sep 24 '18 at 0:08

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The matrix you have is a Hilbert matrix. The $(i,j)$ entries of its inverse are given by $$(-1)^{i+j}(i+j-1){n+i-1 \choose n-j}{n+j-1 \choose n-i}{i+j-2 \choose i-1}^2$$which are clearly integers.

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