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The result of Example 16 suggests that perhaps the matrix

$A=\begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} \ \ \cdots \ & \frac{1}{n} \\ \frac{1}{2}& \frac{1}{3} & \cdots & \frac{1}{n+1}\\ \frac{1}{3} & . & \cdots & .\\. & . & \cdots & .\\\frac{1}{n} & \frac{1}{n+1} & \cdots & \frac{1}{2n-1} \end{bmatrix}$

is invertible and $A^{-1}$ has integer entries. Can you prove that?

Grobber
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    This is all you need: http://en.wikipedia.org/wiki/Hilbert_matrix – DonAntonio Jun 29 '13 at 16:25
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    Well its definately not a homework problem. Basically I'm trying to understand Linear Algebra on my own from the book by Hoffman and Kunje and found this problem left as an exercise. – Grobber Jun 29 '13 at 18:55
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    @Grobber It is always helpful to say what you have tried. People are more likely to answer a question if it seems you have thought about it a bit, and also saying what you have tried helps them to direct their help. – user1729 Jun 29 '13 at 19:48

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