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When I studied Linear Algebra over 55 years ago, we used a book by Hoffman and Kunze. I still cannot solve the last problem of Chapter $1$. I don't have the book any more but I remember the problem. It is: Show that for all $n$ the following matrix has an inverse with integer entries.
\begin{bmatrix} \frac {1}{2} & \frac {1}{3} ...\frac {1}{n+1}\\ \frac {1}{3} & \frac {1}{4} ...\frac {1}{n+2}\\ ...\\ \frac {1}{n+1} & \frac {1}{n+2} ...\frac {1}{2n}\\ \end{bmatrix}
The proof should only use the material that would be reasonably covered in Chapter $1$ of a reasonable Linear Algebra text. That means without determinants. I was able to confirm it for some small $n$; I think up to $6$.

After posing this question, it was alleged that it was a duplicate and it certainly looks that way, superficially. However,

  1. This question asks for an elementary solution. The solution(s) proposed to the alleged duplicate are not elementary. This is the primary difference.
    (To be elementary the solution should not involve determinants. For example, it could show that the linear transformation represented by the matrix is $1-1$ or onto and that the inverse is an integer matrix, or that the form of the inverses as $n$ grows follows a well-defined pattern, or that there is a well defined sequence of elementary row operations which row-reduces the matrix to the identity matrix and the product of which is a n integer matrix.)
  2. The entries in this matrix are $\frac {1}{i+j}$ but in the alleged duplicate they are $\frac {1}{i+j-1}$. Since I am working with an old memory, I am probably mistaken about the exact statement as originally posed in Hoffman and Kunze. But, in any case, the problems are different. It is not clear if this difference is significant or just a quibble.
  3. The title of this question is different than the alleged duplicate. People looking for the solution to this problem may not be aware that the matrix in question is called a Hilbert matrix.
Stephen Meskin
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    "we used a book by Hoffman and Kunze. I still cannot solve the last problem of Chapter 1." I also want to see a elementary solution to that problem. –  Nov 04 '17 at 07:05
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    @StephenMeskin so the solution provided in the linked answer is not elementary because it uses determinants, something that was said not to have appeared in chapter 1 (Haven't read the book myself). – rubikscube09 Nov 04 '17 at 07:16
  • @Lord Shark the Unknown et al I have added a defense of non-duplicity – Stephen Meskin Nov 04 '17 at 15:02
  • @StephenMeskin There's a thread somewhere on meta where you can request that the question be reopened. – Qudit Nov 04 '17 at 22:57
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    I tried to vote to reopen it, and then I got a pop-up saying "you have already voted to reopen it" :) – Harambe Nov 05 '17 at 06:08
  • Now that you know, Stephen, that it's called the Hilbert matrix, I bet you could find a good answer with a few minutes of searching the web. – Gerry Myerson Nov 05 '17 at 06:12
  • @Qudit Thanks, but I went to Meta and couldn't find such a thread. There was advice that I should edit the question. I have done that. – Stephen Meskin Nov 05 '17 at 16:51
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    @Qudit was probably referring to https://math.meta.stackexchange.com/questions/19042/requests-for-reopen-undeletion-votes-etc-volume-01-2015-current-versio – but editing puts it in the review queue, and I see two votes to reopen, so I'd suggest posting to meta only if you don't get what you want after giving the review queue some more time to act. – Gerry Myerson Nov 05 '17 at 22:54
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    @StephenMeskin Why in the title you have $1/(i+j-1)$ not $1/(i+j)$ ? – Widawensen Nov 06 '17 at 16:18
  • @Widawensen Someone must have edited my title without me noticing it. Thanks for the heads up. – Stephen Meskin Nov 06 '17 at 16:54
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    So, Stephen, have you taken up the websearch? – Gerry Myerson Nov 06 '17 at 22:56

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