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,
- 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.) - 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.
- 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.
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