# Prove that a matrix is invertible [duplicate]

This question already has an answer here:

Show that the matrix

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

is invertible and $A^{-1}$ has integer entries.

This problem appeared in Chapter one of my linear algebra textbook so I assume nothing more is required to prove it than elementary row operations. I've been staring at it for a while and considered small values of $n$ but there doesn't seem to be a clever way of easily reducing it to the identity matrix. Could someone give me a hint or a poke in the right direction rather than a full solution as I'm still hoping to work it out myself.

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## marked as duplicate by Potato, Nick Peterson, Brandon Carter, Omnomnomnom, 40 votesJul 29 '13 at 23:24

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Very nice situation. Today I asked the same question and got this page as a link. I saw the question many times previously but today I became very much interested to know the solution. Many student has faced the same several times. – Dutta Dec 26 '13 at 16:28

## 1 Answer

If your linear algebra textbook is Kenneth Hoffmann and Ray Kunze's Linear Algebra book then I was in the same situation you're right now a few years ago, but I was adviced at the moment that this particular exercise wasn't as easy as one might expect at first.

The matrix you have is called the Hilbert matrix and the question you have was already asked a couple of times in math overflow here and here. They have excellent answers so I will just point you to them.

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Thanks! I am using Hoffman and Kunze and according to the comments in your second link, it seems that a simple proof is possible with the concept of a determinants, but they're not presented yet in Hoffman and Kunze. It's nice to know what the matrix is called though! – Herman Chau Jun 22 '11 at 11:05
Thanks, those two links were very helpful. Also, it's nice to know what the matrix is called! – Herman Chau Jun 22 '11 at 11:06
@Herman Chau I'm glad it helped you. And yes, it is good to know that the matrix has a name, since Hoffmann and Kunze's book didn't say anything about it. – Adrián Barquero Jun 22 '11 at 11:10