# Prove that a matrix is invertible [duplicate]

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, Nicholas R. Peterson, Brandon Carter, Omnomnomnom, 40 votesJul 29 '13 at 23:24

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