Q:

Prove that a matrix is invertible

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.

A:

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.

