# The inverse of the matrix $\{1/(i+j-1)\}$ [duplicate]

Let $n$ be a positive integer. Show that the matrix

$$\begin{pmatrix} 1 & 1/2 & 1/3 & \cdots & 1/n \\ 1/2 & 1/3 & 1/4 & \cdots & 1/(n+1) \\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1/n & 1/(n+1) & 1/(n+2) & \cdots & 1/(2n-1) \end{pmatrix}$$

is invertible and all the entries of its inverse are integers. This is an exercise in Hoffman and Kunze's linear algebra book. Any hints will be appreciated!

## marked as duplicate by Theoretical Economist, Paul Frost, Namaste linear-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 24 '18 at 0:08

The matrix you have is a Hilbert matrix. The $(i,j)$ entries of its inverse are given by $$(-1)^{i+j}(i+j-1){n+i-1 \choose n-j}{n+j-1 \choose n-i}{i+j-2 \choose i-1}^2$$which are clearly integers.