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Is there a quick way to invert the $n\times n$ Vandermonde matrix with columns $(1, x, x^2, ..., x^{n-1})$ where $x$ takes values $0,1,...,n-1$ (in ascending order from left to right)?

Perhaps by row operations $(A|I)\to (I|A^{-1})$ where $A$ is the Vandermonde matrix, and $I$ the identity matrix? I don't see how to do it though... or maybe there is another way?

Thank you.

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    See [here](http://www.proofwiki.org/wiki/Inverse_of_Vandermonde's_Matrix).2011-11-19
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    The "quick" ($O(n^2)$) method is to use [divided differences](http://mathworld.wolfram.com/DividedDifference.html) (i.e., the [Björck-Pereyra algorithm](http://dx.doi.org/10.1090/S0025-5718-1970-0290541-1).)2011-11-19
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    Somewhat related, the determinant of such a matrix, for fixed $n$, is $G(n+1)$, where $G$ the [Barnes G-function](http://en.wikipedia.org/wiki/Barnes_G-function)2011-11-19

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