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Let $x_1, \ldots, x_n$ be a collection of distinct elements of a field $\mathbb{F}$ and let $B \in M_n(\mathbb{F})$ be the matrix with $a_{ij} = x^{j-1}_i$. Prove that $\operatorname{det}B= \prod_{i<j}(x_j - x_i)$. And find the matrix $B^{-1}$.

So I know that the matrix has the form

$\begin{pmatrix} 1 & x_1 & x_1^2 &\dots &x_1^{n-1}\\ 1 & x_2 & x_2^2 &\dots &x_2^{n-1}\\ \vdots &&&&\vdots\\ 1 & x_n & x_n^2 &\dots &x_n^{n-1}\\ \end{pmatrix}$

Using row operations this matrix can become

$\begin{pmatrix} 1 & x_1 & x_1^2 &\dots &x_1^{n-1}\\ 0 & x_2-x_1 & x_2^2-x_1^2 &\dots &x_2^n-x_1^{n-1}\\ \vdots &&&&\vdots\\ 0 & x_n-x_{n-1} & x_n^2-x_{n-1}^2 &\dots &x_n^n-x_{n-1}^{n-1}\\ \end{pmatrix}$

Not completely sure what to do from here. And I don't know how to find $B^{-1}$.

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This might be of help. – Git Gud Feb 12 '13 at 7:29
Your last column should have $x_{i}^{n-1}$ to make the matrix square. – Andreas Caranti Feb 12 '13 at 7:40
I'd also recommend this Inverse of Vandermonde's Matrix – Kaster Feb 12 '13 at 7:55
@user60504 Do you still need help with this? – Git Gud Feb 12 '13 at 19:54
No. I think I got it. Thanks for all your help! – user4593 Feb 12 '13 at 20:15

A way to calculate the determinant is to see it as a polynomial in $x = x_{n}$. Clearly it has roots $x_{1}, \dots , x_{n-1}$ (a matrix with two equal rows has determinant zero), Thus the determinant is $$ A \cdot (x-x_{1}) \cdot \dots \cdot (x-x_{n-1}), $$ where $A$ is a polynomial in $x_{1}, \dots , x_{n-1}$. Now notice that $A$ is the coefficient of $x^{n-1}$ in this polynomial, and expanding the determinant with respect to the last row you see that this is the same (Vandermonde) determinant of size one less.

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