Inverse of Vandermonde Matrix

Theorem
Let $V_n$ be the Vandermonde's matrix of order $n$ given by:


 * $V_n = \begin{bmatrix}

x_1 & x_2 & \cdots & x_n \\ x_1^2 & x_2^2 & \cdots & x_n^2 \\ \vdots & \vdots & \ddots & \vdots \\ x_1^n & x_2^n & \cdots & x_n^n \end{bmatrix}$

Then its inverse $V_n^{-1} = \left[{b}\right]_n$ can be specified as:


 * $b_{ij} = \dfrac {\displaystyle \sum_{\stackrel{1 \le k_1 < \ldots < k_{n-j} \le n} {k_1, \ldots, k_{n-j} \ne i}} \left({-1}\right)^{j-1} x_{k_1} \ldots x_{k_{n-j}}} {\displaystyle x_i \prod_{\stackrel {1 \le k \le n} {k \ne i}} \left({x_k - x_i}\right)}$