# Solving Vandermonde equation system

Given the following:

$$\begin{pmatrix} 1 & 1 & 1 & ... & 1 \\ a_1 & a_2 & a_3 & ... & a_n \\ a_1^2 & a_2^2 & a_3^2 & ... & a_n^2 \\ \vdots & \vdots & \vdots & \vdots & \vdots \\ a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & ... & a_n^{n-1}\end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \\ \vdots \\ x_n\end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ 0 \\ \vdots \\ 1 \end{pmatrix}$$

How can this be solved to give $$x_k$$? By Cramer's rule, this should be $$x_k=\frac{\det(A_i)}{\det(A)}$$

How can this result be simplified? Since this is a Vandermonde matrix, $$\det(A)=\prod_{1\leq i. However, can the numerator be written any simpler? In a paper where I have seen this equation system used, the result is given as: $$x_k=\prod_{1\leq j \leq n,\;j \neq i} (a_i - a_j)^{-1}$$ but it's not clear how the simplification has been achieved?

Using Cramer's rule, we know

$$x_k = \frac{\det(A_k)}{\det(A)}$$

Now, as we know the Vandermonde determinant is equal to

$$\det(A)=\prod_{1\leq i

To calculate $$\det(A_k)$$, we replace the kth column with $$b_i$$s which in this case are all $$0$$s and last entry $$1$$. If we calculate the determinant using the kth column, we need only consider the determinant using the last entry as all the other ones will multiply with $$0$$ and become $$0$$. This last entry sub determinant also turns out to be a vandermonde determinant of one lower degree. So calculating the determinant using that

$$\det(A_k) = (-1)^{n+k}\prod_{1\leq i

where both $$i,j\neq k$$

When you divide them, all the terms that include $$a_k$$ are all that are left.

$$x_k = \left[(-1)^{n+k}\left(\prod_{k+1\leq j\leq n}(a_j-a_k) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$

The terms which have greater index than $$a_k$$ come under one umbrella with the greater index term leading and the ones with lower come under a second umbrella with $$a_k$$ leading.

Rearrange all the terms in first umbrella so that $$a_k$$ term is leading and take all the $$-1$$s that come out and multiply it with existing to get your answer .

$$x_k = \left[(-1)^{n+k}\left((-1)^{n-(k+1)}\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$

$$x_k = \left[(-1)^{n+k+n-k-1}\left(\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$

$$x_k = \left[(-1)^{2n-1}\left(\prod_{k+1\leq j\leq n}(a_k-a_j) \prod_{1\leq i\leq k-1}(a_k-a_i)\right)\right]^{-1}$$

$$x_k = \left[\prod_{1\leq i\leq n, i\neq k}(a_i-a_k)\right]^{-1}$$

• There might be a $-1$ multiplication/divide error – Sauhard Sharma Nov 14 '18 at 19:27