# Determinant with Exponents

• Apr 27th 2006, 01:03 PM
ThePerfectHacker
Determinant with Exponents
I was working on a problem, I need to show that a certain type of equation has a unique solution. By Cramer's Rule the problem reduces to:
$\left| \begin{array}{ccccc} 1^0&2^0&3^0&....&n^0\\ 1^1&2^1&3^1&....&n^1\\1^2&2^2&3^2&....&n^2\\....&. ...&....&....&....\\1^n&2^n&3^n&....&n^n \end{array} \right| \not = 0$
I asked my teacher he said there is a name for it, but did not know much about it. Someone know what this is? Or how to prove it?
• Apr 27th 2006, 01:24 PM
Rebesques
• Apr 27th 2006, 03:00 PM
ThePerfectHacker
Thank you :) really helps. Now I can complete the proof.
• Apr 27th 2006, 06:14 PM
ThePerfectHacker
You should have caught my mistake, it must have been,
$
\left| \begin{array}{ccccc} 1^0&2^0&3^0&....&n^0\\ 1^1&2^1&3^1&....&n^1\\1^2&2^2&3^2&....&n^2\\....&. ...&....&....&....\\1^{n-1}&2^{n-1}&3^{n-1}&....&n^{n-1} \end{array} \right| \not = 0
$

Because otherwise it is non-square.
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Can you demonstrate that for the regular Vandermonde Matrix
$\Delta =\prod_{i>j}(x_i-x_j)$
• Apr 27th 2006, 10:34 PM
TD!
A proof can be found here.