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Find the determinant of $$\begin{bmatrix}1 & a & a^2 & a^{3}\\ 1 & b & b^{2} & b^{3}\\ 1 & c & c^{2} & c^{3}\\ 1 & d & d^{2} & d^{3} \end{bmatrix}$$

This seems super ugly to attack directly by using cofactors; is there any way to simplify this?

Rodrigo de Azevedo
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ithisa
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    Please see this. Key word is Vandermonde. – André Nicolas Mar 27 '13 at 20:40
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    This page has several proofs http://www.proofwiki.org/wiki/Vandermonde_Determinant – Quimey Mar 27 '13 at 20:41
  • Proving the general result seems to be overkill. – ithisa Mar 27 '13 at 20:43
  • The general case is not fundamentally harder than the $4\times 4$. Anyway, you can take the idea for a general proof and specialize it to the $4\times $$. – André Nicolas Mar 27 '13 at 20:48
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    If you view the determinant as a function of $a$, evidently it's a polynomial of degree $3$. By inspection--and understanding the basic properties of determinants--you can identify three roots immediately, so you know it up to some multiple that depends only on $b$, $c$, and $d$. Repeating this observation for the other variables pins down the determinant up to a constant. By choosing nice values for $a$, $b$, $c$, and $d$, you can compute that constant--and have thereby reduced the problem to finding the determinant of a single matrix with numerical entries. – whuber Mar 27 '13 at 20:52
  • I don't quite understand. Could you post a detailed answer? Also, note that this is the first assignment after the introduction of the concept of determinant. – ithisa Mar 27 '13 at 21:02
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    Eric: The proof of the general result is not overkill. Chances are that by giving this exercise, your teacher wants you to either find one of the ideas that prove the general Vandermonde determinant, or to get your hands dirty and compute it by brute force after which the general proof will feel like a cakewalk. To be honest, it'd probably be best for your education to do both. :) – darij grinberg Mar 28 '13 at 00:52
  • See: http://math.stackexchange.com/questions/525334/vandermonde-determinant-by-induction and http://math.stackexchange.com/questions/275202/vandermonde-determinant – Martin Sleziak Jan 27 '14 at 17:34

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There is a way to simplify this. You can row reduce it. For starters, add the negative of the first row to every other row to get zeroes along the first column. You can keep doing this until the matrix is upper triangular. By then, you just multiply the diagonal entries.

noobProgrammer
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  • Yes, I know this. I'm not saying that the thing can't be computed; it's just that you end up with a huge polynomial that needs ugly grouping and factoring to get to the one gotten by the Vandermonde formula. Is there a more straightforward way to get a clean formula? – ithisa Mar 28 '13 at 01:49
  • If every element in a row contains the same factor, you can pull that out. If you try the suggested row reduction, and then pull out common factors, the next step of row reduction will not be so bad. You will have to make use of an algebra fact, namely that $x^n-y^n$ factorizes. – Will Orrick Mar 28 '13 at 03:09
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A direct computation, together with a factorization is not difficult at all. We obtain $$ \det\begin{pmatrix}1 & a & a^2 & a^{3}\\ 1 & b & b^{2} & b^{3}\\ 1 & c & c^{2} & c^{3}\\ 1 & d & d^{2} & d^{3} \end{pmatrix}= (a - b)(a - c)(a - d)(b - c)(b - d)(c - d) $$

Dietrich Burde
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