Date: 02/11/2003 at 19:55:04 From: Roger Subject: Determinant Compute the n x n determinant: | 1 1 1 ... 1 | | x1 x2 x3 xn | | x1^2 x2^2 x3^2 ... xn^2 | | . . | | . . | | . . | | x1^(n-1) x2^(n-1) x3^(n-1) ... xn^(n-1) |
Date: 02/13/2003 at 09:11:40 From: Doctor Jacques Subject: Re: Determinant Hi Roger, Let us consider a simple example, with n = 3. The determinant is: | 1 1 1 | D= | x y z | | x^2 y^2 z^2 | If you expand this determinant, you will get a polynomial in x, y, z. Each term in this polynomial is of the form: 1*a*b^2 where a and b are taken from x, y, z. This means that each term is of degree 3. Notice now that, if x = y, the determinant has two equal columns, and is therefore equal to 0. This means that our polynomial is divisible by (x-y). In a similar way, D is divisible by (y-z) and (z-x). We conclude that: D = (x-y)(y-z)(z-x)*p(x,y,z) where p(x,y,z) is an unknown polynomial in x, y, z. However, (x-y)(y-z)(z-x) is already of degree 3, so p(x,y,z) is just a constant, say A. We have thus: D = A(x-y)(y-z)(z-x) It remains to find the value of A. This can be done by looking at the coefficient of a particular term, for example y*z^2. You should have no trouble generalizing this to higher degrees. Note that the degree of the polynomial is n(n-1)/2, and this is just (by chance?) the number of pairs of different variables. This determinant is called a Vandermonde determinant. Please feel free to wrie back if you require more assistance. - Doctor Jacques, The Math Forum http://mathforum.org/dr.math/
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