Determinant of an nxn matrix

1. Sep 13, 2016

grassstrip1

The problem I have is this: Show that

\begin{bmatrix} 1 & 1 & 1 \\ λ_{1} & λ_{2} & λ_{3} \\ λ_{1}^{2} & λ_{2}^{2} & λ_{3}^{2} \end{bmatrix}
Has determinant

$$(λ_{3} - λ_{2}) (λ_{3} - λ_{1}) (λ_{2} - λ_{1})$$

And generalize to the NxN case (proof not needed)

Obviously solving the 3x3 was not hard, I simply expanded the expression for the determinate given and showed it to be the same as the one i calculated using the rule of Sarrus.

However for the nxn case I'm not sure how to proceed. I tried entering bigger and bigger matrices into wolfram but there was no clear pattern. Is there a non computational way to solve this?

2. Sep 13, 2016

Staff: Mentor

Have you tried an induction over $n$?

3. Sep 13, 2016

I don't have a complete answer, but a 4x4 with the 4th row being cubic will result in a 6th power polynomial form. The number of combinations of 4 things taken 2 at a time is 6, making for the 6 factors. (I don't know how to choose the minus or plus sign for the terms, but if it does generalize, this might be helpful.)...editing... One additional observation is if two columns of a matrix are identical, I think the determinant is zero. This is consistent with setting the product of the factors equal to zero.

Last edited: Sep 13, 2016
4. Sep 13, 2016

LCKurtz

5. Sep 14, 2016

grassstrip1

Thank you! The formula for the vandermonde determinant was exactly what I needed!