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Determinant of an nxn matrix

  1. Sep 13, 2016 #1
    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. jcsd
  3. Sep 13, 2016 #2

    fresh_42

    Staff: Mentor

    Have you tried an induction over ##n##?
     
  4. Sep 13, 2016 #3

    Charles Link

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    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
  5. Sep 13, 2016 #4

    LCKurtz

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    Gold Member

  6. Sep 14, 2016 #5
    Thank you! The formula for the vandermonde determinant was exactly what I needed!
     
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