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Proving a formula for the determiante in a special matrix

  1. Feb 24, 2009 #1
    1. The problem statement, all variables and given/known data

    Let a1, a2 are real numbers, where n > 1 show that: determinant of:

    | 1 a1 a21 .... .... an-11 |
    | 1 a2 a22 .... .... an-12 |
    :
    :
    | 1 an a2n .... .... an-1n |


    = [tex]\prod [/tex] (aj - ai)
    1[tex]\leq[/tex]i<j<n

    2. Relevant equations
    if you row reduce a matrix the determinate is the product of the leading diagonal(previous question was finding determinate of matrices by row reducing them)

    3. The attempt at a solution
    tried using induction but get stuck very quickly.
    i got RHS =
    = [tex]\Pi [/tex]0<i<j<n (aj - ai) [tex]\Pi [/tex]0<k<n (an - ak)
     
    Last edited: Feb 24, 2009
  2. jcsd
  3. Feb 25, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi rosh300! :smile:

    But that's the answer, isn't it?

    The first part is all non-identical pairs up to n-1, and the second part is all non-identical pairs of which the higher is n.
     
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