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Determinant of symmetric matrix with non negative integer element

  1. May 14, 2012 #1
    Let \begin{equation*}
    A=%
    \begin{bmatrix}
    0 & 1 & \cdots & n-1 & n \\
    1 & 0 & \cdots & n-2 & n-1 \\
    \vdots & \vdots & \ddots & \vdots & \vdots \\
    n-1 & n-2 & \cdots & 0 & 1 \\
    n& n-2 & \cdots & 1 & 0%
    \end{bmatrix}%
    \end{equation*}.
    How can you prove that det(A)=[(-1)^n][n][2^(n-1)]? Thanks.
     
    Last edited: May 14, 2012
  2. jcsd
  3. May 14, 2012 #2
    Try n=4
     
  4. May 14, 2012 #3

    chiro

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    Science Advisor

    Hey golekjwb and welcome to the forums.

    For the general proof I would use an induction argument. The differences between say n = k and n = k + 1 has to do with evaluating one extra minor determinant for that extra row and you would show that under a simplification that the formula is correct.

    For a specific n=4, just evaluate the determinant for that particular dimension for your particular matrix, expand out and see what you get.
     
  5. May 14, 2012 #4

    tiny-tim

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    Science Advisor
    Homework Helper

    welcome to pf!

    hi golekjwb! welcome to pf! :smile:

    have you tried adding or subtracting rows or columns, to get a simpler matrix?
     
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