Determinant of symmetric matrix with non negative integer element

  • Thread starter golekjwb
  • Start date
  • #1
4
0
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:

Answers and Replies

  • #2
4
0
Try n=4
 
  • #3
chiro
Science Advisor
4,797
133
Try n=4

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.
 
  • #4
tiny-tim
Science Advisor
Homework Helper
25,836
251
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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