# Determinant of symmetric matrix with non negative integer element

1. May 14, 2012

### golekjwb

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. May 14, 2012

### golekjwb

Try n=4

3. May 14, 2012

### chiro

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. May 14, 2012

### tiny-tim

welcome to pf!

hi golekjwb! welcome to pf!

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