Determinant of symmetric Matrix with non negative integer element Let \begin{equation*}
M=%
\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-1 & \cdots & 1 & 0%
\end{bmatrix}%
\end{equation*}
How can you prove that $\det(M)=(-1)^n\cdot n \cdot 2^{n-1}$? 
I just guess the formula in the right hand side by observing the calculation for small n  but I can't prove for arbitrary n. Thanks everyone.
 A: Let's take a $4\times 4$ matrix (I don't want to type much).
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 0 & 1 & 2 \\
2 & 1 & 0 & 1 \\
3 & 2 & 1 & 0
\end{vmatrix} $$
Since adding a row into another does not change determinant values. Add $-i'th$ row into $i+1$'th row.
$$\begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & -1 & -1 & -1 \\
1 & 1 & -1 & -1 \\
1 & 1 & 1 & -1
\end{vmatrix} $$
Repeat the process with columns. 
$$\begin{vmatrix}
0 & 1 & 1 & 1 \\
1 & -2 & 0 & 0 \\
1 & 0 & -2 & 0 \\
1 & 0 & 0 & -2
\end{vmatrix} = \frac{1}{2}\begin{vmatrix}
0 & 1 & 1 & 1 \\
2 & -2 & 0 & 0 \\
2 & 0 & -2 & 0 \\
2 & 0 & 0 & -2
\end{vmatrix} = \frac{1}{2}\begin{vmatrix}
3 & 1 & 1 & 1 \\
0 & -2 & 0 & 0 \\
0 & 0 & -2 & 0 \\
0 & 0 & 0 & -2
\end{vmatrix}$$
Now what you can say about its determinant?
A: You can prove it by method of induction
For n =2, that is 2x2 determinant
det[M] = -1, so the formula is correct
Now assume that the formula is correct for n that is
det [M] = (−1)^n*n*2^n−1
Now consider n+1 x n+1 determinant. Treat the first nxn rows columns as cofactor call it A. Now it is easy to prove for n+1
By induction then the formula is proved.
Thanks  
