Determinant of symmetric Matrix with non negative integer element

Question

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.

Answer 1

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?