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.
Determinant of symmetric Matrix with non negative integer element
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0Sorry, that's my mistake. I just fix it. ā 2012-05-14
2 Answers
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?
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0@Dilawar: How can you find such a briliant idea quickly (I mean the selection of elementary column/row operation)? Is it a routine problem? ā 2012-05-14
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] = (ā1)^n*n*2^nā1 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