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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.

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Are you sure that element at the bottom is $n-2$ and not $n-1$? –  J. M. May 14 '12 at 8:43
    
Sorry, that's my mistake. I just fix it. –  beginner May 14 '12 at 8:47
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2 Answers 2

up vote 5 down vote accepted

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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+1 that's nice $ $ –  draks ... May 14 '12 at 9:05
    
How is the idea to generalize that to arbitrary n –  beginner May 14 '12 at 9:07
    
Repeat the same process with $n \times n$ matrix. –  Dilawar May 14 '12 at 9:09
    
Why the elementary column operation don't change the determinant? –  beginner May 14 '12 at 9:12
    
Adding a column into another does not change the determinant. See the definition of determinant (sum of permutation). Its a standard result, you can look up wikipedia. –  Dilawar May 14 '12 at 9:17
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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

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