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Let $x_1,x_2,\ldots,x_n$ be $n$ real numbers that satisfy $x_1. Define \begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & \cdots & x_{n-1}-x_{1} & x_{n}-x_{1} \\ x_{2}-x_{1} & 0 & \cdots & x_{n-1}-x_{2} & x_{n}-x_{2} \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ x_{n-1}-x_{1} & x_{n-1}-x_{2} & \cdots & 0 & x_{n}-x_{n-1} \\ x_{n}-x_{1} & x_{n}-x_{2} & \cdots & x_{n}-x_{n-1} & 0% \end{bmatrix}% \end{equation*}

Could you determine the determinant of $A$ in term of $x_1,x_2,\ldots,x_n$?

I make a several Calculation: For $n=2$, we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} \\ x_{2}-x_{1} & 0% \end{bmatrix}% \text{ and}\det (A)=-\left( x_{2}-x_{1}\right) ^{2} \end{equation*}

For $n=3$, we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & x_{3}-x_{1} \\ x_{2}-x_{1} & 0 & x_{3}-x_{2} \\ x_{3}-x_{1} & x_{3}-x_{2} & 0% \end{bmatrix}% \text{ and}\det (A)=2\left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right) \left( x_{3}-x_{1}\right) \end{equation*}

For $n=4,$ we get

\begin{equation*} A=% \begin{bmatrix} 0 & x_{2}-x_{1} & x_{3}-x_{1} & x_{4}-x_{1} \\ x_{2}-x_{1} & 0 & x_{3}-x_{2} & x_{4}-x_{2} \\ x_{3}-x_{1} & x_{3}-x_{2} & 0 & x_{4}-x_{3} \\ x_{4}-x_{1} & x_{4}-x_{2} & x_{4}-x_{3} & 0% \end{bmatrix} \\% \text{ and} \\ \det (A)=-4\left( x_{4}-x_{1}\right) \left( x_{2}-x_{1}\right) \left( x_{3}-x_{2}\right) \left( x_{4}-x_{3}\right) \end{equation*} Finally, I guess that the answer is $\det(A)=2^{n-2}\cdot (x_n-x_1)\cdot (x_2-x_1)\cdots (x_n-x_{n-1})$. But I don't know how to prove it.

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    Please remember that in order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are; this will prevent people from telling you things you already know, and help them give their answers at the right level. Also, many users find the use of the imperative ("Prove", "Show", etc) to be rude when asking for help. Please consider rewriting your post.2012-05-14
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    Your guess would not account for the case $n=2$...2012-05-14
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    For $n=2$ is trivial. Maybe for $n>2$2012-05-14
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    I try math induction, but don't know how to connect n by n matrix with (n+1) by (n+1) matrix2012-05-14
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    I'll just note that the matrices being considered by the OP, 1. are very similar to Cauchy matrices; 2. can be expressed as the difference of two [type D matrices](http://math.stackexchange.com/questions/44511), and 3. have inverses that are cyclic tridiagonal (tridiagonal with extra elements in the upper right and lower left corners).2012-05-14
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    Thanks J.M. I will learn it2012-05-14

2 Answers 2