# How to evaluate this determinant?

Can someone give me a hint how to solve $$\left|\begin{array}{ccccc} 1 & 1 & \ldots & & 1\\ 2x_{1} & 2x_{2} & & & 2x_{n}\\ \vdots\\ nx_{1}^{n-1} & nx_{2}^{n-1} & \ldots & & nx_{n}^{n-1}\\ \\ \end{array}\right|$$ ?

I know that I somehow have to use the Vandermonde determinant to do this, but I can't figure out how to get rid of the coefficients. Can someone give me a hint please ?

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Here's a hint: multiplying a matrix on the left by a diagonal matrix does what?

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perfect, thanks – arila Nov 7 '11 at 17:33

Because $\det A = \sum \epsilon_{i_1, i_2, \ldots, i_n} a_{1,i_1} \cdot a_{2,i_2} \cdots a_{n,i_n}$. Since $k x_i^{k-1} = \partial_{x_i} x_i^k$, determinant of your matrix $A$ is $$\det A = \partial_{x_1, x_2, \ldots, x_n} \left( x_1 x_2 \cdots x_n W(x_1, x_2, \ldots, x_n) \right)$$ where $W(x_1, x_2, \ldots, x_n)$ is the determinant of the Vandermonde matrix.

$$\det A = \partial_{x_1, x_2, \ldots, x_n} \left( x_1 x_2 \cdots x_n \prod_{i < j} (x_i - x_j) \right)$$

It actually looks like the derivative can be found in closed form, with the result of $$\det A = n! \prod_{i < j} (x_i - x_j)$$