How can I prove that this matrix is nonsingular?

How can I prove that $\begin{vmatrix} \frac{1}{(x_1-y_1)^2} & \frac{1}{(x_1-y_2)^2} & \cdots & \frac{1}{(x_1-y_n)^2} \\ \frac{1}{(x_2-y_1)^2} & \frac{1}{(x_2-y_2)^2} & \cdots & \frac{1}{(x_2-y_n)^2} \\ \vdots & \vdots & \ddots & \vdots \\ \frac{1}{(x_n-y_1)^2} & \frac{1}{(x_n-y_2)^2} & \cdots & \frac{1}{(x_n-y_n)^2} \end{vmatrix} \ne 0$?

For $x_i$ and $y_i$ are all distinct.

• Is there any unspecified assumptions about $x_i$ and $y_i$, such as $x_i$ are all distinct? Commented Nov 14, 2015 at 14:52
• If $x_1=x_2$ and $y_1=y_2$ then you fail for $n=2$. Commented Nov 14, 2015 at 15:09
• Usually, what one does is to fix the values for $y$ (all distinct because otherwise the matrix is singular), clear your fractions, find the degree of the corresponding polynomial in the $x_i$'s and try to see if there are enough obvious answers to to the polynomial to account for all the easy solutions. Commented Nov 14, 2015 at 15:11
• We need all $x_i$ and all $y_j$ to be distinct. If $x_i =x_k$ for some $i,k$, then two of the rows would be identical and so the determinant would be zero. The same follows for the $y_j$'s except that two or more columns would be identical. Commented Nov 14, 2015 at 15:13

Even if all the $$x_i$$ and $$y_j$$ are distinct, this is not true.
For instance, if $$x_1=2/3$$, $$x_2=2$$, $$y_1=1$$, $$y_2=0$$, we have $$\begin{vmatrix}\frac1{(x_1-y_1)^2}&\frac1{(x_1-y_2)^2}\\\frac1{(x_2-y_1)^2}&\frac1{(x_2-y_2)^2}\end{vmatrix} =\begin{vmatrix} \frac1{(2/3-1)^2}&\frac1{(2/3-0)^2}\\ \frac1{(2-1)^2}&\frac1{(2-0)^2} \end{vmatrix} =\begin{vmatrix} 9&9/4\\ 1&1/4 \end{vmatrix}=0$$