We know that the Vandermonde determinant of order $n$ is the determinant defined as follows:


$$\begin{vmatrix}
1&x\_1&x\_1^2&\dots&x\_1^{n-1}\\
1&x\_2&x\_2^2&\dots&x\_2^{n-1}\\
\ldots&\ldots&\ldots&\ldots&\ldots\\
1&x\_n&x\_n^2&\dots&x\_n^{n-1}\\
\end{vmatrix}=\prod\limits\_{1\leq i<j\leq n}(x\_j-x\_i).$$


For any or some special nonempty subset $S\subseteq \{(i,j)\mid1\leq i<j\leq n\}$，does there exist some matrix $A\_S$ which is similar to the Vandermonde matrix (for example, every element of $A\_S$ is something just like ${x\_i}^j$) such that $$\begin{vmatrix}
A\_S
\end{vmatrix}=\prod\limits\_{(i,j)\in S}(x\_j-x\_i)$$
$$\text{or}$$$$\begin{vmatrix}
A\_S
\end{vmatrix}\neq 0 \quad\text{if and only if}\prod\limits\_{(i,j)\in S}(x\_j-x\_i)\neq 0\quad?$$



A question about the Vandermonde determinant
It is not what you are looking for, I'm sure, but there is a matrix which is $N \times N$ for $N=2|S|.$ It is almost entirely zeros except that on the main diagonal there are $2 \times 2$ blocks $$\begin{vmatrix}
1&x\_j\\
1&x\_i \end{vmatrix},$$ one for each desired term $x\_i-x\_j.$


There are certainly more compact solutions for some choices of $S$, such as the Vandermonde case of all pairs.


