Let $f\in \mathbb{R}[x_1,x_2,x_3,x_4]$ defined by $$f_a(x_1,x_2,x_3,x_4)=\prod_{1\leqslant i<j\leqslant4}(x_i-x_j)^{2a_{ij}}$$ where $a=(a_{12},a_{13},a_{14},a_{23},a_{24},a_{34})\in \mathbb{N}^6$. Define a $4\times4$ matrix $A_f$ as follow: $$A_{f_a}=\begin{bmatrix} L(1) & L(\frac{x_2}{x_1}) & L(\frac{x_4}{x_3}) & L(\frac{x_2x_4}{x_1x_3})\\ L(\frac{x_1}{x_2}) & L(1) & L(\frac{x_1x_4}{x_2x_3}) & L(\frac{x_4}{x_3})\\ L(\frac{x_3}{x_4}) & L(\frac{x_2x_3}{x_1x_4}) & L(1) & L(\frac{x_2}{x_1})\\ L(\frac{x_1x_3}{x_2x_4}) & L(\frac{x_3}{x_4}) & L(\frac{x_1}{x_2}) & L(1) \end{bmatrix}$$ where $L(\frac{x_{i_1}\cdots x_{i_k}}{x_{j_1}\cdots x_{j_k}})$ denotes the coefficient of $$\left(\prod_{1\leqslant i<j\leqslant4}(x_ix_j)^{a_{ij}}\right)\frac{x_{i_1}\cdots x_{i_k}}{x_{j_1}\cdots x_{j_k}}$$ in the expansion of $f$.

For example, when $a_1=(a,0,0,0,0,b)$, \begin{align} f_{a_1}(x_1,x_2,x_3,x_4)&=(x_1-x_2)^{2a}(x_3-x_4)^{2b},\\ \\ (-1)^{a+b}A_{f_{a_1}}&= \begin{bmatrix} \binom{2a}{a}\binom{2b}{b} & -\binom{2a}{a-1}\binom{2b}{b} & -\binom{2a}{a}\binom{2b}{b-1} & \binom{2a}{a-1}\binom{2b}{b-1}\\ -\binom{2a}{a-1}\binom{2b}{b} & \binom{2a}{a}\binom{2b}{b} & \binom{2a}{a-1}\binom{2b}{b-1} & -\binom{2a}{a}\binom{2b}{b-1}\\ -\binom{2a}{a}\binom{2b}{b-1} & \binom{2a}{a-1}\binom{2b}{b-1} & \binom{2a}{a}\binom{2b}{b} & -\binom{2a}{a-1}\binom{2b}{b}\\ \binom{2a}{a-1}\binom{2b}{b-1} & -\binom{2a}{a}\binom{2b}{b-1} & -\binom{2a}{a-1}\binom{2b}{b} & \binom{2a}{a}\binom{2b}{b} \end{bmatrix}. \end{align} It is not difficult to verify that the determinant of $A_{f_{a_1}}$ is nonzero. In fact,$$\det(A_{f_{a_1}})=\binom{2a}{a}^4\binom{2b}{b}^4\frac{(2a+1)^2(2b+1)^2}{(a+1)^4(b+1)^4}.$$ Moreover, I verified that $\det(A_{f_a})\neq0$ for many simple $a\in \mathbb{N}^6$.

**My question**

Does $\det(A_{f_a})\neq0$ hold for all $a\in \mathbb{N}^6$? Is there any significance for the matrix $A_{f_a}$? Any idea is welcome!