I recently find a curious and unexplainable(as seems to me) equation on determinant as follows.
$$3\begin{vmatrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
a_1a_2a_3 & b_1b_2b_3 & c_1c_2c_3 & d_1d_2d_3
\end{vmatrix}$$
$$=\begin{vmatrix}
a_1(b_1+c_1+d_1) & b_1(a_1+c_1+d_1) & c_1(a_1+b_1+d_1) & d_1(a_1+b_1+c_1) \\
a_2 & b_2 & c_2 & d_2 \\
a_3 & b_3 & c_3 & d_3 \\
a_2a_3 & b_2b_3 & c_2c_3 & d_2d_3
\end{vmatrix}$$
$$+\begin{vmatrix}
a_1 & b_1 & c_1 & d_1 \\
a_2(b_2+c_2+d_2) & b_2(a_2+c_2+d_2) & c_2(a_2+b_2+d_2) & d_2(a_2+b_2+c_2) \\
a_3 & b_3 & c_3 & d_3 \\
a_1a_3 & b_1b_3 & c_1c_3 & d_1d_3
\end{vmatrix}$$
$$+\begin{vmatrix}
a_1 & b_1 & c_1 & d_1 \\
a_2 & b_2 & c_2 & d_2 \\
a_3(b_3+c_3+d_3) & b_3(a_3+c_3+d_3) & c_3(a_3+b_3+d_3) & d_3(a_3+b_3+c_3) \\
a_1a_2 & b_1b_2 & c_1c_2 & d_1d_2
\end{vmatrix} $$
Although I have verified this equation by explicit computation (which involves many terms and I did it by computer), I wonder if there is a proof of it (either conceptual or tricky). If one can not find a "proof", it seems that there might be something beyond linear algebra underlying the equation (possibly algebraic geometry?).
Best Answer
Not that I am happy with the forthcoming proof, but at least it is human.
As Iosif Pinelis notes, both sides of your equation are multiaffine. Also, for every $i=1,2,3$ they are homogeneous of degree 2 with respect to $R_i:=(a_i,b_i,c_i,d_i)$. Thus, every monomial which may appear after expanding the determinants has 2 variables from each of sets $R_1,R_2,R_3$. How to get out a coefficient of monomial $\prod_{i\in S} x_i$ in the homogeneous multiaffine polynomial $p(x_1,x_2,\ldots,x_n)$, where $S\subset \{1,\ldots,n\}$? Just put $x_i=1$ for $i\in S$ and $x_i=0$ for $i\notin S$, the value of this polynomial is the desired coefficient.
Thus, it suffices to check your identity when two guys in each $R_i$ are equal to 1 and two guys are equal to 0. In particular, this yields $a_i(b_i+c_i+d_i)=a_i$ etc. So, your three determinants in LHS now surprisingly have the same first three rows $R_1,R_2,R_3$ as the determinant in RHS. Using the linearity of the determinant with three fixed rows as the function of the fourth row, we may rewrite now your identity as
$$\begin{vmatrix} a_1 & b_1 & c_1 & d_1 \\ a_2 & b_2 & c_2 & d_2 \\ a_3 & b_3 & c_3 & d_3 \\ a_4 & b_4 & c_4 & d_4 \end{vmatrix}=0,\, \text{where}\, a_4=a_1a_2+a_2a_3+a_3a_1-3a_1a_2a_3,\,\text{etc}.$$ Note that $a_4=0$ unless exactly two entries of $A:=(a_1,a_2,a_3)$ are equal to 0 (in this case $a_4$ equals 1), analogously for other columns. Therefore, if no column and no row of our $4\times 4$ matrix is 0, each sequence $A,B,C,D$ must contain at least one 1, and at least one of them must contain exactly two 1's. Since totally they contain 6 1's, we conclude that two of them (say, $A$ and $B$) contain two 1's. There exists $i\in \{1,2,3\}$ such that $a_i=b_i=1$. Then the $i$-th and 4-th lines of our matrix coincide.