Let $\left\{\Delta_1,\Delta_2,.....,\Delta_n\right\}$ be the set of all determinants of order 3 that can be made with the distinct real numbers Let $\left\{\Delta_1,\Delta_2,.....,\Delta_n\right\}$ be the set of all determinants of order 3 that can be made with the distinct real numbers from the set $S=\left\{1,2,3,4,5,6,7,,8,9\right\}$.Then prove that $\sum_{i=1}^{n}\Delta_i=0$

I know that the total number of determinants that can be formed by using distinct real numbers from the given set is $9!$.But i dont know how to prove that their sum is zero.Please help me.Thanks
 A: A symmetry argument suffices.  Note that swapping, say, the first two rows of each of the $n$ matrices causes the determinants to change sign.  If we sum those up we will surely get $-\sum_{i=1}^n \Delta_i$.  On the other hand, this row swap operation is an involution on the set of matrices formed from $S$: so we just get the same set of $n!$ determinants, but in a different order.  Thus
$$-\sum_{i=1}^n \Delta_i = \sum_{i=1}^n \Delta_i,$$
so both sides are $0$.
A: We have $det(A) = a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} -a_{11}a_{23}a_{32} - a_{13}a_{22}a_{31} - a_{12}a_{21}a_{33}$. There are $9!$ such sums. So each term in this sum will be the same (eg. $\Sigma a_{11}a_{22}a_{33} = \Sigma a_{12}a_{23}a_{31}$) , and there are exactly equal number of positive and negative terms, so the result is zero. 
A: Note that if you swap two columns of a matrix, its determinant is multiplied by $-1$.  So let's say that two matrices $A$ and $B$ made up of the numbers in $S$ are equivalent if either $A=B$ or $A$ is obtained from $B$ by swapping the first two columns.  This is an equivalence relation, and each equivalence class has two elements.  The sum of the determinants within each equivalence class is $0$.  Thus by partitioning the entire sum into these equivalence classes, we find that the total sum is $0$.
