How to prove this result?

Question

Let {$\Delta_1,\Delta_2,\Delta_3\cdots\cdots\cdots\cdots\Delta_n$} be the set of all determinants of order 3 that can be made with the distinct real numbers from set $S=\{1,2,3,4,5,6,7,8,9\}$.

Then prove that -

$$\sum_{i=1}^n\Delta_i=0$$

Answer 1

I think you are talking about $3\times 3$ matrices. Take one matrix $A$. You have $6$ possibls permutation on the rows. $3$ even and three odds. Hance you have 3 matrices with determinants $\lambda$ and 3 matrices with determinants $-\lambda$. Take it from here.

Answer 2

We have $n=9!$ determinants: the number of permutations of $9$ components of a $3\times 3$ matrix and let $P$ the set of these determinants. Let $\Delta$ a given determinant and $\Delta'$ the determinant obtained from $\Delta$ by interchanging its second and third row then we have $\Delta'=-\Delta$. Notice that we can create a partition of $\frac{9!}2$ sets $P_\Delta=\{\Delta,\Delta'\}$ of $P$ and then

$$P=\cup P_\Delta$$ so we see easily that $$\sum_{\Delta\in P}\Delta=\sum_{\Delta\in P_\Delta}\Delta=0$$