(image is attached for those who think I have changed the statement of the question while copying from the book)
Chose any 9 distinct integers. These 9 integers can be arranged to from 9! determinants each of order 3. The sum of these 9! determinants is?
My approach
For any Δ, there exist -Δ in arrangements
∴ sum = 0
I am looking for another approach!
You can pair up the determinants: For any matrix $M$, define $M'$ by flipping the first and second rows. Then $\det(M)+\det(M')=0$. To avoid duplications, sum over all $M$ such that $m_{11} \lt m_{21}$. Then each matrix appears exactly once as either $M$ or $M'$.