summation of determinants of $3\times3$ matrices I have an algebra problem but no idea how to solve it. The problem is: "you can create 9! matrices the elements of which lie in a set $ \{1,2,3,...,9\} \subset \mathbb N$ so that their elements do not repeat, i.e. e.g.
$$
\begin{pmatrix}1&2&9\\3&5&7\\6&4&8 \end{pmatrix}
$$
Find the sum of the determinants of all these matrices."
Could you give me a hint how to solve it? Thank you.

see if we interchange rows or columns the determinant becomes $(-1)^n$ of the original determinant where $n$ = number of interchanges so here we have three even arrangements while three odd arrangements implying  sum of all these $9!$ determinants as they are even and odd pairs is $0$. eg the example you given has determinant value $-114$ if we change 1 row then it becomes $+114$ so now as we have three odd combinations and three even so we get $3(-114)=3(114)=0$ also note that if in a particular row or column if elements are in AP then the determinant value is $0$ hope its clear.
