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.
 A: Observe all such matrices $X = \begin{bmatrix} x_{ij} \end{bmatrix}$ where $x_{11} < x_{12}$. Let's call that set $S_1$.
Now, observe the rest of them and call that set $S_2$. Those have $x_{11} > x_{12}$. Notice that those two sets are bijective. What is the bijection (denote it with $f$)?
This means that your sum is equal to
$$\sum_{X \in S_1} \det X + \sum_{X \in S_2} \det X = \sum_{X \in S_1} \det X + \sum_{X \in S_1} \det f(X) = \sum_{X \in S_1} (\det X + \det f(X)).$$
If you understand what $f$ is, you'll notice that $\det f(X) = -\det X$.
Works for all $n > 1$.
A: Hint: Consider the permutation you have in the example. There are six permutations of the rows, three odd and three even. So the sum of the determinants of these six matrices is …
A: Given that others have answered, and the basic trick is the same, here is another way of thinking about the problem.
If I exchange the first and second rows of every matrix, I change the sign of every determinant and hence the sign of the sum. On the other hand, I get exactly the same set of matrices, so the sum must stay the same. This leaves me with only one possibility.
I mention it because it deals with the problem globally, and that kind of global reasoning is sometimes very useful. The local information is hidden in "I get exactly the same set of matrices" and if you write down the detail of what makes that obvious you will find yourself mirroring the other suggestions people have posted.
A: The number of those matrices is even, and note that the sum of the determinants of the pair $\begin{pmatrix}a&b&c\\d&e&f\\g&h&i\end{pmatrix}$ and $\begin{pmatrix}b&a &c\\e&d&f\\ h&g& i\end{pmatrix}$ is 0. If we pair up these matrices in this way we can see that the required sum is 0.
A: 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.
