Modular Arithmetic over a Matrix What are the rules for modular arithmetic when multiplying two matrices?  I want to calulate $C = AB \mod{n}.$  Aside from the obvious way of performing the modulo after the multiplication, when and where can i safely perform the modulo during the multiplication algorithm?

[1] Normally:
$C_{ij}=\displaystyle\sum\limits_{k=0}^m A_{ik}B_{kj}$
Can I take each of these summands $A_{ik}B_{kj} \mod{n}$, as follows?
[2]
$C_{ij}=\displaystyle\sum\limits_{k=0}^m [ A_{ik}B_{kj}\pmod{n} ]$

Here is an example:
$A =
\left(\begin{array}{cc}
9 & 2 \\
10 & 10 \\
\end{array}\right)$
$B =
\left(\begin{array}{cc}
7 & 3 \\
1 & 6 \\
\end{array}\right)$
$C =
\left(\begin{array}{cc}
65 & 39 \\
80 & 90 \\
\end{array}\right)$
$C \equiv
\left(\begin{array}{cc}
2 & 4 \\
3 & 6 \\
\end{array}\right) \mod{7}$

edit:
using [2]
$C \mod 7=
\left(\begin{array}{cc}
2 & 11 \\
3 & 6 \\
\end{array}\right)$
This doesn't result in the same matrix.

Once you take pass into modular arithmetic, you're stuck there: $C \mod{7}$ has values in the integers $\mod{7}$, not in the integers themselves.
But modulo 7, $\left(\begin{matrix}2&11\\3&6\end{matrix}\right)=\left(\begin{matrix}2&4\\3&6\end{matrix}\right)$
simply because $11\equiv 4 \mod{7}$.
And in general, yes, you can apply $[2]$; you can even get 
$C\mod{7}\equiv(A\mod{7})(B\mod{7})\mod{7}$.
All this follows from the facts that $$ab \mod{n}\equiv (a\mod{n}b\mod{n}) \mod {n}$$ 
and
$$a+b \mod{n}\equiv (a\mod{n}+b\mod{n}) \mod{n}$$
