I didn't get this thing. Let $R, S$ be two relations from $A \to A$ with $A$ being an arbitrary set. And $M_R$ and $M_S$ their relation matrixes defined as:
$(M_R)_{ij}=\left\{ \begin{array}{l l} 1 & \quad \text{if $iRj$}\\ 0 & \quad \text{otherwise}\\ \end{array} \right.$
$M_S$ and $M_{R \circ S}$ are defined analogue. Why it holds that $M_{R \circ S}=M_R \cdot M_S$ (no, not exactly, but I don't know how I can express this in a nice way)? This is my incomplete sort of proof. Can you give me some advice how to improve it?
Now $iR \circ Sj \Leftrightarrow \exists x : iRx \wedge xSj$. Suppose $\exists x : iRx \wedge xSj \Rightarrow (M_R)_{ix}=1, (M_S)_{xj}=1 \Rightarrow (M_R)_{ix} \cdot (M_S)_{xj} = 1$.
On the other hand, suppose $\neg(\exists x : iRx \wedge xSj) \Leftrightarrow \forall x: \neg iRx \vee \neg xSj \Rightarrow (M_R)_{ix} = 0 \vee (M_S)_{xj} = 0 \Rightarrow (M_R)_{ix} \cdot (M_S)_{xj} = 0$.
Therefor, if we take the product it will sum up all those things. If there exists such an $x$ for any $i,j$, there will turn up a number greater than $0$. Otherwise, if there does not exists such a number it will be a $0$. If we replace all those integers greater than $0$ by $1$, it is the matrix of the productrelation $R \circ S$.