Suppose you have
[tex]\mathbf A=\begin{bmatrix}1&2&3\\2&4&6\end{bmatrix}[/tex]
[tex]\mathbf B=\begin{bmatrix}1&0\\0&2\\1&2\end{bmatrix}[/tex]
Then the matrix product is
[tex]\mathbf{AB}=\begin{bmatrix}\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}1\\0\\1\end{bmatrix}&\begin{bmatrix}1&2&3\end{bmatrix}\begin{bmatrix}0\\2\\2\end{bmatrix}\\\\\begin{bmatrix}2&4&6\end{bmatrix}\begin{bmatrix}1\\0\\1\end{bmatrix}&\begin{bmatrix}2&4&6\end{bmatrix}\begin{bmatrix}0\\2\\2\end{bmatrix}\end{bmatrix}[/tex]
In this step what we did was take all combinations of one row vector of [tex]\mathbf A[/tex] and one column vector of [tex]\mathbf B[/tex]. There are 2 rows in [tex]\mathbf A[/tex] and 2 columns in [tex]\mathbf B[/tex] - these numbers *must* match in order for the notion of the matrix product to make sense - so there will be 4 total scalar products we need to compute.
Now, when I say "scalar product", I'm referring to what you *may* already know as the "dot product". We take the corresponding components of the row and column vector factors, multiply them, then add up all the products:
[tex]\mathbf{AB}=\begin{bmatrix}1\cdot1+2\cdot0+3\cdot1&1\cdot0+2\cdot2+3\cdot2\\2\cdot1+4\cdot0+6\cdot1&2\cdot0+4\cdot2+6\cdot2\end{bmatrix}[/tex]
[tex]\mathbf{AB}=\begin{bmatrix}4&10\\8&20\end{bmatrix}[/tex]
