Not an answer, but might be interesting.
Let $A$ and $B$ be real matrices, then $(AB)_{ij}=\sum_{k}a_{ik}b_{kj}=\left\langle
R_i(A), C_j(B)^T \right\rangle$ where $R_i(A)$ denotes the $i$-th row of $A$ and $C_j(B)$ denotes the $j$-th column of $B$ and $\left\langle -,-\right\rangle$ denotes the standard inner product on $\mathbb{R}^n$. (So indeed, a matrix product is nothing but a bunch of inner-products).
Now suppose $A$ is a real $n\times n$-matrix and $AA^T=I$. Then $(AA^T)_{ij}=\delta_{ij}$. Now notice that $C_i(A^T)^T=R_i(A)$ by definition of $A^T$. Thus $(AA^T)_{ij}=\left\langle R_i(A), R_j(A) \right\rangle=\delta_{ij}$. It follows that the rows of $A$ form on orthonormal basis of $\mathbb{R}^n$. This also explains why a square matrix satisfying $AA^T=I$ is called orthogonal.
This shows that this way of thinking about matrix multiplication can be interesting. (For example: try to find the analogues for complex matrices).