Given two square $n\times n$ matrices $A$ and $B$, how would you construct a $3n\times 3n$ block upper triangular matrix $C$ whose inverse would somehow give the product $A\cdot B$? ($C^{-1}$ is obviously not equal to the product matrix, it just needs to contain its coefficients somewhere).
The upper triangular matrix has to contain $A$ and $B$
Try to multiply
$$\begin{bmatrix} I & A & 0 \\ 0 & I & B \\ 0 & 0 & I\end{bmatrix}\begin{bmatrix} I & -A & AB \\ 0 & I & -B \\ 0 & 0 & I\end{bmatrix}$$
Try to understand the implication of this result rather than just a maths puzzle. It shows us a relationship between computational complexity of matrix inversion and matrix multiplication.