- $(m, n) (m', n') = (m + n \cdot m', nn')$

$\begin{bmatrix}
1 & 0 \\ m & n
\end{bmatrix}
\begin{bmatrix}
1 & 0 \\ m' & n'
\end{bmatrix} =
\begin{bmatrix}
1 & 0 \\ m + n \cdot m' & n \times n'
\end{bmatrix}$

This way of writing down semidirect products as matrices makes many things
immediately clear:
- The semidirect product is some kind of "shear" transform, since that's what a shear transformation looks like, matrix-wise.
- The resulting monoid $M \ltimes_{\phi} N$ has identity $(0_M, 1_N)$, since for the matrix to be identity, we need the 2nd row to be $(0, 1)$.
- The inverse operation if $(M, N)$ were groups would have to be such that

$\begin{bmatrix} 1 & 0 \\ m + n \cdot m' & n \times n' \end{bmatrix} =
\begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$

Hence:
- $nn' = 1$ implies that $n' = 1/n$.
- $m + n m' = 0$ implies that $m' = -m/n$.