Exercise 2.3.30

(a) $E=A^{-1}=\left[\begin{array}{cc}\hfill 1& \hfill 0\\ \hfill -1& \hfill 1\end{array}\right]$ will reduce row 2 of $\mathit{EM}$ to $\left[\begin{array}{cc}2\hfill & 3\hfill \end{array}\right]$.

(b) Then $F=B^{-1}=\left[\begin{array}{cc}\hfill 1& \hfill -1\\ \hfill 0& \hfill 1\end{array}\right]$ will reduce row 1 of $\mathit{FEM}$ to $\left[\begin{array}{cc}1\hfill & 1\hfill \end{array}\right]$.

(c) Then $E=A^{-1}$ twice will reduce row 2 of $\mathit{EEFEM}$ to $\left[\begin{array}{cc}0\hfill & 1\hfill \end{array}\right]$.

(d) Now $\mathit{EEFEM}=B$. Move $E$ ’s and $F$ ’s to get $M=\mathit{ABAAB}$. This question focuses on positive integer matrices $M$ with $\mathit{ad}-\mathit{bc}=1$. The same steps make the entries smaller and smaller until $M$ is a product of $A$ ’s and $B$ ’s.