Generators of Special Linear Groups Linear algebra and special-linear group experts please help:
I learn that in principle one can generate this $M$ matrix form the $B_1$ and $B_2$ matrix below. Here
$$
M=\begin{pmatrix} 0& 1& 0\\ -1& 1& 0\\ 0& 0& 1 \end{pmatrix}
$$
from:
$$
B_1=\begin{pmatrix} 0& 0& 1\\ 1& 0& 0\\ 0& 1& 0 \end{pmatrix},
\text{ 
and   }\;\; 
B_2=\begin{pmatrix} 1& 1& 0\\ 0& 1& 0\\ 0& 0& 1 \end{pmatrix}.
$$

Question: How to generate $M$ from $B_1$ and $B_2$? i.e. So what is the exact expression to make $M=\dots B_1 \dots B_2$ as a product of $B_1$ and $B_2$ matrices?

I learned the basics about those generators and SL(N,Z) from the book Coxeter and Moser on "Generators and relations for discrete groups" (published by Springer, 1957).
Thank you! :o)

I'm not sure what you are looking for, but here is an exact expression:
$M=B_2B_1B_2B_1B_2^{-1}B_1^{-1}B_2^{-1}B_1B_2B_1$
also 
$M=(B_2B_1^2)^2(B_2B_1)(B_1B_2)^{-2}$
The subgroup $H$ of $G$ generated by a set $X$ is the intersection of all subgroups of $G$ that contain $X$, so $H$ is the smallest subgroup of $G$ that contains $X$. Since $X$ is a subgroup it is closed under inversion, and so must contain the inverses of every element of $X$. Hence one is generally allowed to use inverses of generators (or equivalently, both positive and negative powers). In this particular case, $B_1^3$ is the identity, so $B_1^{-1} = B_1^2$ and we can avoid inverses of $B_1$ if we don't mind multiplying by two copies of $B_1$. However, it does not appear that $B_2^{-1}$ can be written only using positive powers of $B_1$ and $B_2$ (I verified that any such expression has the powers adding up to more than $29$).
