Algorithm to generate an arbitrary matrix of special linear group $SL(2,\mathbb{Z})$ I have a given $2\times 2$ special linear matrix, for example
\begin{equation}
m=\begin{pmatrix} 55 & 8469 \\ 1 & 154 \end{pmatrix}
\end{equation}
and I would like to get the generating form of it from the s and t matrices which are these:
\begin{equation}
t=\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}\qquad\qquad\qquad s=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}
\end{equation}
I don't know if exist an algoritmh for this problem or any proceedings which could help me.
 A: The key is that the second matrix rotates by 90 degrees.
$$\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\
\end{array}
\right)^a.\left(
\begin{array}{cc}
 0 & -1 \\
 1 & 0 \\
\end{array}
\right).\left(
\begin{array}{cc}
 1 & 1 \\
 0 & 1 \\
\end{array}
\right)^b $$
Gives the result
$$\left(
\begin{array}{cc}
 a & a d-1 \\
 1 & d \\
\end{array}
\right)$$
A: Disclaimer: The ideas for the following answer come from a pdf I found by Keith Conrad at the University of Connecticut. 
The idea of the following is to deal with a simple case (one of the entries is $0$) and then reduce to this simple case using the Euclidean algorithm.
Let our arbitrary matrix $m$ be written as
$$m=\left(\begin{matrix}a&b\\c&d\end{matrix}\right)$$
Simple Case: First suppose that $c=0$. 
As $\det(m)=1$ this means we must have $a=d=\pm1$. If $a=d=1$ we can directly write $m=t^b$, while if $a=d=-1$ we can write $m=s^2t^{-b}$ (note that $s^2=-I$).
General Case: Now suppose $|c|>0$. 
We describe an iterative process to reduce this matrix to the simple case. First suppose that $|a|<|c|$. Then let $m'=sm$ (so now $m'$ has $|a'|\ge|c'|$ since we have just switched them up to a sign). Second suppose that $|a|\ge |c|$, and use the Euclidean algorithm to write $a=cq+r$ with $0\le r<c$. Now take $m'=t^{-q}m$. This matrix has $a-cq=r<c$ as the top left entry and $c$ as the bottom left entry. 
Applying this process eventually leads to $c=0$ (using the switching operation after the reducing operation always leads to a $c'$ of strictly lesser magnitude than $c$), and from there we fall into our simple case.
A: This rather small example can also be done by hand without any algorithms.
Note that
$$
\begin{pmatrix}
a & b \\ 
c & d
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\ 
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
a & a+b \\ 
c & c+d
\end{pmatrix}
$$
While $S$ switches the columns and puts a minus sign in.
Then by guesstimating powers, we reduce the size of scalars quite efficiently.
For example:
$$
\begin{pmatrix}
55 & 8469 \\ 
1 & 154
\end{pmatrix}
\begin{pmatrix}
1 & -153 \\ 
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
55 & 54 \\ 
1 & 1
\end{pmatrix}.
$$
Multiply by $S^3=S^{-1}$:
$$
\begin{pmatrix}
55 & 54 \\ 
1 & 1
\end{pmatrix}
\begin{pmatrix}
0 & 1 \\ 
-1 & 0
\end{pmatrix}=
\begin{pmatrix}
-54 & 55 \\ 
-1 & 1
\end{pmatrix}
$$
This look quite close to a power of $T$. So mulitply it by $T$:
$$
\begin{pmatrix}
-54 & 55 \\ 
-1 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\ 
0 & 1
\end{pmatrix}=
\begin{pmatrix}
-54 & 1 \\ 
-1 & 0
\end{pmatrix}
$$
This we recognize as $T^{54}S$. Working backwards, this all means that
$$
MT^{-153}ST=T^{54}S.
$$
Hence $M=T^{54}ST^{-1}S^{-1}T^{153}$. (this might be simplified though)
