A free subgroup of GL(2,Z)? Is the subgroup of $GL(2,\mathbb Z)$ generated by the matrices 
$$ \left( \begin{array}{cc}
1 & 1  \\\
1 & 0 \end{array} \right) \ \ \text{and} \ \
\left( \begin{array}{cc}
2 & 1  \\\
1 & 0 \end{array} \right)
$$
free of exponential growth? More generally, how does one find all the relations between two matrices? 
I am sure this is well known, so any relevant references will be appreciated. 
My motivation comes from dynamical systems where these matrices specify two automorphisms of the 2-torus; I am interested in studying the orbits of their joint action. 
 A: A subgroup of $SL(2,\mathbf{Z}$) is free iff it's torsion-free; this is a useful trick (and it's not an immediately obvious fact: it's because $SL(2,\mathbf{Z})$ acts on a tree with finite stabilisers). This is a useful trick in situations like this.
However, if your matrices are denoted $a$ and $b$, then random mucking about gives me that $ba^{-2}$ has order 2, so the subgroup you're asking about cannot possibly be free.
EDIT: this was an answer to the original question "is the group free" and goes nowhere towards answering the revised question.
A: I interpret the question "how does one find all the relations between the matrices" as "find a set of defining relations for the group generated by the two matrices".
To do that we need a presentation of $GL(2,\mathbb{Z})$. I found one in the paper:
T. Brady, Automatic structures on Aut$F_2)$, Arch. Math. 63, 97-102 (1994).
$\langle p,s,u \mid p^2=s^2=(sp)^4=(upsp)^2=(ups)^3=(us)^2=1 \rangle$
where $p= \left(\begin{array}{cc}0&1\\1&0\end{array}\right)$, $s= \left(\begin{array}{cc}-1&0\\0&1\end{array}\right)$, $u= \left(\begin{array}{cc}1&1\\0&1\end{array}\right)$.
Denoting your two matrices by
$a= \left(\begin{array}{cc}1&1\\1&0\end{array}\right)$,
$b= \left(\begin{array}{cc}2&1\\1&0\end{array}\right)$ 
we have $a=up$, $b=u^2p$.
Putting $H = \langle a,b \rangle$ and using coset enumeration in Magma, it turns out that $H={\rm GL}(2,\mathbb{Z})$. So your two matrices actually generate all of GL$(2,\mathbb{Z})$.
In fact, denoting $a^{-1}$ and $b^{-1}$ by $A$ and $B$, we have 
$p=aBa$, $u=a^2Ba$, $s=abaBAbabA$.
Using the modified coset enumeration algorithm, we can compute a presentation of $H$, which came out as the not particularly enlightening
$H = \langle a,b \mid (aBa)^2, (AbaBA)^2, aBabABAbAbABAbAbABAbAbAB, (abABabaB)^3 \rangle$.
A: No, the subgroup is not free. You can check that if $A$ is your first matrix, $B$ is the second matrix, then $[A,B]^6=1$, where $[A,B]=ABA^{-1}B^{-1}$.
 Update 1.  There is a paper by Klimenko and Kopteva which might be useful: 
All discrete ${\scr{RP}}$ groups whose generators have real traces. 
Internat. J. Algebra Comput. 15 (2005), no. 3, 577–618. 
It might be in the arXiv, I did not check. 
 Update 2.  The subgroup does have exponential growth. The matrices $B^2$ and $A^6$ belong to the Sanov subgroup. This is a free subgroup of $SL(2,Z)$ consisting of matrices of the form $$\left\[\begin{array}{ll} 4k+1 & 2n \\\ 2m & 4l+1\end{array}\right\].$$ Since $A^6$ and $B^2$ do not commute, they generate a free subgroup of rank 2. The definition of Sanov subgroup can be found in the standard group theory book by Kargapolov and Merzlyakov, 14.2.1 and the Exercise 14.2.3 (I refer to the third edition). 
