Trivial product of two matrices? Is it possible to find two matrices $A$ and $B$, so that there does not exists a product of matrices $A$,$A^{-1}$,$B$,$B^{-1}$ that is equal to $Id$, under the condition that the product is irreducible, that is, it is  is not trivial as a word (e.g. $AA^ {-1}B^{-1}B$).
If there is no simple answer I would be happy with some references to where I should start.
 A: A brief comment, really, on R W's answer. Sanov's theorem states that the matrices 
$$\begin{pmatrix} 1 & a \\ 0 & 1\end{pmatrix}$$ and $$\begin{pmatrix}1 & 0 \\ a & 0\end{pmatrix}$$ generate a free group as long as $a\geq 2,$ though $a>2, a \in \mathbb{N}$ gives a free group of infinite index in $SL(2, \mathbb{Z}).$
A: If I understand correctly, you will find the answer in a good exposition of the Banach-Tarski paradox. Finding two rotation matrices in $\mathbb{R}^3$ that generate the free group in two generators is a usually the biggest part of the construction.
A: As far as I understand, the OP asks about existence of free subgroups with 2 generators in matrix groups. The simplest example (if one does not require the matrices to be orthogonal which the OP did not) is probably the free group generated by the matrices 
$$\left[ \begin{array}{cc} 1 & 2 \\\ 0 & 1 \end{array} \right]$$ 
and 
$$\left[ \begin{array}{cc} 1 & 0 \\\ 2 & 1 \end{array} \right]\;,$$
see The free group $F_2$ has index 12 in  SL(2,$\mathbb{Z}$) 
