TITLE: Trivial product of two matrices?
QUESTION [6 upvotes]: 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.

REPLY [11 votes]: 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,$\mathbb{Z}$)

REPLY [7 votes]: 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 $SL(2, \mathbb{Z}).$