show/hide this revision's text 2 question edited so answer edited to clarify.

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.
show/hide this revision's text 1

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.