Your representation $p$ is *not* faithful, since we have
$$
  (ABA^{-1}BA^{-1}BAB^{-1})^3 \ = \ 1.
$$
In particular, this means that
$$
  (aba^{-1}ba^{-1}bab^{-1})^3 \ = \
  \left(\begin{array}{rr}%
  -24587&42408\\%
  15048&-25955\\%
  \end{array}\right)
$$
lies in the kernel of $p$.