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

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 $p$.

Your representation $p$ is not faithful, since we have $$ ABA^{-1}BA^{-1}BAB^{-1}ABA^{-1}BA^{-1}BAB^{-1}ABA^{-1}BA^{-1}BAB^{-1} \ = \ 1. $$

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