The Problem: Prove that the subgroup of $SL_2(\mathbb{F_3})$ generated by $\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}$ and $\begin{pmatrix}1&1\\1&-1\\\end{pmatrix}$ is isomorphic to the quaternion group of order $8$.
Source: Abstract Algebra, $\mathit{3^{rd}}$ edition by Dummit and Foote.
My Attempt: Let $A=\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}$, $B=\begin{pmatrix}1&1\\1&-1\\\end{pmatrix}$, then $A^2=\begin{pmatrix}-1&0\\0&-1\\\end{pmatrix}$, $A^3=\begin{pmatrix}0&1\\-1&0\\\end{pmatrix}$, $A^4=\begin{pmatrix}1&0\\0&1\\\end{pmatrix}$, $B^2=\begin{pmatrix}2&0\\0&2\\\end{pmatrix}$, $B^3=\begin{pmatrix}2&2\\2&-2\\\end{pmatrix}$, $AB=\begin{pmatrix}-1&1\\1&1\\\end{pmatrix}$, $BA=\begin{pmatrix}1&-1\\-1&-1\\\end{pmatrix}$, thus the cardinality of the group generated by $\begin{pmatrix}0&-1\\1&0\\\end{pmatrix}$ and $\begin{pmatrix}1&1\\1&-1\\\end{pmatrix}$ is greater than $8=|Q_8|$, and the claim fails.
Where exactly did things go wrong? Any hint would be greatly appreciated.