These two matrices generate a free group:
$$
A=\left(
\begin{array}{ccc}
 \frac{1}{3} & \frac{2 \sqrt{2}}{3} & 0 \\
 -\frac{2 \sqrt{2}}{3} & \frac{1}{3} & 0 \\
 0 & 0 & 1
\end{array}
\right),
B=\left(
\begin{array}{ccc}
 1 & 0 & 0 \\
 0 & \frac{1}{3} & \frac{2 \sqrt{2}}{3} \\
 0 & -\frac{\sqrt{2}}{3} & \frac{1}{3}
\end{array}
\right).
$$
To see that they do, note that the entries of $3A$ and $3B$ live in the ring $\mathbb{Z}[\sqrt{2}]$, which admits a surjective homomorphism to the field $\mathbb{F}_3(i)$.  This map induces a map on the matrix rings, under which $A$ and $B$ become

$$
A' =\left(
\begin{array}{ccc}
 1 & - i & 0 \\
  i & 1 & 0 \\
 0 & 0 & 0
\end{array}
\right),
B'=\left(
\begin{array}{ccc}
 0 & 0 & 0 \\
 0 & 1 & - i \\
 0 & i & 1
\end{array}
\right).
$$
If there were some non-trivial reduced group word in $A$ and $B$ that gave the identity, then a similar word in $3A$, $3A^{\top}$, $3B$ , and $3B^{\top}$ would be a multiple of the identity matrix.  However, one may check that any monoid word in the generators $A'$, $A'^{\top}$, $B'$, $B'^{\top}$ that evaluates to a multiple of the identity must contain the subword $A'A^{\top}$ or some similar forbidden subword.