No, the subgroup is not free. You can check that if $A$ is your first matrix, $B$ is the second matrix, then $[A,B]^6=1$, where $[A,B]=ABA^{-1}B^{-1}$.

Update 1. There is a paper by Klimenko and Kopteva which might be useful: All discrete ${\scr{RP}}$ groups whose generators have real traces. Internat. J. Algebra Comput. 15 (2005), no. 3, 577–618. It might be in the arXiv, I did not check.

Update 2. The subgroup does have exponential growth. The matrices $B^2$ and $A^6$ belong to the Sanov subgroup(this . This is a free subgroup of $SL(2,Z)$ consisting of matrices of the form $$\left[\begin{array}{ll} 4k+1 & 2n \\ 2m & 4l+1\end{array}\right].$$ Since $A^6$ and $B^2$ do not commute, they generate a free subgroup of rank 2. The definition of Sanov subgroup can be found in the standard group theory book by Kargapolov and Merzlyakov, 14.2.1 and the Exercise 14.2.3 (I refer to the third edition).

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No, the subgroup is not free. You can check that if $A$ is your first matrix, $B$ is the second matrix, then $[A,B]^6=1$, where $[A,B]=ABA^{-1}B^{-1}$.

Update 1. There is a paper by Klimenko and Kopteva which might be useful: All discrete ${\scr{RP}}$ groups whose generators have real traces. Internat. J. Algebra Comput. 15 (2005), no. 3, 577–618. It might be in the arXiv, I did not check.

Update 2. The subgroup does have exponential growth. The matrices $B^2$ and $A^6$ belong to the Sanov subgroup (this is a free subgroup of $SL(2,Z)$ consisting of matrices of the form $$\left[\begin{array}{ll} 4k+1 & 2n \\ 2m & 4l+1\end{array}\right].$$

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No, the subgroup is not free. You can check that if $A$ is your first matrix, $B$ is the second matrix, then $[A,B]^6=1$, where $[A,B]=ABA^{-1}B^{-1}$.