Submonoid of $SL(2,\mathbb{Z})$ generated by two matrices. I'm trying to figure out which kind of matrices can be written as $A^{k_{1}}B^{k_{2}}...A^{k_{n}}B^{k_{n}}$ where 
A=$\begin{pmatrix} 1&1\\0&1\end{pmatrix}$ and B=$\begin{pmatrix} 2&1\\1&1\end{pmatrix}$.
I did it for $C=\begin{pmatrix} 1&2\\0&1\end{pmatrix}$ and $D=\begin{pmatrix} 1&0\\2&1\end{pmatrix}$ and got that any word on $\{C,D\}$ appears as $\begin{pmatrix} a&2b\\2c&d\end{pmatrix}$with $a$ and $d$ odd and obviously $2b$ and $2c$ even. I can show this for $\{C,D\}$, but can't find an expression for $B^{k}$ so I can't go on and look for something similar.

Regarding $B^k$:
$B$ is Hermitian and thus diagonalizable.  See this for details.  So, we have that:
$$B = \overbrace{\begin{pmatrix}
\frac{1+\sqrt{5}}{2} & \frac{1-\sqrt{5}}{2} \\ 1 & 1
\end{pmatrix}}^{P}
\overbrace{\begin{pmatrix}
\frac{3+\sqrt{5}}{2} & 0 \\ 0 & \frac{3-\sqrt{5}}{2}
\end{pmatrix}}^{D}
\overbrace{\begin{pmatrix}
\frac{1}{\sqrt{5}} & \frac{-1+\sqrt{5}}{2\sqrt{5}} \\ -\frac{1}{\sqrt{5}} & \frac{1+\sqrt{5}}{2\sqrt{5}}
\end{pmatrix}}^{P^{-1}}
$$
So, we can write $B = PDP^{-1}$ where $D$ is diagonal.  It follows that $B^k = PD^kP^{-1}$, which you should be able to calculate.
