Show, that if $\mathbf{A}= \left( \begin{array}{cc} 1&1\\ 0&1 \end{array} \right)$, $\mathbf{B}= \left( \begin{array}{cc} 0&1\\ -1&0 \end{array} \right)$ and $\mathrm{SL}(2, \mathbb{Z}) := \{ \mathbf{C}\in\mathrm{M}(2\times 2;\mathbb{Z})\, |\, \det(\mathbf{C}) = 1\}$ then $\langle\mathbf{A}, \mathbf{B}\rangle = \mathrm{SL}(2, \mathbb{Z})$.
I found this exercise in a textbook for linear algebra in a chapter about the determinant, so it should be solved rather elementarily and without any deeper understanding of group theory ($\mathrm{SL}(2, \mathbb{Z})$ is defined only for the exercise). Showing $\langle\mathbf{A}, \mathbf{B}\rangle \subseteq \mathrm{SL}(2, \mathbb{Z})$ was easy but I got stuck with the opposite direction. Any help would be appreciated.