<s>[INST] Prove a generator of $\mathrm{SL}_2(\mathbb{R})$ Definitions
Let $\mathrm{SL}_2(\mathbb{R}) := \left \{M := \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathbb{R}^{2 \times 2}: \det{M} = 1 \right \}$ with matrix multiplication be a group of $2 \times 2$ matrices.
Let 
\begin{align}
  A_\lambda :&= \begin{pmatrix} \lambda & 0\\ 0 & \lambda^{-1}\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R}) \text{ with }\lambda \in \mathbb{R} \setminus \{0\}\\
  B_t :&= \begin{pmatrix} 1 & t\\ 0 & 1\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R}) \text{ with } t \in \mathbb{R}\\
  C :&= \begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})
\end{align}
be elements of $\mathrm{SL}_2(\mathbb{R})$.
Question
Do $A_\lambda, B_t$ and $C$ generate $\mathrm{SL}_2(\mathbb{R})$?
My try
I think they do. When I could show that 
$$\begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})$$
can be multiplied with a combination of the three matrices $A, B, C$ in such a way that the result is the neutral element of this group
$$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$
then I could generate the matrix by multiplying the inverse matrices. 
The inverse matrices can be generated because:
\begin{align}
  A_\lambda^{-1} &= A_{\frac{1}{\lambda}}\\
  B_t^{-1}       &= B_{-t}\\
  C^{-1}         &= C^3
\end{align}
Now the only thing left is to get from 
$$M = \begin{pmatrix} a & b\\ c & d\end{pmatrix} \in \mathrm{SL}_2(\mathbb{R})$$
to
$$\begin{pmatrix} 1 & 0\\ 0 & 1\end{pmatrix}$$
Ok. So let's try it.
Case 1: $a = 0$
As $\det(M)=1 = ad - bc = 0d-bc$ we know that $bc=-1$. Especially is $c \neq 0$.
$$\begin{pmatrix} 0 & 1\\ -1 & 0\end{pmatrix} \cdot \begin{pmatrix} a & b\\ c & d\end{pmatrix} = \begin{pmatrix} c & d\\ -a & -b\end{pmatrix}$$
Continue with Case 2.
Case 2: $a \neq 0$
Normalize ($M \cdot A_{\frac{1}{a}}$):
$$\begin{pmatrix} a & b\\ c & d\end{pmatrix} \cdot \begin{pmatrix} \frac{1}{a} & 0\\ 0 & a\end{pmatrix} = \begin{pmatrix} 1 & ab\\ \frac{c}{a} & ad\end{pmatrix}$$
Continue with Case 3.
Case 3: $a=1$
Get a $0$ with $M \cdot B_{-b}$:
$$\begin{pmatrix} 1 & b\\ c & d\end{pmatrix} \cdot \begin{pmatrix} 1 & -b\\ 0 & 1\end{pmatrix} = \begin{pmatrix} 1 & 0\\ c & d-bc\end{pmatrix}$$
We know that $\det{M} = 1 = ad - bc = d - bc$.
Continue with Case 4.
Case 4: $a=1$, $b=0, d=1$
At this stage we have matrices that look like
$$\begin{pmatrix} 1 & 0\\ c & 1\end{pmatrix}$$
This is where I'm stuck. Can you help me?
 [/INST] In Case 4 you have $d=1$ since the determinant is still $1$, then
$$
A_{-1} C B_c C \pmatrix{1 & 0 \\ c & 1} = \pmatrix{1 & 0 \\ 0 & 1}.
$$
</s>