Particular generators of GL(2,R)

Question

I want to prove that $$ \text{GL}_{2}( \mathbb{R} ) = \left\langle \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}, \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}; \ a \in \mathbb{R} \setminus \{0\} \right\rangle $$ I know that $$ \begin{pmatrix} a & 0 \\ 0 & 1 \end{pmatrix}^{k} = \begin{pmatrix} a^{k} & 0 \\ 0 & 1\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 0 & a \end{pmatrix}^{l} = \begin{pmatrix} 1 & 0 \\ 0 & a^{l} \end{pmatrix},$$$$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{m} = \begin{pmatrix} 1 & m \\ 0 & 1\end{pmatrix}, \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}^{h} = \begin{pmatrix} 1 & 0 \\ h & 1\end{pmatrix}. $$ Since the determinant of these matrices isn't zero $ \forall k,l,m,h \in \mathbb{Z}, a \neq 0$, the determinant of a product of any permutation of these matrices isn't zero. Therefore this generates invertible matrices and for $k=l=m=h=0$ we get identity matrix. Is there an efficient way to prove that this generates all invertible two by two matrices over $ \mathbb{R}$ other than checking all possible permutations?

Answer 1

As noted by @LouisHainaut, these four matrices enable you to do the elementary row operations. Since it is possible to get to and from the identity matrix to any invertible matrix by doing elementary row operations, the result follows.

The first two allow you to multiply a row by a scalar. The last two, to add a multiple of a row to another.