I am working on some exercise of Riemannian Geometry. I am proving that $G=\left\{\begin{pmatrix} 1& 0 & 0 \\ x&1&0 \\ y&x&1 \end{pmatrix}\mid x, y\in \mathbb R\right\}$ is a Lie Group. For the proof I am going to prove that the mapping that associates for each element its inverse is differentiable. This mapping is $I:G\longrightarrow G$ defined as $I\left(\begin{pmatrix} 1& 0 & 0 \\ x&1&0 \\ y&x&1 \end{pmatrix}\right)=\begin{pmatrix} 1& 0 & 0 \\ -x&1&0 \\ x^2-y&-x&1 \end{pmatrix}$
I have seen that some people, when they calculate the differential of $I$, considerer the embedding of $G\longrightarrow \mathbb R^9,\ X=\begin{pmatrix} 1& 0 & 0 \\ x&1&0 \\ y&x&1 \end{pmatrix}\mapsto \begin{pmatrix} 1& 0 & 0 & x&1&0 & y&x&1 \end{pmatrix}^t$
And they write the differential of I as: $Df_{(X)}(u,v)=\begin{pmatrix} 0& 0 \\0& 0 \\0& 0 \\-1& 0 \\0& 0 \\0& 0 \\2x& -1 \\-1& 0 \\0& 0 \\ \end{pmatrix} \begin{pmatrix} u\\ v \end{pmatrix}$
But I wonder if it is not correct, or even better, to write it as: $Df_{(X)}(u,v)=\begin{pmatrix}\begin{pmatrix} 0& 0 & 0 \\ -1&0&0 \\ 2x&-1&0 \end{pmatrix}& \begin{pmatrix} 0& 0 & 0 \\ 0&0&0 \\ -1&0&0 \end{pmatrix}\end{pmatrix}\begin{pmatrix} u\\ v \end{pmatrix}$
I think that writing the function mapping $G$ to $\mathbb R^9$ it is for trivially see that $I$ is differentiable, but in the other way it is trivial too. Am I right? What do you think? Thanks.