Is this matrix injective/surjective? Is this matrix injective or surjective? $M=\begin{pmatrix} 1 & 2 & -3\\  2 & 3 & -5 \end{pmatrix}$
I need to calculate rank.
$M^{T}= \begin{pmatrix}
1 & 2\\ 
2 & 3\\ 
-3 & -5
\end{pmatrix}$. If we form it with Gauss, we get (it's formed correctly): $\begin{pmatrix}
6 & 12\\ 
0 & 5\\ 
0 & 0
\end{pmatrix}$
$\Rightarrow rank(M)= 2 \Rightarrow$ not surjective because $M \in \mathbb{R}^{3}$ but $rank(M) \neq 3$
Is it injective? No because $dim(Ker(M)) = 3-2= 1 \neq 0$

Is it good? If not please explain not too complicated. I think there can be trouble at beginning when I transposed?
Edit: I'm not sure if $M$ is really in $\mathbb{R}^{3}$, I said that because we have $2$ lines but $3$ columns. That's fine?

This matrix expresses a transformation $M: \mathbb{R}^3 \to \mathbb{R}^2$, so it can't be injective. To prove that it's surjective, though, you just need to find two vectors in $\mathbb{R}^3$ whose images are not scalar multiples of each other (this means that the images are linearly independent and therefore span $\mathbb{R}^2$). $u = (1, 0, 0)$ and $v = (0, 1, 0)$ work for this: $Mu = (1, 2)$ and $Mv = (2, 3)$.
