If we identify $M_{2 \times 3}$ with $\Bbb{R}^6$ and similarly with the other vector space, we see that $L$ is a linear transformation from $\Bbb{R}^6$ to $\Bbb{R}^9$. Now I claim that the kernel of $L$ is trivial. Indeed, suppose there is a matrix
$A = \left[\begin{array}{ccc} a & b & c & \\ d & e & f \end{array}\right]$
such that $L(A) = 0$. Then we get the following system of equations:
$\begin{eqnarray*} 2a - d &=& 0 \\ 2b - e &=& 0 \\ 2c - f &=& 0 \\ a + 2d &=& 0 \\ b + 2e &=& 0 \\ c + 2f&=& 0 \\ 3a + d &=& 0\\ 3b + e &=& 0 \\ 3c + f &=& 0. \end{eqnarray*}$
In other words we are trying to find the null space of the matrix
$\left[\begin{array}{cccccc} 2 & 0 & 0 & -1 & 0 & 0 \\ 0& 2 & 0 & 0 & -1 & 0 \\ 0& 0& 2 & 0 & 0 & -1 \\ 1 & 0 & 0 & 2 & 0 & 0 \\ 0& 1 & 0 & 0 & 2 & 0 \\ 0& 0& 1 & 0 & 0 & 2\\ 3 & 0 & 0 & 1 & 0 & 0 \\ 0& 3 & 0 & 0 & 1 & 0 \\ 0& 0& 3 & 0 & 0 & 1 \end{array}\right].$
Upon row reduction, this matrix in reduced row echelon form is
$\left[\begin{array}{cccccc} 1 & 0 & 0 & 0 & 0 & 0 \\ 0& 1 & 0 & 0 & 0 & 0 \\ 0& 0& 1 & 0 & 0 & \\ 0 & 0 & 0 & 1 & 0 & 0 \\ 0& 0 & 0 & 0 & 1 & 0 \\ 0& 0& 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 \end{array}\right].$
You can see that there are no free variables, so that the dimension of the kernel if zero. By rank nullity, we have
$\begin{eqnarray*} \dim \textrm{ran} L &=& \dim \Bbb{R}^6 - \dim \textrm{ker} L \\ &=& 6 - 0 \\ &=& 0 \end{eqnarray*}$
from which it follows that the range of $L$ has dimension 6.