Rank-nullity theorem on an example

Rank. $R_2\leftarrow R_2-2R_1$:
$$\begin{pmatrix} 1 & 2 & 3 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}.$$
$R_3\leftarrow R_3-R_2$:
$$\begin{pmatrix} 1 & 2 & 3 & 1 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
There are two nonzero rows, so $\operatorname{rank}(f)=2$. (Consistently, the columns $c_2=2c_1$ and $c_3=3c_1$ are redundant, while $c_1=(1,2,0)^T$ and $c_4=(1,3,1)^T$ are not proportional, so $\operatorname{Im} f=\operatorname{span}(c_1,c_4)$ has dimension $2$.)

Kernel. The reduced system is $x_4=0$ and $x_1+2x_2+3x_3+x_4=0$, i.e. $x_4=0$ and $x_1=-2x_2-3x_3$. Therefore
$$\ker f=\{(-2s-3t,\ s,\ t,\ 0): s,t\in\mathbb{R}\}=\operatorname{span}\big((-2,1,0,0),\,(-3,0,1,0)\big).$$
These two vectors are independent (look at the 2nd and 3rd coordinates), so $\dim\ker f=2$.

Conclusion. $\operatorname{rank}(f)=2$ and $\dim\ker f=2$, and $2+2=4=\dim\mathbb{R}^4$: the rank-nullity theorem is verified on this example.