Rank-nullity on a 3×4 matrix

Row reduction. $L_2\leftarrow L_2-2L_1$, $L_3\leftarrow L_3-L_1$:
$$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 \end{pmatrix}.$$
$L_3\leftarrow L_3-L_2$:
$$\begin{pmatrix} 1 & 2 & 3 & 4 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$
Rank is $2$.

Kernel dimension. By rank-nullity, $\dim\ker A = 4-2 = 2$.

Kernel. Solve $Ax=0$: $x_1+2x_2+3x_3+4x_4=0$ and $x_4=0$. So $x_1+2x_2+3x_3=0$. Free variables $x_2, x_3$; basis $(-2,1,0,0)$, $(-3,0,1,0)$.

Conclusion. $\operatorname{rank}(A)=2$ and $\dim(\ker A)=2$.