Let $\bf{A}$ be a matrix acting on $\bf{r}$ $= (x_1,y_1,z_1)^T$ to produce a new vector $\bf{v}$ $=(x_2,y_2,z_2)^T:$ $$\bf{A.r} = \begin{pmatrix} 1 & 2 & 3\\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 \\z_1 \end{pmatrix}= \begin{pmatrix} x_2 \\ y_2 \\z_2 \end{pmatrix} $$ Clearly this matrix has rank 1 i.e. we can write
$$\begin{pmatrix} 1 & 2 & 3\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} x_1 \\ y_1 -2x_1 \\z_1 -3x_1 \end{pmatrix}$$ I have a few questions:
What would it mean to "solve" the system of equations (i.e., what are we doing mathematically and what will it tell us at the end)? (If this sounds confused, it's because it is. I can attempt further clarification if it would help.)
The dimension of $\bf{ker\ A}$ is 2 since any vector that we feed in that satisfies $x_1 + 2(y_1-2x_1) + 3(z_1 -3x_1) = 0$ (i.e. a plane of vectors) will give the null vector. What does this imply about the structure of our solution? That is has dimension 1?
A matrix can represent a system of linear equations. In your example, the equations are:
$x\cdot x_1 +2y \cdot y_1 + 3z \cdot z_1=x_2$
$2x\cdot x_1 +4y \cdot y_1 + 6z \cdot z_1=y_2$
$3x\cdot x_1 +6y \cdot y_1 + 9z \cdot z_1=z_2$
After you did gauss elimination you got the equations:
$x\cdot x_1 +2y \cdot (y_1 - 2x_1) + 3z \cdot (z_1-3x_1)=x_2$
$0x\cdot x_1 +0y \cdot (y_1 - 2x_1) + 0z \cdot (z_1-3x_1)=y_2$
$0x\cdot x_1 +0y \cdot (y_1 - 2x_1) + 0z \cdot (z_1-3x_1)=z_2$
Solving the system means get all $(x,y,z)$ satisfying the equations.
$dim(ker A) = 2$ meaning you will have infinite solution (if no contradiction occurs i.e $0\cdot x+0 \cdot y + 0 \cdot z = a \neq 0$), and the degree of freedom is 2.