Finding the kernel of a linear map Our exercise is to find all solutions to the equation $Ax = 0$, among others for the following matrix  
$$A =\begin{pmatrix} 6 & 3 & -9 \\ 2 & 1 & -3 \\ -4 & -2 & 6 \end{pmatrix}.$$
This amounts to finding the kernel, and obviously, the rows of the matrix are multiples of each other, so we can reduce the equations to:
$$6x_1 + 3x_2 - 9x_3 = 0.$$
Choosing $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$ would fulfill the equation in my opinion.
However, the official solution is described as the following set:
$$\lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}$$
with $\mu, \lambda \in$  R.
My questions now:
Where does the vector (and its multiples) $\lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ come from?
Also, since after the reduction we are left with only one row, the dimension of the kernel should accordingly be $1$ - however $\left[\begin{smallmatrix} 0 \\ 3 \\ 1  \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 1 \\ -2 \\ 0  \end{smallmatrix}\right]$ (from $x_3 = 0$, $x_1 = 1$ and $x_2 = -2$) seem to be two different basis vectors - what have I done or understood wrongly?
Many thanks

First, note that 
$$\begin{pmatrix} 1 \\ -2 \\ 0  \end{pmatrix} =
\begin{pmatrix} 1\\1\\1\end{pmatrix} - \begin{pmatrix} 0 \\ 3 \\ 1 \end{pmatrix}, $$
So your missing solution 
$\left[\begin{smallmatrix} 1 \\ -2 \\ 0  \end{smallmatrix}\right] $ 
is, in fact, a special case of the general solution 
$ \lambda \left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right] + \mu \left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right] $ with $\lambda =1$ and $\mu = -1$.
Second, the general solution can be obtained from reduced equation $6x_1 + 3x_2 - 9x_3 = 0$ by imposing certain assumptions:


*

*For example, if we assume $x_1 = 0$, we will get the following
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies \begin{cases}x_1 = 0 \\ 3x_2 - 9x_3 = 0 \end{cases} 
\implies \begin{cases}x_1 = 0 \\ x_2 = 3 x_3\end{cases} 
$$
Assuming parametrization $x_3 = \mu$, we get 
$$
\begin{cases} x_1 = 0 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies 
\begin{cases}x_1=0 \\ x_2 = 3 x_3 \\ x_3 = \mu & - \operatorname{parameter}\end{cases} 
\iff 
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} 
$$
We can rewrite this solution in the vector form:
$$
\begin{cases}x_1 = 0 \\ x_2 = 3\mu \\ x_3 = \mu \end{cases} 
\iff
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} 
= \begin{bmatrix} 0 \\ 3\mu \\ \mu \end{bmatrix} 
\iff 
x
= \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu
$$

*Similarly, imposing condition $x_1 = x_2$, we  write
$$
\begin{cases} x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0\end{cases} 
\implies \begin{cases}x_1 = x_2 \\ 6x_1 + 3x_2 - 9x_3 = 0 \end{cases}  
\implies \begin{cases}x_1 = x_2 \\ x_1 = x_3\end{cases} 
$$
Assuming parametrization $x_1 = \lambda$, we get 
$$
\begin{cases} 
x_1 = \lambda  & -\operatorname{parameter}\\ x_2 = x_1 \\ x_3 = x_1
\end{cases} 
\implies 
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} 
$$
The vector form of the solution then looks like 
$$
\begin{cases} x_1 = \lambda \\ x_2 = \lambda \\ x_3 = \lambda \end{cases} 
\implies 
x = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} 
= \begin{bmatrix} \lambda \\ \lambda \\ \lambda \end{bmatrix} 
\iff 
x= \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda
$$
Third, Since parametrized solutions $x = x(\lambda)$ and $x = x(\mu)$ are linearly independent, and since the equation  $ 6x_1 + 3x_2 - 9x_3 = 0$ is linear, the general solution can be written as the sum of two independent parametrized particular solutions, i.e.
$$
x = x_\lambda + x_\mu = \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda + \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu
$$

Finally, Note that we could have imposed different assumptions on $x$, and that would result in different parametrized solutions. 
However, as long as these solutions will be linearly independent, the will still span the same two-dimentional space of solutions which we have now. 
Thus, the kernel of $A$, which is also a space of all solutions of the system $Ax = 0$, is the linear span of vectors $\left[\begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix}\right]$ and $\left[\begin{smallmatrix} 0 \\ 3 \\ 1 \end{smallmatrix}\right]$:
$$
\ker A 
= \operatorname{span} \left(\; \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \;\right) 
= \left\{\; \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}, \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix} \; \right\}
= \left\{\ \vec v = \begin{bmatrix} 1 \\ 1  \\ 1\end{bmatrix}  \lambda + \begin{bmatrix} 0 \\ 3  \\ 1\end{bmatrix}  \mu 
\mathrel{\bigg|}
\lambda, \mu \in \mathbb R\ \right\}
$$
