Solving $A \vec{x} = \vec{0}$ Suppose that A is a 3 x 3 matrix and that the solution to
$A\vec{x} = \left[\begin{array}{ccc}8\\10\\5\end{array}\right]$ is
$\vec{x} = \left[\begin{array}{ccc}4\\-1\\0\end{array}\right] + s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right], s \in \mathbb R$

The problem:
Is this enough information to find the solution to $A \vec{x} = \vec{0}$?
If yes, what is the solution?
If no, explain why not.
The solution:

1) Yes. Solution to $A \vec{x} = \vec{0}$ is $\vec{x} = s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right], s \in \mathbb R$
  2) If the solution to $A\vec{x} = \left[\begin{array}{ccc}8\\10\\5\end{array}\right]$ is $\vec{x} = \left[\begin{array}{ccc}4\\-1\\0\end{array}\right] + s\left[\begin{array}{ccc}2\\3\\-1\end{array}\right]$, then
  3) $\left[\begin{array}{c|c}&8\\A&10\\&5\end{array}\right] \xrightarrow[\text{}]{\text{rref}} \left[\begin{array}{ccc|c}1&0&-2&4\\0&1&3&-1\\0&0&0&0\end{array}\right]$ then
  4) $\left[\begin{array}{c|c}&0\\A&0\\&0\end{array}\right] \xrightarrow[\text{}]{\text{rref}} \left[\begin{array}{ccc|c}1&0&-2&0\\0&1&3&0\\0&0&0&0\end{array}\right]$

My question:
How am I supposed to start this question?
(the solution is given, but I do not know how to achieve even the first step)
 A: This actually does not require much work, just a bit of thought. Remember that the matrix transformation is linear meaning if we multiplied $A$ by the sum of 2 vectors $v+ w$ then we get $Av + Aw$ more concretely
$$A(v+w) = A(v) + A(w)$$
Now we are told that the solution to your equation involves 2 vectors, lets call them $v$ and $w$. Now the solution is $x = v+ sw$ where $s$ can be any real number. But why can $s$ be any real number, simple because $Aw = 0$. Let me show you. Let $s$ be substiuted by 2 arbitrary numbers $1$ and $2$
$$Ax = A(v+ sw) = Av + sAw = Av + 1(Aw) =  Av + 2(Aw)=  \begin{pmatrix}8 \\ 10 \\5 \end{pmatrix}$$
$$Av + Aw = Av + 2Aw$$
$$Aw = 0$$
As a result, we can see why it does not matter what $s$ is.
$$Ax = A(v+ sw) = Av + sAw = Av + s(0) = \begin{pmatrix}8 \\ 10 \\5 \end{pmatrix}$$
So it didnt matter what $s$ was as $Aw = 0$. So what is the solution to $Ax = 0$ then?
A: If the solution of $A\vec{x} = b$ is of the form $\vec x=\vec x_1+s\vec x_2$, then
$$A\vec x_1+sA\vec x_2=b.$$
But for this relation to be true for any $s$, we must have 
$$A\vec x_2=0$$ and $x=s\vec x_2$ is a solution of $A\vec x=0$.
