How to find matrix from general solution? 
Find a $2 \times 3$ system (two equations with three unknowns) such that its general solution has form $\begin{pmatrix}1\\1\\0 \end{pmatrix}+s\begin {pmatrix}1\\2\\1\end{pmatrix},\ s \in \Bbb R.$ 

I tried thinking that $s\begin {pmatrix}1\\2\\1\end{pmatrix}$ is solution to kernel of asked matrix, and tried matrix $\begin {pmatrix}1&-1&1\\1&-1&1\end{pmatrix}$, but it also includes $\begin{pmatrix}1\\1\\0 \end{pmatrix}$ in its kernel!
 A: Let $$A\mathbf{x}=\mathbf{b}\tag{1}$$ be the system in matrix form, where $A$ is the coefficient matrix for the asked system in matrix form. Where
$\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}$
Notice that $\ker(A)=\{k\mathbf{v}|k\in\mathbb{R}\}$, where
$\mathbf{v}=\begin{pmatrix}1\\2\\1\end{pmatrix}$
You can take any matrix $A$ as above, for instance 
$$A=\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}$$
since $\mathbf{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}$ is a particular solution we have $\mathbf{b}=\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$. Hence the system 
$$\begin{pmatrix}2&-1&0\\0&1&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}1\\1\\0\end{pmatrix}$$
satisfies the problem.
A: One can translate the first two components of $u= (x,y,z)^T= (s+1, 2s+1, s)^T$ into the following system $Au=b$, the third component stays free and acts as $z=s$:
$$
x = s + 1 = z + 1 \iff x - z  = 1 \\
y = 2s + 1 = 2 z  + 1 \iff y - 2 z = 1 \\
$$
Or written as augmented matrix
$$
(A\mid b) =
\left(
\begin{array}{rrr|r}
1 & 0 & -1 & 1 \\
0 & 1 & -2 & 1 \\
\end{array}
\right)
$$
