Particular, special and general solutions For a given augmented matrix
$
A=\left(\begin{matrix}
4 & 6 & -2 & 2 \\
2 & 0 & 4 & 2 \\
\end{matrix}\right)
$, how would you find the particular, special, and genera solution? I am independently studying Linear Algebra but got stuck on this problem because I have no one to ask from. For the textbook I am using, I believe special is meant as the null space vector. 
 A: I'll model how we can find the general solution here:
We have the augmented matrix, where : $A = [A'\mid b\,]$ and $A'$ is a $2 \times 3$ coefficient matrix in the system $$A'X = {\bf b,\quad} \text{with}\;X = \begin{bmatrix}  x \\ y \\ z\end{bmatrix}\;\text{and}\;{\bf b} = \begin{bmatrix} 2\\ 2 \\ 0 \end{bmatrix}.$$ So we have 
$$A = [A'\mid b\,] = 
\begin{pmatrix}
4 & 6 & -2 & 2 \\
2 & 0 & 4 & 2 \\
0 & 0 & 0 & 0
\end{pmatrix}$$
Writing the system this way helps to remind us that we'll have a "free variable" to account for; writing $A$ this way adds no information, and takes away no information, from your posted augmented matrix:
$$A = 
\begin{pmatrix}
4 & 6 & -2 & 2 \\
2 & 0 & 4 & 2 \\
\end{pmatrix}$$
So we proceed to get the matrix in row-echelon form. 
Swapping rows and dividing each row by $2$ gives us $\begin{pmatrix}
1 & 0 & 2 & 1 \\
2 & 3 & -1 & 1 \\
0 & 0 & 0 & 0
\end{pmatrix}\quad$
And taking $\;(-2 R_1 + R_2 \rightarrow R_2)\;$ gives us: $\;\begin{pmatrix}
1 & 0 & 2 & 1 \\
0 & 3 & -5 & -1 \\
0 & 0 & 0 & 0
\end{pmatrix}$
I "carried" the zero row along just to make evident after reducing our matrix a bit, that our system has no unique solution, but infinitely may solutions: The coefficient matrix has rank $2 < 3$.
So we equate the "free" variable $z$ to some parameter, say $z = t.$ Then we read off what this tells us about $x, y$ with respect to the parameter $z = t$.
$$\displaystyle x + 2z = 1 \iff x = 1 - 2t;\quad 3y - 5z = -1 \iff y = \frac53 t - \frac 13$$
Then the general solution is given by $\;\displaystyle X = \begin{bmatrix} x \\ y\\ z \end{bmatrix} \; = \;\begin{bmatrix} 1 - 2t \\ \frac 53t - \frac 13\\ t \end{bmatrix}$
Now, can you take it from there, what a particular solution might be? (Hint: set $t$ equal to the simplest value you can think of to simplify the particular solution.)
