How can I find the rank of a matrix $A$ when the solution sets to $Ax=b$ is given? 
Suppose the complete set of solutions to $Ax = b$ where $b = [2, 4, 2]^T$  is given by
  $$
 x = [2, 0, 0]^T + c[1, 1, 0]^T + d[0, 0, 1]^T.
$$


This is from a YouToube Video of Gilbert Strang's lecture on Linear Algebra. He said that the rank of the matrix $A$ was $1$. I understand that $b$ is a column vector so its rank is $1$. How did the students identify the dimension of the null space of $A$ to be $2$? This is something that baffles me. 
$$A = \begin{bmatrix}1 & -1 & 0 \\ 2 & -2 & 0 \\ 1 & -1 & 0 \end{bmatrix}$$
Firstly, why is it that if $[0, 0, 1]^T$ is in the null space, the third column is a zero vector. 
Assuming $c = 1$, $d = 1,$ 
and $A$ is the matrix as proposed, in the first row of $B$, $1\times 2 - 1 \times 1 + 1  \times 0$ gives $1$ instead of $2.$ 
 A: I wonder if the easiest way to look at this is a follows:
$[2,0,0]^\top$ is a particular solution to $Ax=b$. Why? Because it is true that it doesn't matter what $c$ and $d$ values we pick: for any $c$ and $d$ we get a vector, $x = [2, 0, 0]^\top + c[1, 1, 0]^\top + d[0, 0, 1]^\top$ that satisfies $Ax=b$. Therefore, we can pick $c=0$ and $d=0$, and we end up with $x = [2, 0, 0]^\top.$
This is tantamount to saying that $c[1, 1, 0]^\top + d[0, 0, 1]^\top$ are really non-contributory. Or in other words, that they are in the null space of $A.$ If a system of three equations with three unknowns can be solved in more than one way is because the system is under-determined, and there are free variables.
$c[1, 1, 0]^\top + d[0, 0, 1]^\top$ represents all linear combinations of two linearly independent vectors in the null space, and any vector space (or subspace) observes closure under addition and scalar multiplication. Hence, $c[1, 1, 0]^\top + d[0, 0, 1]^\top=0.$
A: From the rank-nullity theorem, the following relation satisfies for an $m \times n$ matrix.
$$\text{rank}(A)+\text{nullity}(A)=n$$
The dimension of the null space is called the nullity of A.
A: Watch the video from 10:38 again. 
First of all he points out that the size of the matrix $A$ is $3\times 3$. Secondly, the assumption of the complete sets of solutions tells you that
$$
Null(A)=2.
$$
Now the rank-nullity theorem tells you the rank of $A$ is given by
$$
Rank(A)=3-Null(A)=1.
$$
A: The homogeneous equation $A x = 0$ has as set of solutions $\mathcal{O}$ the set of differences of the solutions of the original equation (as $A x = A y = b$ implies $A (x - y) = 0$, and conversely if $Ax = b$ and $A u = 0$ then $A (x+u) = b$), so this set is
$$
\mathcal{O} = \{ a\,[1, 1, 0]^T + b\,[0, 0, 1]^T : a, b \in \mathbb{R}\},
$$
a $2$-dimensional space.
Therefore the nullity of $A$ is $2$ and the rank is $3 - 2 = 1$.
