linear system consistent For a matrix $$A =
\begin{bmatrix}
1 & -1& 2 & -1 & 0\\
-2 & 2 & -3& 0 & 1\\
3&-3&8&-7&2\\
 \end{bmatrix}$$
which reduces to...
\begin{bmatrix}
1 & -1& 0 & 3 & -2\\
0 & 0 & 1& -2 & 1\\
0&0&0&0&0\\
 \end{bmatrix}
How can I find all the possible vectors $b$, for which the linear system $Ax = b$ is consistent?
I have obtained the basis for $col(A)$.
Could someone explain the concept in a layman's term?
Thanks.
 A: Just augment your matrix by a vector $(x,y,z)^T$.  This yields to
$$\begin{pmatrix}1&-1&2&-1&0&x\\
0&0&1&-2&1&2x+y\\
0&0&0&0&0&-7x-2y+z\end{pmatrix}.$$
From here we see that $b$ must be
$$\begin{pmatrix}x\\ y\\ 7x+2y\end{pmatrix}.$$
Edit (much easier): since the dimension of the image of $A$ is $2$, any vector $b$ that solves the system must lie in that image.  So $b$ must lie in the span of any two linear independent  column vectors of $A$.  In the case of my first solution these are the fourth and the fifth vector.
A: As the space $col(A)$ is the space spanned by the columns of $A$, it is very helpful to understand the solution behaviour of $Ax=b$:
If you take a Vector $x = (x_1, x_2,x_3, x_4, x_5)^T$ and compute $Ax$, you will get a vector which is the first column of $A$ with multiplied by $x_1$, the second column multiplied by $x_2$, and so on, i.e. $Ax$ is a linear combination of the columns of $A$, and the other way around: Every vector from $col(A)$ can be written as $Ax$ for some $x$. Finding a basis of $col(A)$ means that you're looking for a least redundant way to describe such linear combinations.
Consistency means that $b$ is in a way that we can find some $x$ such that $Ax=b$, i.e. the system is solvable. In view of the paragraph above, this means that this is precisely the case when $b$ lies in $col(A)$, i.e. it can be written as a linear combination of a basis of $col(A)$. In this particular case, $b=(0, 0, 1)^T$ doesn't lead to a consistent system: You can't find a linear combination of the columns of A such that the first two rows are zero, but the third isn't. You don't see this immediately with $A$, but the reduced form (or the basis you found) helps to see it.
So when looking for all consistent $b$s, you are looking for all vectors in $col(A)$.
I'm aware that this answer is not precisely in layman's terms, but I still hope that it helps you to put your pieces together.
A: The reduced form of $A$ shows that $A$ has rank 2 (because it contains an invertible $2\times2$ submatrix). As a consequence, the vectors
$$V=A\mathbf R^5=\{y\in\mathbf R^3|\,\exists x\in\mathbf R^5,\; Ax=y\}$$
form a two-dimensional vector space. If $b\in V$, there is a solution to the system $Ax=b$, otherwise ($b\in\mathbf R^3\setminus V$), there is none.
A: $b$ is a solution if there is $x$ with $Ax = b$. To find all possible $b$ you therefore apply $A$ to every $x \in X$: 
$$ B = \{ Ax \mid x \in X\}$$
$X= \mathbb R^5$. It is enough to instead of applying $A$ to every $x$ to apply it to any basis $b_1,\dots , b_5$. This is because the images of the basis span the image. Let $b_i$ be the standard basis.  Then the images of $b_i$ are the columns of $A$. You can see that some are linearly dependent. Now it is easy to find the image as a span of linearly independent columns of $A$.
