b such that Ax = b has no solution having found column space $A:=\begin{bmatrix}
  2 & 6 & 0 \\
  3 & 1 & 3 \\
  1 & 0 & 0 \\
  4 & 8 & 1
 \end{bmatrix}$
I've found the basis for the column space by doing row reduction (i.e. basis is just the columns vectors of A in this case), and the null space only has the trivial solution.
Question
Find a basis for $B = \{b \in R^4 \ | \ Ax = b \ has \ no\  solution\}$.
Is there a quick way of doing this? I know I can augment A with $b = (b_1, b_2, b_3, b_4)^T$, do row reduction, and then look at the row of zeros, but that seems quite laborious?
 A: When you multiply a matrix by a (column) vector, you get a vector that is a linear combination of the columns of the matrix, where the coefficients/weights come from the components of the column vector. I'm too lazy to illustrate with a $4\times 3$ example, so here's a $2\times 2$ example: $$Ax=\begin{pmatrix} \mathbf{A_1} & \mathbf{A_2}\end{pmatrix}\begin{pmatrix}x_1 \\ x_2\end{pmatrix}=x_1\mathbf{A_1}+x_2\mathbf{A_2}$$
where $\mathbf{A_1}$ is the first column of the matrix $A$ and $\mathbf{A_2}$ is the second column. 
In other words, $Ax$ is a linear combination of the columns of $A$. But the column space of $A$, i.e. the image of $A$, is just the span of the columns of $A$. So the equation $Ax=b$, which says that $b$ is a linear combination of the columns of $A$, will have a solution if and only if $b$ is in the column space of $A$.
A: Denote $A\in\mathbb{F}^{m\times n}$ by 
$$
A=\begin{bmatrix}a_{11} & a_{12} & . & . & . &  &  & a_{1n}\\
a_{21} &  &  &  &  &  &  & a_{2n}\\
. &  & . &  &  &  &  & .\\
. &  &  & . &  &  &  & .\\
. &  &  &  & . &  &  & .\\
\\
\\
a_{m1} & a_{m2} & . & . & . &  &  & a_{mn}
\end{bmatrix}
$$
and 
$$
x=\begin{bmatrix}x_{1}\\
x_{2}\\
.\\
.\\
.\\
\\
\\
x_{n}
\end{bmatrix}
$$
then 
$$
Ax=\begin{bmatrix}a_{11}x_{1}+a_{12}x_{2}+...+a_{1n}x_{n}\\
a_{21}x_{1}+a_{22}x_{2}+...+a_{2n}x_{n}\\
.\\
.\\
.\\
\\
\\
a_{m1}x_{1}+a_{m2}x_{2}+...+a_{mn}x_{n}
\end{bmatrix}=\begin{bmatrix}a_{11}\\
a_{21}\\
.\\
.\\
.\\
\\
\\
a_{m1}
\end{bmatrix}x_{1}+\begin{bmatrix}a_{12}\\
a_{22}\\
.\\
.\\
.\\
\\
\\
a_{m2}
\end{bmatrix}x_{2}+...+\begin{bmatrix}a_{1n}\\
a_{2n}\\
.\\
.\\
.\\
\\
\\
a_{mn}
\end{bmatrix}x_{n}
$$
Note that column matrices that are multiplied by the $x_{i}$ are
the columns of $A$ hence any vector $b$ of the form $b=Ax$ is a
linear combination of the columns of $A$ and thus in the span of
the columns of $A$.
On the other hand, denote the columns of $A$ as $C_{1},C_{2},...,C_{n}$.
If $b$ is a linear combination of the columns of $A$ this precisely
means that there are scalars $\alpha_{1},\alpha_{2},...\alpha_{n}$
s.t 
$$
b=\alpha_{1}C_{1}+\alpha_{2}C_{2}+...+\alpha_{n}C_{n}
$$
and we see (using the above about $Ax$) that choosing $x_{i}=\alpha_{i}$
would give us $Ax=b$.
We conclude that the span of the columns of $A$ is precisely the
set of solutions $Ax=b$, thus you are looking for all the vectors
in the space that are not in that spanned subspace 
Note: Since $0\not\in B$ (since $Ax=0$ have a solution, $x=0$)
then $B$ is not a subspace and thefore we can't talk about a basis
for this space. 
