Show $Ax = e_{m+1}$ is consistent. $A = \binom{C}{a^{T}}$. $C$ is $mxn$ matrix. $a$ is $n$-vector that is lin. indep. of rows of $A$. $C$ is an $mxn$ matrix of any shape, shape, rank, etc.  $a$ is a $n$-vector that is linearly independent of the rows of $C$. Let $A = \binom{C}{a^{T}}$.  Let $e_{m+1}$ denote the $m+1$ Identity vector.  Show that $Ax=e_{m+1}$ is consistent. 

I don't know where to start.
Note that $Ax=e_{m+1}$ means that you should find an $x$ with $Cx=0$ and $a^Tx=1$. Because you can scale $x$ appropriately if necessary, it suffices to find $x\in \operatorname{ker}(C)$ with $x\notin \operatorname{ker}{a^T}$. This is possible because $\operatorname{ker}(C)\subseteq \operatorname{ker}{a^T}$ implies that $a^T$ is linearly dependent to the rows of $C$, see e.g. this math.stackexchange question with $L_i$ taken to be the functionals given by multiplying with the rows of $C$ and $L$ is taken to be the functional given by multiplying with $a^T$.
This is going to be hard to prove because it is false. 

Consider 
$$
C = \left( \begin{array}{ccc} 2 &amp; 2 &amp; 2 \\ 1 &amp; 1 &amp; 1
\end{array}\right) \\
a^T = ( -1, 0, 1 )
$$
which is not linearly dependent on the rows of $C$.

Then 
$$
A = \left( \begin{array}{ccc} 2 &amp; 2 &amp; 2 \\ 1 &amp; 1 &amp; 1 \\ -1 &amp; 0 &amp; 1
\end{array}\right)  
$$ and $ Ax = e_3$ is
$$
2x_1 + 2x_2 + 2x_3 = 1 \\x_1 + x_2 + x_3 = 1 \\ -x_1 + x_3 = 1
$$
and those first two equations are inconsistent with each other.