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