How many solutions to this under-determined equation I have a system of equations that looks as such:

$$A=\begin{pmatrix}-3&amp;-3&amp;2\\ \:-9&amp;-9&amp;6\end{pmatrix}$$
$$b=\begin{pmatrix}2\\ \:-4\end{pmatrix}$$

How many solutions exist to this matrix? I tried to answer this using RREF, which gives me:

=\begin{pmatrix}-9&amp;-9&amp;6\\ 0&amp;0&amp;0\end{pmatrix}

Thereby, since I have 2 free variables, I have a plane of infinitely many solutions. Is this fair?
Your notation is vague. If you mean "How many solutions are there to 
$$A\begin{bmatrix}x\\y\\z\end{bmatrix}=b?"$$ the answer is "none". Sophisticated reason: rank of coefficient matrix=1 $\ne$2=rank of augmented matrix. Basic reason: multiplication of 1st given equation by 3 gives $$-9x-9y+6z=6$$ a contradiction to the second given equation which is $$-9x-9y+6z=-4.$$
No, there is no solution. Let $A\in\mathbb R^{m\times n}$ and $b\in\mathbb R^m$. The main idea is to reduce the augmented coefficient matrix $(A\vert b)$ to row echolon form via Gaussian elimination. This does not change the solution set. We get
$$
\begin{pmatrix}
a_{11} &amp; a_{12} &amp; \dots &amp; a_{1k} &amp; \dots &amp; a_{1n} &amp; b_1\\
0 &amp; a_{22} &amp; \dots &amp; a_{2k} &amp; \dots &amp; a_{2n} &amp; b_2\\
\vdots &amp; \ddots &amp; \ddots &amp; \vdots &amp; \ddots &amp; \vdots &amp; \vdots\\
0 &amp; \dots &amp; 0 &amp; a_{kk} &amp; \dots &amp; a_{kn} &amp; b_k\\
\vdots &amp; \ddots &amp; \vdots &amp; 0 &amp; \dots &amp; 0 &amp; \vdots\\
0 &amp; \dots &amp; 0 &amp; 0 &amp; \dots &amp; 0 &amp; b_m
\end{pmatrix}
$$
Now, if $b_i$, $k+1\leq i\leq m$ is non-zero, then the linear equation system has no solution. This is equivalent to saying that the rank of $A$ is smaller than the rank of $(A\vert b)$. So, we have at least one solution if
$$\operatorname{rank}(A) = \operatorname{rank}(A\vert b).$$
It is exactly one solution if
$$\operatorname{rank}(A) = n$$
and infinitely many solutions if
$$\operatorname{rank}(A) = r &lt; n.$$
In the latter case the solution set has the dimension $n-r$.