Is $Cx\geq b$ solvable if and only if $C$ is invertible? Let's say that we have a system of linear inequalities:
$$
\begin{bmatrix}
c_{1,1} & c_{1,2} & \ldots & c_{1,n} \\
c_{2,1} & c_{2,2} & \ldots & c_{2,n} \\
\vdots & \vdots & \ddots & \vdots \\
c_{m,1} & c_{m,2} & \ldots & c_{m,n}
\end{bmatrix}
\times
\begin{bmatrix}
x_1 \\
x_2 \\
\vdots \\
x_n
\end{bmatrix}
\geq
\begin{bmatrix}
b_1 \\
b_2 \\
\vdots \\
b_m
\end{bmatrix}
$$
It can be represented in a matrix form:
$$\mathbf{C}\mathbf{x} \geq \mathbf{b}$$
Does it hold that:
$$\mathbf{x} \geq \mathbf{C}^{-1}\mathbf{b} $$
and that $\mathbf{C}$ is invertible if and only if the whole system is solvable?
P.S. All the numbers $x_i, b_i, c_{i,j}$ are real. Would restricting them to be integers change the answer?
EDIT 1: for matrices $\mathbf{x}$ and $\mathbf{y}$ it holds that $\mathbf{x} \geq \mathbf{y}$ if and only if every element of $\mathbf{x}$ is $\geq$ to corresponding element in $\mathbf{y}$.
EDIT 2: the $x_i$ are bounded to $[-2,2]$.
 A: First of all, no, C need not be invertible, because if $m \gt n$ then you have an overdetermined system (not a square matrix, so an inverse would not exist), but it can still hold as an inequality.  
Also, two rows could be identical, in which case the size would not be consequential, it would only matter that the inequality holds. 
It might help to conceptualize matrix multiplication as application of the dot product between the rows of your matrix, and the column vector $\vec{x}$. If $v_i \cdot{x} \ge b_i$, then your condition is met. 
If $C^{-1}$ exists then the condition $\mathbf{x} \geq \mathbf{C}^{-1}\mathbf{b}$ must be true, where '$\ge$' is an elementwise comparison of the vectors.
A: Your question is the problem of feasibility in linear programming.
You should check this question.
This feasibility problem $Cx \leq b$ is equivalent to the following optimization problem
\begin{align*}
max. \quad &t \\
c_ix + t \leq &b_i,\ i = 1,\ldots, m
\end{align*}
where $c_i$ are the rows of $C$.
