I continue to come across $Ax \leq b$ or $Ax= b$ in optimization problem, but I am having trouble interpreting the meaning of this. Does this have a similar meaning to the following (Cramer's Rule) from Linear Algebra?
$Ax = b \Leftrightarrow x = A^{-1} b = \frac{1}{det(A)}[adj(A)]b$
Where $A$ is a matrix and I assume $x$ and $b$ are vectors? $x$ being the solution? Essentially, this is stating give me $x$ such that $Ax$ is $\leq$ to $b$?
I remember encountering the same confusion as you when presented with such inequalities in optimization problems.
$x = (x_1, \dotsc, x_n)^T$ is a column vector of variables which are usually under our control in the problem, but subject to certain constraints. For instance, you might encounter the constraint $x \geq 0$. In the vector terms you may be used to, this is not very meaningful, but in this context it means that $x$ is entrywise greater than $0$: Explicitly, $x_1 \geq 0, \dotsc, x_n \geq 0$. you can see why this shorthand is helpful when there are a lot of variables!
The statement $Ax \geq b$, is similarly shorthand for a system of linear inequalities with coefficients given by $A$ via usual matrix multiplication, and $b$ a fixed column vector.
For example, if $$A = \left(\begin{array}{cc} 1&2\\ -1&6 \end{array}\right)$$ and $b = (2, 3)^T$, the inequality $Ax \leq b$ represents the system:
$$\begin{array}{rcrcr} x_1 &+& 2x_2 &\leq& 2\\ -x_1 &+& 6x_2 &\leq& 3 \end{array}$$
These inequalities specify further constraints on the variable $x_1, \dotsc, x_n$. Note that $A$ need not be a square matrix, so it will not rarely be possible to solve with equality by inverting $A$ to obtain $x = A^{-1}b$, and furthermore this is usually not useful: the aim of the problem is to extremize an expression in terms of the variables $x_1, \dotsc, x_n$, and typically only some of the inequalities in this system will hold with equality at an optimal solution.