For $\det(A)=0$, how do we know if $A x = b$ has no solution or infinitely many solutions? If the determinant $\det(A)$ of the matrix $A$ of a non-homogeneous system of equations is $0$, then how do we know if it has no solutions or infinitely many solutions?
And while we are at it, kindly answer the following "sub-questions" arising from it. I shall be really grateful to you as it will be crucial to my understanding of the whole thing:
a) Since the determinant being zero means that a situation of "Division by zero" arises (using Cramer's Rule), the "no solution" option is understandable as division by zero is not defined. But it confuses me how then, in any circumstance, the system can have infinitely many solutions. I mean, won't we encounter division by zero in all cases when determinant is zero? So please give me an intuitive and insightful explanation to it.
b) Will I be wrong to assume that, in a case when determinant is equal to zero, there are infinitely many solutions if and only if it's a homogeneous system of equations? Please explain why or why not.
And kindly don't forget the main question--"for determinant $=0$, how to know if there are no or infinitely many solutions?"
 A: For the case of a linear system of non-homogeneous equations, you need to consider the augmented matrix and compare its rank to the rank of the coefficient matrix of the system.
A: There are two cases actually: 
If the vector b is not in the column space of the matrix A, it will have no solutions. 
If the b is in the column space of A, and since det(A)=0, then it will have infinitely many solutions. 
Hoping this can be a good starting point for you. 
A: For any square linear system $\,A\vec x=\vec b\,$ over some field, there exists a unique solution iff $\,\det A\neq 0\,$ , as then we can use the inverse matrix:
$$A\vec x=\vec b\Longleftrightarrow A^{-1}A\vec x=A^{-1}\vec b\Longleftrightarrow A^{-1}\vec b=\vec x $$
As for (a) and your "main question": if $\,\det A=0\,$ one still may have to check whether there are no solutions or infinite solutions (assuming we're working on an infinite field). For example, if the system is homogeneous (over an infinite field) it must have infinite solutions, whereas if the system is  non-homogeneous it may have no solutions or several:
$$\begin{cases}x+y=1\\x+y=1\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{1}\longrightarrow\,\,\text{infinite solutions}$$
$$\begin{cases}x+y=1\\x+y=0\end{cases} \Longleftrightarrow \begin{pmatrix}1&1\\1&1\end{pmatrix}\binom{x}{y}=\binom{1}{0}\longrightarrow\,\,\text{no solutions at all}$$
and, of course, in both cases above we have $\,\det A=0\,$
A: Let be the system of equations:
$$
\left\{
\begin{array}{l}
ax+by=e \\
cx+dy=f
\end{array}
\right. $$
If we multiply the determinant by $x$ and use some properties of determinants we get:
$$x\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
ax & b \\
cx & d \\
\end{array}
\Rightarrow$$
$$\Rightarrow
x\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
e-by & b \\
f-dy & d \\
\end{array}
\Rightarrow$$
$$\Rightarrow
x\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
e & b \\
f & d \\
\end{array}+
\begin{array}{|cc|}
-by & b \\
-dy & d \\
\end{array}
\Rightarrow$$
$$ \Rightarrow x\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
e & b \\
f & d \\
\end{array}+ 0
\Rightarrow$$
$$\Rightarrow x\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
e & b \\
f & d \\
\end{array}
$$
If we have a system where $\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}=0$ and $\begin{array}{|cc|}
e & b \\
f & d \\
\end{array}=0$ then $x$ can be any real number.
We can do the same with $y$, but in that case we get:
$$\Rightarrow y\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}= \begin{array}{|cc|}
a & e \\
c & f \\
\end{array}
$$
Again if $\begin{array}{|cc|}
a & b \\
c & d \\
\end{array}=0$ and $\begin{array}{|cc|}
a & e \\
c & f \\
\end{array}=0$, $y$ can assume many values depending on $x$ values or vice versa.
A: I've got a geometric interpretation:
The $2 \times 2$ matrix $\bf{A}$ transforms a vector $\bf{x}$ in the plane to another vector $\bf{b}$.
If $\text{det }\bf{A}=0$, this transformation is, in fact, a flattening (the geometric interpretation of the determinant is that it is the area produced by the transformation of the unit square):
Any vector will be transformed into a vector which will be aligned along the same line going through the origin, whatever the vector $\bf{x}$.
So, if you have $\bf{}Ax=b$ and $\text{det }\bf{A}=0$:

*

*If $\bf{b}$ is aligned along this particular line, you can indeed choose any $\bf{x}$. That is : infinite solution.

*But if  $\bf{b}$ is not aligned along this particular line, it's impossible to find a vector $\bf{x}$: The flattening by $\bf{}A$ always produces a vector aligned along this certain line, so $\bf{b}$ can never be a result of the transformation of $\bf{x}$ by $\bf{}A$.

You can reason the same with homogeneous equation: It's a transformation by $\bf{A}$ of a vector $\bf{x}$ into a single point: the origin $O\,(0,\quad 0)$:
Hence if $\text{det }\bf{A}=0$, that is always true, since the flattening produces vectors aligned along a particular line always going through the origin.
