Taking $b=4$ ensures that the matrix $A$ is singular, i.e $det(A)=0$.
Showing a system of linear equations is not solvable (has no solutions) is, by definition, the same thing as showing that the system of linear equations is "inconsistent".
Here, we need to show solvable (consistent solution) and by luck here we got infinite solutions .
So what I've understood is whether the system has any solution depends on the left hand side of the equations
Only, in case of $det(A) = 0$, we can say either no or infinite in case of non homogenous equations
and
only and only infinite in case of homogenous equations .
An ex :-
Take the system of linear equations (Non homogenous)
$$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=0,\end{cases}$$
Here, $det(A) =0$, and solving the complete system we have infinite solutions .
and now
Take the same system of linear equations
$$\begin{cases}x+3y-z=4\\4x-y+2z=8\\2x-7y+4z=-3,\end{cases}$$
Here, also $det(A) =0$, and solving the complete system we have no solutions .
Note :- I just changed $0$ to $-3$ in last equation RHS .