Understanding linear systems with infinite solutions I came across following excerpt:

Consider linear system $Ax=v$. $A$ is a transformation matrix.
(a) If $|A|=0$, then there will be infinite vectors $x$ which transform to v.
(b) If $|A|\neq 0$, then there will be single vector $x$ which transform to v.

I am not able to visualize (a).
I know following

*

*$|A|=0$ if rank of $A\lt n$, where $n$ is dimension of system

*If rank of $A$ is $R$, then we have $R$ number of $(n-1)$ dimensional planes. That is in 3D, if rank of $A$ is 2, then we will have $R=$ 2 planes each $3-1=2$ dimensional:
 
Fig.1 ($u + v + w = 2, 2u + 2v +2w = 4, 3u + v + 4w = 6$)
If rank of $A$ is 1 then we will have single 2 dimensional plane:
 
Fig.2 ($u + v + w = 2; 2u + 2v +2w = 4; 3u + 3v + 3w = 6$)
In column pictures, we will have vectors beloging to the same plane as can be seen in above blue vector images.

Q1. In above two examples, I have given $A$ and $v$. Can someone please explain with examplpe, in each of these two cases, how we can have infinite number of vectors $x$, such that $Ax=v$?
Update:
I want to add two more questions:
I also read:

$|A|=0$ if $A$ squishes the whole space into lower dimension.

Q2. What does that mean in above two examples? Does it mean that if we multiply $A$ of fig 1 by some vector $x$, we will get $v$ belonging to same plane and in case of fig 2, to the same line?
(Also I guess, above we have to make assumption that $v$ belong to same plane as vectors of $A$ for infinite solution, as otherwise there will be no solution. I wanted to discuss infinite solution scenarios.)
 A: Let $A$ be a square matrix. The following are equivalent:

*

*$\det A = 0$

*$A$ has full rank

*$A$ is invertible

So if $\det A = 0$, $A$ doesn't have full rank. By the rank-nullity theorem, the null space has nonzero dimension, so it has infinitely many vectors. So if the system $Ax = v$ has a solution $x'$, it has infinitely many solutions, because we can always add on another vector $u$ in the null space to get another solution $x' + u$: $A(x' + u) = Ax' + Au = Ax' + 0 = v$.
Also, if $\det A = 0$, then "$A$ squishes the whole space into lower dimension" because $A$ not having full rank means that the dimension of the column space is less than the dimension of the domain (again, this follows from the rank-nullity theorem). If $A$ is an $n \times n$ matrix, we can think of it as a mapping from $n$-dimensional space to $n$-dimensional space. The column space of $A$, i.e. the subspace of all vectors that can be reached by $A$, has dimension less than $n$, so $A$ "squishes" $n$-dimensional space into something of a lower dimension.
Conversely, if "$A$ squishes the whole space into lower dimension", this means some vectors aren't hit by $A$, so $A$ is not invertible, so $\det A = 0$.
