When you know one solution of the system $Ax=b$, call it $x_0$ and all solutions of the homogeneous system $Ax=0$, then you know all solutions of $Ax=b$.
Indeed, if $Ay=0$, then $A(x_0+y)=Ax_0+Ay=b+0=b$.
Conversely, if $x_1$ is a solution of $Ax=b$, then
$$
Ax_1=b=Ax_0
$$
What can you do with this? Observe that $b$ can be removed and you can write
$$
Ax_1=Ax_0
$$
Therefore $A(x_1-x_0)=0$, telling you that $y=x_1-x_0$ is a solution of $Ax=0$. But then $x_1=x_0+y$ has the desired form.
In particular, if $Ax=b$ has infinitely many solutions, also $Ax=0$ has.
Think to the equation of a line, $y=mx+q$. If $(x_0,y_0)$ and $(x_1,y_1)$ are points of the line, then $(x_1-x_0,y_1-y_0)$ is a point on the line $y=mx$. Indeed
$$
y_1-y_0=(mx_0+q)-(mx_1+q)=m(x_1-x_0)
$$
The points of the line $y=mx+q$ can be obtained by translating all the points of $y=mx$ by the same point.
Now let's see your particular problem. There is another fact to keep in mind:
if a linear system has two distinct solutions, then it has infinitely many solutions
This is because only the following cases can happen for a system: it has
- no solution, or
- exactly one solution, or
- infinitely many solutions
For, assume $Ax=b$ has two distinct solutions $x_1$ and $x_2$; then, by the same reasoning as above, $y=x_1-x_2$ is a solution of $Ax=0$. But then also $\alpha y$ is a solution of $Ax=0$ for all scalars $\alpha$, so the system $Ax=b$ has infinitely many solutions, because every vector of the form $x_1+\alpha y$ is a solution.
You now should see that showing that a system has infinitely many solutions is equivalent to showing it has two distinct solutions.
Now, solvability of the system $Ax=c$ is equivalent to $c\in\operatorname{colspace}(A)$; so, assume $Ax=c$ is solvable and $Ax=b$ has infinitely many solutions. Then
- $Ax=b$ has infinitely many solutions, so
- $Ax=0$ has infinitely many solutions, so
- $Ax=c$ has infinitely many solutions.
First, we know that $Ax_1-b$ and $Ax_2-b$ are in the left nullspace of $A$ (i.e. the orthogonal complement of the image of $A$). This is because, if $Ax_1-b$ was not in the left nullspace of $A$, then the residual $\|Ax_1-b\|$ would not be minimized and $x_1$ would not be a least-squares solution.
Now, notice that:
$$b=Ax_1-(Ax_1-b)=Ax_2-(Ax_2-b)\implies Ax_1-Ax_2=(Ax_1-b)-(Ax_2-b)$$
The left side of this equation is clearly is in the image of $A$, since $Ax_1$ and $Ax_2$ are both in the image of $A$. Meanwhile, the right side of this equation is in the left nullspace of $A$, since $Ax_1-b$ and $Ax_2-b$ are both in the left nullspace of $A$. Thus, this quantity is in both the image and the left nullspace of $A$, but since these two subspaces are orthogonal complements of each other, the only way a vector can be in both of them at the same time is if that vector is $\vec 0$.
Therefore:
$$Ax_1-Ax_2=\vec 0\implies Ax_1=Ax_2$$
Best Answer
$Ax = b$ has unique solution that means $A$ has full rank. Hence $Ax = c$ has unique solution. (Here $A$ is square also)