linear algebra, basic questions $ b= (1,1,1),c= (1,2,1) , 0=(0,0,0) , A-matrix$
I need help to prove or disprove the following:


*

*If $Ax=b$ has infinite solutions then $Ax=c$ has infinite solutions.

*if $Ax=b$ has single solution then $Ax=0$ has single solution.


I have been looking at this for an houre and I got no clue what information $Ax=b$ gives me.
 A: 
  
*
  
*If $Ax=b$ has infinite solutions then $Ax=c$ has infinite solutions.
  

If $Ax=b$ has infinite solutions, then the nullspace of $A$ is nontrivial.  Easy proof: let $x_1$, $x_2$ be two solutions where $x_1-x_2\ne 0$ then $A(x_1-x_2) = b-b=0$. $\ \ \square$
If the nullspace of $A$ is nontrivial then $Ax=c$ will either have infinitely many solutions (if $c$ is in the column space of $A$) or none (otherwise).  JMoravitz provides an example of the latter in the comments.


  
*if $Ax=b$ has single solution then $Ax=0$ has single solution.
  

If $Ax=b$ has a single solution, then the nullspace of $A$ is trivial. Easy proof: Assume $y\ne 0$ is in the nullspace of $A$ and $x_1$ is a solution to $Ax=b$. Then $A(x_1+y) = b+0 = b$ so $x_1+y\ne x_1$ is another solution to $Ax=b$.  Contradiction.$\ \ \square$
Then $Ax=$ anything will have a single solution.  Proof: Say $x_1$, $x_2$ are two solutions to $Ax=d$ for some $3\times 1$ matrix $d$ where $x_1\ne x_2$.  Then $A(x_1-x_2) = d-d = 0$.  But then $x_1-x_2\ne 0$ is in the nullspace.  Contradiction.$\ \ \ \square$
