> 
> If $\mathbf{Ax = b}$ has infinitely many solutions, why is it impossible for $\mathbf{Ax = c}$  
> 
>  (where $\mathbf{c}$ is a new right side) to have only one solution? 
> 
> 
> 

**Proof :** Take two solutions of $\mathbf{Ax = b} :$ $\mathbf{Ax\_1 = b}$ and $\mathbf{Ax\_2 = b} \implies \mathbf{\color{green}{A(x\_1 - x\_2) = 0}}$.  

Thus, if $\mathbf{Ax\_0 = c}$ then add the homogenous solution in green to $\mathbf{x\_0}$: $\mathbf{Ax\_0 + \color{green}{A(x\_1 - x\_2)} = c + \color{green}{0} }$ so $\mathbf{x\_0}$ is not unique.  

If $\mathbf{c}$ is not in $colspace(A)$, then no solution to $\mathbf{Ax = c}$.

$\Large{{1.}}$ I accept the proof but I don't perceive the intuition? Would someone please explain/uncloak it?  

I recollect: If $\mathbf{c}$ is not in $colspace(A)$, then no solution to $\mathbf{Ax = c}$. 

$\Large{{2.}}$ How would you divine/previse the strategy of subtracting two solutions to $\mathbf{Ax = b} $ and then adding this homogeneous solution to $\mathbf{Ax = c}$ ?

$\Large{{3.}}$ Does the above prove that $\mathbf{Ax = c}$ has infinitely many solutions also,  

beyond $\mathbf{x\_0}$ and $\mathbf{x\_0 + \color{green}{(x\_1 - x\_2)}}$?