If $Ax=b$, for $b\ne 0$, has more than one solution, then $Ax=0$ does as well. T or F. Prove this.

Question

I get that this is true, because there's one free variable, so no matter what the augmented matrix is, there always will be an infinite amount of solutions. Right? But how to I explain this as a proof?

Answer 1

If $x_1$ and $x_2$ are both solutions of $Ax=b$ then what can you say about $A(x_2-x_1)$?

Answer 2

$$\rm\begin{eqnarray} ax = &\rm b&\rm = ax',\ \ x\ne x'\\ \rm \Rightarrow\ a(x\!-\!x') = &0&\rm = a\,0,\ \ \,x -x'\ne 0\end{eqnarray}$$