How to prove, that solution of system $Ax = b$ exists only if there is no solution of $A^T y = 0$ and $b^T y = 1$?

Question

I have a little linear algebra problem here:

How can I prove, that there is a solution of system $Ax = b$ only if there is no solution of $A^T y = 0$ and $b^T y = 1$?

Answer 1

Assume $A^Ty=0$ and $b^Ty=1$. Then $$1=b^Ty=(Ax)^Ty=x^TA^Ty=x^T0=0.$$

Answer 2

By the fundamental theorem of linear algebra, $\mathrm{Im}(A)\perp\mathrm{Ker}(A^T)$. If $Ax=b$ is solvable and hence $b\in\mathrm{Im}(A)$, we have $b^Ty=0$ for all $y\in\mathrm{Ker}(A^T)$ (solutions of $A^Ty=0$).