If $Ax=0$ has infinite solutions, can $Ax=b\not=0$ have exactly one solution > Let $A\in\mathbb{R}^{4\times 2}$ and $b\in\mathbb{R}^4$ with $b\not=0$. > > Is it possible that: > > $Ax=b$ has exactly one solution AND $Ax=0$ has infinite solutions. I think not, but I am bogged down on this. Infinite solutions for $Ax=0$ means that every single line of $A$ is potentially "canceling"or (0 0) to begin with. But what means $Ax=b$ in all its consequences? Is there a lemma I overlook here?

No. Take $x_1$ to be a solution of $Ax = b$ and $x_2 \
eq 0$ to be a solution of $Ax = 0$. Then, $x_3 = x_1 + x_2$ is also a solution of $Ax = b$, since $$Ax_3 = A(x_1+x_2) = Ax_1 + Ax_2 = b + 0 = b.$$