If $A$ is a square matrix and $Ax = b$ has a unique solution for some $b$, is $A$ necessarily invertible? 
Let $A$ be a square matrix. Suppose that $A x = b$ has a unique solution for some $b$. Is $A$ necessarily invertible?

I said no because the invertible matrix theorem states that $A x = b$ has a unique solution for each $b$. Is this correct or does the wording not make a difference?
 A: Yes. Let $x$ be the unique solution of $Ax=b$. If $y$ is a nonzero solution of $Ay=0$, then $A(y+x)=b$ and $y+x\neq x$, a contradiction.
A: One has "$Ax=b$ has a unique solution" $\implies$ "$Ax=0$ has $x=0$ as unique solution" $\equiv$ "$\ker(A)=\{0\}$" $\implies$ "the linear map defined by $A$ is injective" $\implies$ "the linear map defined by $A$ is bijective" $\implies$ "$A$ is invertible" $\implies$ "for all vectors $v$, $Ax=v$ has a unique solution" $\implies$" $Ax=b$ has a unique solution". Since this chain ends where it started, all statements it contains are equivalent.
Note that the only hard implication in the chain is "the linear map defined by $A$ is injective" $\implies$ "the linear map defined by $A$ is bijective", which uses the fact that the mentioned linear map has spaces of equal finite dimension at departure and at arrival. This suggest that the equivalence established might fail in infinite dimension, which indeed it does. In infinite dimension, a linear equation whose left hand side is given by a linear operator on the space (which corresponds to "square" for a finite matrix equation) might have a unique solution for some value of its right hand side, but no solution for other values. For instance in the vector space of polynomials, the equation $(X-1)P=X^4-X$ is linear in the unknown polynomial $P$ (the powers if $X$ are just fixed vectors here), and has a unique solution, namely $P=X^3+X^2+X$, but the equation $(X-1)P=X^3-2$ with the same left hand side has no solutions (and the linear map is therefore not invertible). So you were right to question whether this equivalence really holds, but in the (finite dimensional) matrix case it does.
A: Hint It's perhaps easier to see the contrapositive here: Suppose $A$ is not invertible, so that there is some nonzero element $x_0 \in \ker A$. Given one solution to $A x = b$, can we use $x_0$ to construct another?
