Finding a unique solution with a vector Suppose $A$ is a $3 \times 3$ matrix and $y$ is a vector in $\mathbb R^3$ such that the equation $Ax = y$ does not have a solution. Does there exist a vector $z$ in $\mathbb R^3$ such that the equation $Ax = z$ has a unique solution?
My thought is if it has no solution, $A$ must have a row with all $0$s equaling a non-zero in $y$. So my thought is, the answer is no, because even though I can have $z$ have a $0$ element in the all $0$ row in $A$, the fact that I have an all $0$ row means I have a free variable meaning even though it is now consistent, it has infinite solutions and not a unique one. Am I correct on this, or no? 
 A: Your reasoning is correct.  However the way you've written it is a little hard to understand.  So before turning it in, I'd suggest rephrasing it.  Here is a slightly more lucid way of expressing your argument:
If there exists a $y$ such that $Ax=y$ has no solution, then the RREF of the augmented matrix $\left[\,A\mid y\,\right]$ must contain a row of the form $\begin{bmatrix}0 & 0 & 0 & | & c\end{bmatrix}$ where $c\ne 0$.  Note that because the RREF of a matrix is unique and the steps to row reduce $A$ is exactly the same as those to row reduce $\left[\,A\mid y\,\right]$ (except possibly a few at the end which only affect the $y$ column), the RREF of $A$ alone must have a row of all $0$'s.
Now consider a $z$ such that $Ax=z$ has a solution.  Knowing that the RREF of $A$ has a row of all zeros we see that the RREF of $\left[\,A\mid z\,\right]$ must have a row of all zeros as well.  Because $A$ is a square matrix, that means that there is a column without a pivot and thus this system will have a free variable.  Therefore there are infinitely many solutions.$\ \ \ \square$
