I am struggling on the following problem statement:
Given a real matrix $A$, the equation $Ax = \begin{bmatrix}0 & 1 & 0\end{bmatrix}^\top$ has multiple solutions, while the equations $$Ax = \begin{bmatrix}1 \\ 0 \\ 0\end{bmatrix} \quad \mathrm{and} \quad Ax = \begin{bmatrix}0\\ 0 \\ 1\end{bmatrix}$$ both have no solutions. What is the rank of $A$?
From the first equation, it is clear that $r < n$ therefore the rank has to be less than 3. However I don't understand how we can precisely determine the rank of $A$ given the second and third equation.
Any help on how to start this problem would be greatly appreciated.
I will try to convey some intuition on that problem. If you have multiple solutions with $\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}$ you have infinitely many solutions. With that in mind, consider what that means. It means that your matrix $A$ or your system of equations at least has the dimension $1$ because you have every possible solution concerning your second component like a line in the $\mathbb{R}^3$ (or every linear combination of the second component). However, your other components give no solutions. That means, your solutions do not lie in the other two dimensions. All in all, the solution has dimension $1$ which implies that the rank of your matrix is also $1$.
EDIT: Actually that answer is not true. Look down in the comment. There you find an counterexample by sinray. The only thing one can say is that the rank has to be at least $1$ and less than $3$.