Consider the following task:
If $A$ is a $5 \times 5$ matrix and the equation $Ax = b$ is consistent for every $b$ in $\mathbb{R}^5$, is it possible that the equation has more than one solution? Why or why not?
I answered the following, but I am not sure if my statements are true. If they are false, I ask you to point it out and elaborate. If they are true, I ask you to elaborate as to why they are true. I will numerate each statement I am insecure about.
Seeing as $A$ is quadratic, it may or may not be invertible. If it is invertible, we have $A^{-1}(Ax) = A^{-1}b = x$ as a unique solution. (1) (This follows directly from an invertible matrix being row equivalent to the identity matrix, correct?) If $A$ is not invertible, we know that the equation has several solutions, seeing as $A$ is not row equivalent to the identity matrix, and thus may or may not have free variables. (2)
I am really insecure about (2). We are told that the equation $Ax = b$ is consistent for every $b \in \mathbb{R}^5$. Does this imply that we have only one of the following two alternatives:
- $A$ is invertible and we have the unique solution $A^{-1}b$
- $A$ is not invertible and we have infinitely many solutions.
If so, why?
Best Answer
In fact, case 2 can never happen. If you have at least one solution for all $b$, then $A$ must be invertible. You can solve $Ax=b$ for a basis and use linearity to prove this. Conversely, if $A$ is not invertible, its image is less than the full space, its kernel is more than the origin, and some $b$ will not permit a solution of $Ax=b$