What must be true for $Ax = b$ to imply that $x = bA^{-1}$? Assume that $A$ is a matrix, and $x$ and $b$ are column vectors.
1) Is A singular or nonsingular?
From the given (implied) equation, it seems safe to assume that A has an inverse, in which case A must be nonsingular.
2) Is A a square matrix?
To have an inverse, a matrix needs to be a square matrix, so yes.
3) Is the nullspace of A zero or 1 dimensional?
Following from the reasonings above, A is 0 dimensional.
4) Is the matrix A 1x1?
It works, so I think I can say yes.
5) Or "none of the above".
This choice then can be ruled out.
Am I right in thinking this way?
The problem I have with thinking that A may indeed be invertible, is that b is a single column vector. Hence $A^{-1}$ will probably have to be a 1xn matrix for $bA^{-1}$ to work. But then if so, since it's not a square matrix, it actually won't have an inverse. Does that mean "none of the above" (which apprently isn't the correct answer,at least by itself)?
I'm not too confident about my answers because usually, $x=A^{-1}b$, not $x=bA^{-1}$, and it's confusing me a lot. Any help would really be appreciated!
Update: just to be clear, there are no typos!
If we are meant to take the question literally, then the answer is as follows.
In the case that $A$ is a scalar (i.e. a matrix with one entry), then clearly the statement holds when $A \neq 0$.
If $x,b$ are column-vectors of size greater than $1$ and $A$ is a square matrix for which the product $Ax$ is conformable, then the expression $bA^{-1}$ is undefined, which means that the statement $x = bA^{-1}$ is always false. In order for the implication to be true, $Ax = b$ must always be false.
It is not clear whether the statement refers to a fixed vector $b$ or arbitrary vectors $b$ (i.e. the statement should hold for all choices of $x$ and all choices of $b$). If $x,b$ are both arbitrary vectors, then $Ax = b$ is always sometimes true since we can take $x = b = 0$.
If $b$ is meant to be a fixed vector, then the statement $Ax = b$ will always be false (for all $x$) iff $Ax = b$ has no solution, which holds iff $A$ fails to be invertible and $b$ lies outside the column space of $A$.