If $BA = I_n$, then why can't we do $BAx = Bb$ to solve $Ax=b$?

Question

Let $A$ and $B$ be matrices such that $BA = I_n$. As an exercise I am to explain why the following argument is invalid:

To solve $A\vec x=\vec b$ with this initial condition, multiply both sides by $B$, and nothing that $BA = I_n$, we have $\vec x = B\vec b$

I don't see how there's anything wrong with this. $B$ clearly seems to be $A^{-1}$. I feel like the fault in my understanding why the above statement isn't necessarily true would be that $B$ isn't necessarily isn't $A^{-1}$, but I don't know how this is not the case given $BA = I_n$.

To those who don't see anything wrong with it, I'll take a picture of the problem for reference:

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Answer 1

The issue is that $Ax = b$ might not have a solution.

It is correct to say that if $Ax = b$ then $x = Bb$. However, if $b$ is not in the range of $A$, then $Bb$ is not a solution to $Ax = b$.

Answer 2

This is a correct procedure if $A$ and $B$ are both square matrices. However, if the matrices are not square, $B \vec b$ is not necessarily a solution.

Take for example $A =\begin{pmatrix} 1 \\0 \end{pmatrix}$ with $B = \begin{pmatrix} 1 & 0 \end{pmatrix}$ and $\vec b = \begin{pmatrix} 0 \\1 \end{pmatrix}$. Then $BA = I_1$, but $B \vec b = 0$ is not a solution of the equation.

Answer 3

This is a subtle but important point:

You're assuming that the vector $x$ exists when you do the above. If you set $x = Bb$ and then check $A(Bb)$, this is only equal to $b$ if $AB = I$. This doesn't need to be true! For instance, take $A = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$ and $B = \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \end{bmatrix}$, and try this with $b = \begin{bmatrix} 2 \\ 1 \end{bmatrix}$.