Invertible $ABA=B^2$

Question

I cannot figure out why the latter answer is the correct answer

Let $A$ and $B$ be square matrices: apply $ABA$=$B^2$ so the following is certain: (pick one)

1. If $A$ is invertible then so is $B$

2. If $B$ is invertible then so is $A$

Answer 1

Assume $B$ is invertible, than $B^2$ is invertible and from the equation so is $ABA$.

The rest follows as a consequence of multiplication of invertible matrices

Answer 2

Hint

$$A\text{ is invertible}\iff |A|\ne 0$$ $$|A|^2\cdot |B|=|B|^2$$

Answer 3

More fundamentally, if $X$ is the inverse of $B$, then $(X^2AB)A=X^2B^2=I$. Therefore $A^{-1}=X^2AB$. This proof can be modified to work on a monoid, so that if $aba=b^2$ and $bx=xb=1$, we can prove that $a$ has both a left inverse $y$ and a right inverse $z$ and in turn, $y=z$ is the inverse of $a$.