Given two $n \times n$ invertible square matrices $A$ and $B$ such that $(A-B)(A+B) = A^2-B^2$. Which statement is correct? Given two $n \times n$ invertbible matrices $A$ and $B$ such that $(A − B)(A + B) = A^2 − B^2,$ which of the following statements is correct?

*

*$B = ABA^{-1}B$

*None of the given options

*$B = A^{-1}BA$

*$B = BABA^{-1}$

*$A$ is the inverse of matrix $B$
This is a multiple choice questions, which I cant seem to understand. From what I do understand:

*

*Given two matrices $A$ and $B$, the product $AB$ is not equal to $BA$.

*Since they said $(A − B)(A + B) = A^2 − B^2$. It means that $AB$ is equal to $BA$ ie: they commute

Option 1 and Option 4 seems like the same term, but the left hand side is the terms are rearranged im not sure about them. I would choose option 3, since $A^{-1}A = I$ leaving $B=B$ is this correct?
 A: $1.\ \  B = ABA^{-1}B \Leftrightarrow I = ABA^{-1} \Leftrightarrow A = AB \Leftrightarrow I = B$, so this is only the case when $B$ is the identity matrix.
$2.$ False, as we can only find out after doing the rest of the parts.
$3.\ \ B = A^{-1}BA \Leftrightarrow AB = BA$, which is something you have already shown is true.
$4.\ \ B = BABA^{-1} \Leftrightarrow BA = BAB \Leftrightarrow A = AB \Leftrightarrow I=B$, so again, this only holds when $B$ is the identity matrix.
$5.$ This is a solution, but not the only solution, since non-inverse matrices can commute. For example,
$$\begin{bmatrix}2&0\\1&-1\end{bmatrix}\begin{bmatrix}3&0\\1&0\end{bmatrix} = \begin{bmatrix}6&0\\2&0\end{bmatrix} = \begin{bmatrix}3&0\\1&0\end{bmatrix}\begin{bmatrix}2&0\\1&-1\end{bmatrix},  $$ but clearly the matrices aren't inverses.
A: Following the merit of the question, assume that none of $A$ and $B$ is $I_n$.
As deducted by you, $AB=BA$,
$(1)\not\implies AB=BA$, first post-multiply the equation by $B^{-1}$ and then again post-multiply by $A$.
Similarly $(4)\not\implies AB=BA$. First pre-multiply by $B^{-1}$ and then post-multiply by $A$.
$(3)$ is obviously same as you have deducted from the given relation.
$(5)$ is not always true, since if $A=B^{-1}$, then you need $AB=BA=I_n$.
