How to show that $B^{-1}\cdot A^{-1}=B\cdot A$?

Question

I can't find the solution of this problem:

Given two $n\times n$ square matrices $A,B$ such that $A^2\cdot B^2=I_n$, show that $B^{-1}\cdot A^{-1}=B\cdot A$.

Thanks in advance.

Answer 1

Since $A^{2}B^{2}=I_{n}$ note that taking $A^{-1}=ABB$ we have that $A\cdot A^{-1}=I_n$. For square matrices this is enough to show $A$ is invertible. The same follows for $B$ by taking $B^{-1}=AAB$. Using the facts that $A^{-1}\cdot A=I_{n}$ and $B\cdot B^{-1}=I_{n}$ we get that \begin{eqnarray*} A^{2}B^{2}=I_{n} & \iff & AB^{2}=A^{-1}I_{n},\mbox{ by left multiplication}\\ & \iff & AB=A^{-1}I_{n}B^{-1},\,\mbox{ by right multiplication}\\ & \iff & AB=A^{-1}B^{-1}. \end{eqnarray*} Taking inverses of both sides and using the property $(AB)^{-1}=B^{-1}A^{-1}$ gives the desired result.

Answer 2

$A^2 \cdot B^2 = I \Rightarrow A, B$ are invertible. The set of all invertible matrices forms a group. So we also have $B^2 \cdot A^2 = I.$ Hence $B \cdot A = B^{-1}\cdot A^{-1}.$

I'm assuming that the entries of the matrices are from a field.

EDIT: $A^2 \cdot B^2 = I \Rightarrow \text{det}(A^2) \cdot \text{det}(B^2) = \text{det}(A^2 \cdot B^2) = 1 \Rightarrow \text{det}(A^2) \neq 0.$ Now $(\text{det}(A))^2 = \text{det}(A^2) \Rightarrow \text{det}(A) \neq 0.$ Similarly for $B.$