<s>[INST] How to show that $B^{-1}\cdot A^{-1}=B\cdot A$? 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.
 [/INST] $A^2 \cdot B^2 = I \Rightarrow A, B$ 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 $B.$
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