Equality of matrix equations

Question

Let A and B be square matrices of the same type. Does the below equality hold in general? $$(A \cdot (A^{-1}\cdot B)^{-1} \cdot A^{-1} \cdot B^2)^T=B^T \cdot A^T$$

How can this be simplified to figure this out?

Answer 1

Hint:

Begin with simplifying what's inside the parentheses, then transpose. Remember that $$(AB)^{-1}=B^{-1} A^{-1},\qquad(AB)^{\intercal}=B^{\intercal} A^{\intercal}.$$

Answer 2

$$(A \cdot (A^{-1}\cdot B)^{-1} \cdot A^{-1} \cdot B^2)^T=(A \cdot B^{-1}\cdot A \cdot A^{-1} \cdot B^2)^T=(A \cdot B^{-1}\cdot B^2)^T=(A \cdot B)^T=B^T \cdot A^T$$

Answer 3

Of course it doesn't hold in general. In general, matrices don't have inverses.

However, assuming that $A$ and $B$ are invertible, then it is true:\begin{align}(A.(A^{-1}.B)^{-1}.A^{-1}.B^2)^T&=(A.B^{-1}.A.A^{-1}.B^2)^T\\&=(A.B^{-1}.B^2)^T\\&=(A.B)^T\\&=B^T.A^T.\end{align}