Question about square matrices

Question

Given two invertible square matrices A, B is it always true that $$(ABA^{-1})^{-1}=AB^{-1}A^{-1}$$

I think it's false since in matrix multiplication order matters but i can't find a counterexample

Answer 1

If B is invertible then:

$(ABA^{-1})^{-1} = AB^{-1}A^{-1}$

To show this is true, show that $(ABA^{-1})(AB^{-1}A^{-1}) = (AB^{-1}A^{-1})(ABA^{-1}) = I$

Since matrix multiplication is associative we can drop the parentheses.

$ABA^{-1}AB^{-1}A^{-1}$

And we have an $A$ next to an $A^{-1}$ and they cancel each other out (or multiply to $I$)

$ABB^{-1}A^{-1}$

And we again have a pair of inverses together, and after we cancel those we will still have a pair of inverses together.

Answer 2

IF $A$ and $B$ are both invertible square matrices, then $$(ABA^{-1})^{-1} = ((AB)A^{-1})^{-1}=(A^{-1})^{-1}(AB)^{-1}=AB^{-1}A^{-1}$$ If fact, one can show that, given $A,B,C$ invertible matrices, then

• $(AB)^{-1}=B^{-1}A^{-1}$
• $(ABC)^{-1}=C^{-1}B^{-1}A^{-1}$