Question 1109397
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Since your formulation is far from to be perfect, I will edit the formulation:



        Prove:

        If  A  is invertible matrix, then {{{(A^n)^(-1)}}} = {{{A^(-n)}}}.


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I will prove it for n = 2, and you will see from there how to do it in the general case.

We are given that  {{{A*A^(-1)}}} = I.


Then  {{{A^2*A^(-2)}}} = {{{A*A*(A^(-1))*(A^(-1))}}} = {{{A*(A*(A^(-1)))*(A^(-1))}}} = {{{A*I*(A^(-1))}}} = {{{A*(A^(-1))}}} = I.


QED.
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