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What are the restrictions on $A$ and/or $B$ for which the following holds? (in addition to the requirements that A and B should be invertible): $$ (A\cdot B)^{-1} = A^{-1}\cdot B^{-1} $$

Given that for all square invertible matrices: $$ (A\cdot B)^{-1} = B^{-1}\cdot A^{-1} $$ Thanks!

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  • $\begingroup$ $A^{-1} = B^{-1}$ $\endgroup$ – quester Apr 21 '20 at 12:04
  • $\begingroup$ @quester, that can be true for any two invertible matrices that commute. $\endgroup$ – Invisible Apr 21 '20 at 14:37
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Because of the two identities you wrote, what we want is essentially: $$ A^{-1} B^{-1} = B^{-1} A^{-1} $$

We already assumed that A and B are invertible, thus $ \det(A) \neq 0 $ and $ \det(B) \neq 0 $, so $A^{-1} B^{-1}$ should also be invertible, because $\det(A^{-1} B^{-1}) = \det(A^{-1}) \det(B^{-1}) = \det(A)^{-1} \det(B)^{-1} \neq 0 $

Then, if we can invert both the left and the right hand side of the identity above:

$$ (A^{-1} B^{-1})^{-1} = (B^{-1} A^{-1})^{-1} \iff $$ $$ (B^{-1})^{-1} (A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} \iff $$ $$ B A = A B $$

So if the two matrices commute when multiplied (which is generally not true for arbitrary matrices), then it also holds that $(AB)^{-1} = A^{-1} B^{-1}$

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