# In which cases does $(A \cdot B)^{-1} = A^{-1} \cdot B^{-1}$?

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!

• $A^{-1} = B^{-1}$ – quester Apr 21 '20 at 12:04
• @quester, that can be true for any two invertible matrices that commute. – Invisible Apr 21 '20 at 14:37

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 $$\det(A^{-1} B^{-1}) = \det(A^{-1}) \det(B^{-1}) = \det(A)^{-1} \det(B)^{-1} \neq 0$$
$$(A^{-1} B^{-1})^{-1} = (B^{-1} A^{-1})^{-1} \iff$$ $$(B^{-1})^{-1} (A^{-1})^{-1} = (A^{-1})^{-1} (B^{-1})^{-1} \iff$$ $$B A = A B$$
So if the two matrices commute when multiplied (which is generally not true for arbitrary matrices), then it also holds that $$(AB)^{-1} = A^{-1} B^{-1}$$