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If $A$ and $B$ are invertible $n \times n$ matrices, then $(AB)^{-1}$ equals:
A$A^{-1} B^{-1}$
B$B^{-1} A^{-1}$
C$(A B)^{-1} = \dfrac{1}{|AB|}\operatorname{adj}(AB)$ only
DDoes not exist in general
Answer & Solution
Correct answer: B. $B^{-1} A^{-1}$
$(AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AIA^{-1} = AA^{-1} = I$, and similarly on the left. Therefore $(AB)^{-1} = B^{-1}A^{-1}$, with the order reversed (the 'shoes-and-socks' rule). Option C is just the formula for the inverse and does not contradict B.
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