TITLE: Inverse of a Product of Matrices: Why is $(AB)^{-1} = B^{-1} \cdot A^{-1}$?
QUESTION [6 upvotes]: I'm having a bit of difficulty conceptualizing a rule for the inverse of a product of matrices and I'd appreciate any input on it.
Suppose I let:
$A^{-1} = \begin{bmatrix}1 & 2 \\ 3 & 4 \end{bmatrix}$ and $B^{-1} = \begin{bmatrix}5 & 6 \\ 7 & 8 \end{bmatrix}$
From what I understand of the rule:
$(AB)^{-1} = B^{-1} \cdot A^{-1}$
However while I understand matrix multiplication isn't necessarily commutative, I'm unclear on why this must be the case.
It intuitively seems that simplifying the exponent would give me:
$(AB)^{-1} = A^{-1} \cdot B^{-1}$
So why isn't this the case?

REPLY [4 votes]: Because if you put on your socks and then your shoes, in that order, and now you want to take them off again, you have to take off your shoes before you take off your socks.
(In any group, $(AB)^{-1} = B^{-1}A^{-1}$.)