Despite the funny name this is a relatively important finding from linear algebra*, and it is exceedingly simple to use and prove. It takes two square matrices A and B, which have *n* rows and *n* columns. If these two matrices are invertible, i.e. their rank = *n*, then we know this about the inverse of their product: (AB)^{-1} = B^{-1}A^{-1}.

This finding is referred to as the shoe-sock theorem because it resembles the steps needed to invert (that is, undo) the action of putting on socks and shoes. If A is for socks, B is for shoes, and ^{-1} is the process of taking them off, to take them both off (AB)^{-1} one must first take off the shoes B^{-1}, then the socks A^{-1}. Hey, I said it was a funny name, not a particularly *insightful* one...

One proof is nice and easy, though I'm sure harder ones can be devised without much difficulty. This proof relies on the fact that we need to prove B^{-1}A^{-1} to be the inverse of AB. We'll do this by showing

(B^{-1}A^{-1})(AB) = I_{n}

and

(AB)(B^{-1}A^{-1}) = I_{n}.

In other words, we'll show that multiplying AB on either side will produce the identity matrix of size *n* (that is, I_{n}), which is necessary and sufficient proof that the inverse of AB is B^{-1}A^{-1}.

**First:** (B^{-1}A^{-1})(AB) = B^{-1}(A^{-1}A)B = B^{-1}I_{n}B = B^{-1}B **= I**_{n} -- cool.

**And also:** (AB)(B^{-1}A^{-1}) = A(BB^{-1})A^{-1} = AI_{n}A^{-1} = AA^{-1} **= I**_{n} -- very cool, statement proven.

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User zfcorey reminded me that this rule is true of *any* abstract algebra with an identity.