On pg. 95 of Howard Anton's Elementary Linear Algebra, it asks to simplify
$$(AB)^{-1}(AC)^{-1}(D^{-1}C^{-1})^{-1}D^{-1}$$
The answer is supposed to be $B^{-1}$ but I end up with $B^{-1}A^{-1}A^{-1}$
First, I undid all the brackets which led to $B^{-1}A^{-1}C^{-1}A^{-1}CDD^{-1}$
I "cancelled out" $CC^{-1}$ and $DD^{-1}$ using the associative property of matrix multiplication but not sure how to simplify it any further.
I guess $A^{-1}A^{-1} =(AA)^{-1} = (A^2)^{-1}$ (not 100% here).
However, I still have $B^{-1}(A^2)^{-1}.$
The answer isn't $B^{-1}$. It isn't $B^{-1}A^{-2}$ either.
You can't cancel out $CC^{-1}$ since there isn't a $CC^{-1}$ there ($CA^{-1}C^{-1}$ may not equal $CC^{-1}A^{-1}$).