Show by direct computation that
(ABC)^-1 = c^-1b^-1a^-1

Question

Show by direct computation that 
 (ABC)^-1 = c^-1b^-1a^-1

anonymous · Answer

Left hand side = (2*4*6)^-1 = (48)^-1 = 1/48

right hand side = (6)^-1 * (4)^-1 * (2)^-1 = 1/6 * 1/4 * 1/2 = 1/48

anonymous · Answer

you do this by showing that 
$$(abc)(c^{-1}b^{-1}a^{-1})=e$$ if you are talking about groups or 1 if your identity is 1. that makes 
$$(c^{-1}b^{-1}a^{-1})=(abc)^{-1}$$ by definition of inverse

anonymous · Answer

the computation is rather straightforwards since each element will pair with its inverse, giving the identity one by one

anonymous · Answer

and i am assuming you are working in the group setting where things do not necessarily commute, which is why the inverse shows up "backwards"
that is 
$$(ab)^{-1}=b^{-1}a^{-1}$$

anonymous · Answer

matrices for example, or functions work the same way.