Question

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

Answer 1

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

Answer 2

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

Answer 3

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

Answer 4

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
\[(ab)^{-1}=b^{-1}a^{-1}\]

Answer 5

matrices for example, or functions work the same way.