Let A,B,C,D & E be five real square matrices of the same order such that ABCDE=I where I is the unit matrix . Then,
(a)$B^{-1}A^{-1}=EDC$
(b)$BA$ is a nonsingular matrix
(c)$ABC$ commutes with $DE $
(d)$ABCD=\frac{1}{det(E)}AdjE$
More than one option may be correct .
Also , taking the special case A=B=C=D=E=I states all these options be true , but answer key states (a) is incorrect, how ?
First, a special case can only show you, that an answer is wrong, but not that is true in general. Regarding the options:
(a) We have $$ 1 = ABCDE \iff A^{-1} = BCDE \iff B^{-1}A^{-1} = CDE $$ so choosing $C$, $D$, $E$ such that $CDE \ne EDC$ will give you an example for (a) being wrong.
(b) If $BA$ were singular, then $$ 1 = \det(ABCDE) = \det(AB)\det(CDE) = \det(BA)\det(CDE) = 0. $$
(c) We have $$ ABCDE = 1 \iff (ABC)^{-1} = DE $$ and every matrix commutes with its inverse.
(d) We have $E^{-1} = ABCD$ and $E^{-1} = \frac 1{\det E} \mathrm{adj}\, E$.
