Which of the statements are true if $ABCDE = I$? 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 which of the following are true?

(A) $B^{−1}A^{−1}= EDC$ 

(B) $BA$ is a nonsingular matrix

(C) $ABC$ commutes with $DE$ 

(D) $ABCD = {1\over detE}$ Adj $E$.
Hint: the multiplicativity of the determinant implies that all $5 $ mattices are invertible,  now it easily follows that (B), (C) and (D) are true. About (A), you know that $ B^{-1} A^{-1}=(AB)^{-1}=CDE $, so just find an example where $ CDE\neq EDC $ and you'll have a counterexample to (A).
Option B and D is correct. Note that $det(AB)=det(A)det(B)$. Take determinant of both side of $ABCDE=I$. Since $det(I)=1\neq0$, none of the matrix in left hand side is singular. So does $BA$

For Option D, Since $ABCDE=I$, $ABCDEE^{-1}=IE^{-1}=E^{-1}=Adj(E)/det(E)$. It is correct also (since E is non singular)

For Option A, $ABCDE=I$, $$A^{-1}ABCDE=A^{-1}I=A^{-1}$$$$BCDE=A^{-1}$$ $$B^{-1}BCDE=B^{-1}A^{-1}$$$$B^{-1}A^{-1}=CDE\neq EDC$$

EDIT: Option C is is also correct. Note that $(AB)^{-1}=B^{-1}A^{-1}$ and $(AB)C=A(BC)$. From DKal's answer, one knows that $DE=(ABC)^{-1}$. So $ABC(ABC)^{-1}=(ABC)^{-1}ABC$