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 $E$.

  • $\begingroup$ What thoughts do you have on each? Do you have qualms about any of them? Do any immediately jump out at you as being wrong or right? $\endgroup$ – Cameron Williams May 14 '14 at 2:57
  • $\begingroup$ (D) seems right. I am doubtful about (A). $\endgroup$ – Rudstar May 14 '14 at 3:00
  • $\begingroup$ Why does (D) jump out to you as being right? (A) is not true. Do you know why? $\endgroup$ – Cameron Williams May 14 '14 at 3:01
  • $\begingroup$ Can anybody explain what ABC commutes with DE means? $\endgroup$ – Prem kumar May 28 '15 at 11:59

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 $ABC(ABC)^{-1}=(ABC)^{-1}ABC$

  • $\begingroup$ I think this is a homework problem based on the formatting of the question. I feel like you should remove some of the details since you basically give away all of the answers. $\endgroup$ – Cameron Williams May 14 '14 at 3:11
  • $\begingroup$ Oh sorry... The examination point of this question is that whether one note that all of the matrixes are non singular. I shouldn't put too much details... $\endgroup$ – Y.H. Chan May 14 '14 at 3:15
  • $\begingroup$ As I wrote in my answer, I think that (C) is correct because $ DE $ is the inverse of $ ABC $. $\endgroup$ – DKal May 14 '14 at 3:16
  • $\begingroup$ Seems that you are right. $\endgroup$ – Y.H. Chan May 14 '14 at 3:18

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