Question

Let A, B, C, D be (not necessarily square) real matrices such that, ${{A}^{T}}=BCD;{{B}^{T}}=CDA$; ${{C}^{T}}=DAB$ and ${{D}^{T}}=ABC$. For the matrix S = ABCD, consider the two statements,
(i) ${{S}^{3}}=S$
(ii) ${{S}^{2}}={{S}^{4}}$
(A) (ii) is true but not (i)
(B) (i) is true but not (ii)
(C) both (i) and (ii) are true
(D) both (i) and (ii) are false

Answer 1

Apply the formula: - ${{\left( MN \right)}^{T}}={{N}^{T}}.{{M}^{T}}$ and multiply ${{D}^{T}},{{C}^{T}},{{B}^{T}},{{A}^{T}}$ to get ${{\left( ABCD \right)}^{T}}$ in the L.H.S. Now, substitute ABCD = S as given in the question to check whether the conditions given in statement (i) and (ii) are correct or not.

Complete step by step answer:
We have been provided with the relations: -

& {{A}^{T}}=BCD \\\ & {{B}^{T}}=CDA \\\ \end{aligned}$$ $${{C}^{T}}=DAB$$ $${{D}^{T}}=ABC$$ Here, $${{A}^{T}},{{B}^{T}},{{C}^{T}},{{D}^{T}}$$ represents the transpose of A, B, C, D respectively. Now, multiplying $${{D}^{T}},{{C}^{T}},{{B}^{T}},{{A}^{T}}$$ in this order, we get, $${{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}=ABC.DAB.CDA.BCD$$ This can be simplified as: - $$\Rightarrow {{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}=\left( ABCD \right).\left( ABCD \right).\left( ABCD \right)$$ $$\Rightarrow {{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}=S.S.S$$, since ABCD = S $$\Rightarrow {{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}={{S}^{3}}$$ - (1) Now, we know that, $${{\left( MN \right)}^{T}}={{N}^{T}}{{M}^{T}}$$, $$\begin{aligned} & \Rightarrow {{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}={{\left( ABCD \right)}^{T}} \\\ & \Rightarrow {{D}^{T}}{{C}^{T}}{{B}^{T}}{{A}^{T}}={{S}^{T}} \\\ \end{aligned}$$ Substituting this value in equation (1), we have, $$\Rightarrow {{S}^{T}}={{S}^{3}}$$ We can see that in the question we haven’t been informed about $${{S}^{T}}$$, whether it is equal to S or not. Therefore, we can conclude that statement (i), $${{S}^{3}}=S$$ is false. Now, considering statement (ii), we have, $$\Rightarrow {{S}^{4}}={{S}^{3}}.S$$ $$\Rightarrow {{S}^{4}}={{S}^{T}}.S$$, since $${{S}^{T}}={{S}^{3}}$$ is shown above. Now, $$\Rightarrow {{S}^{T}}\ne S\Rightarrow {{S}^{T}}.S\ne {{S}^{2}}$$. So, $${{S}^{4}}\ne {{S}^{2}}$$, therefore statement (ii) is also false. **So, the correct answer is “Option D”.** **Note:** One may note that we have found $${{S}^{T}}={{S}^{3}}$$ and not $$S={{S}^{3}}$$, so statement (i) was considered false. If we were provided with the information that S is a symmetric matrix then we would have used the relation $${{S}^{T}}=S$$ and statement (i) would have turned out to be true. Once statement (i) gets true statement (ii) would also get true. So, remember that we need information about S whether it is symmetric or not.