Question

Let $A,B,C,D$ be (not necessarily square) four matrices such that $A^T=BCD;$ $B^T=CDA;$ $C^T=DAB$ and $D^T=ABC.$ If $S=ABCD,$ then

• A. $S=S^3$
• B. $S=S^4$
• C. $S=S^2$
• D. $S=S^6$

Answer 1

The correct option is A $S=S^3$

$S=ABCD=A\left(BCD\right)=A{A}^T\Rightarrow S=S^T$

Now
$S^T=(ABCD{)}^T$
$=D^T{C}^T{B}^T{A}^T$
$=\left(ABC\right)\left(DAB\right)\left(CDA\right)\left(BCD\right)$
$=SSS$
$=S^3$
$\therefore S^T=S^3=S$