TITLE: Proving $S= S^3$
QUESTION [5 upvotes]: Let $\mathrm{A,B,C,D}$ be (not necessarily square) real matrices such that $\mathrm{A^T = BCD , B^T= CDA, C^T = DAB, D^T = ABC}$ for the matrix $\mathrm{S= ABCD}$ prove that $\mathrm{S= S^3}$

Attempt: 
$\mathrm{S= ABCD \implies S^T = D^TC^T B^T A^T = (ABC)(DAB)(CDA)(BCD)\\ (S^T)^T = (D^TC^TB^TA^T)^3 \implies S^3 = (S^T)}$ (using reversal law)
Where have I gone wrong? Or is it that $\mathrm{S= S^T}$, how can we show that?

REPLY [6 votes]: From $S = ABCD$ and $A^\top = BCD$, you get $S = AA^\top$.