First slide
Transpose of matrix
Question

Let A, B.C, D be (not necessarily) real matrices such that AT=BCD;BT=CDA;CT=DAB and DT=ABC for the matrix S = ABCD, the least value of k such that Sk = S is 

Difficult
Solution

S=ABCD=A(BCD)=AAT S3=(ABCD)(ABCD)(ABCD)=(ABC)(DAB)(CDA)(BCD)=DTCTBTAT=(BCD)TAT=ATTAT=AAT=S S3=S

Hence, least value of k is 3. 

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