Given, AT=A,BT=−B,CT=−C Let M=A13B26−B26A13 Then, MT=(A13B26−B26A13)T =(A13B26)T−(B26A13)T =(BT)26(AT)13−(AT)13(BT)26 =B26A13−A13B26=−M Hence, M is skew symmetric Let, N=A26C13−C13A26 then, NT=(A26C13)T−(C13A26)T =−(C)13(A)26+A26C13=N Hence, N is symmetric. ∴ Only S2 is true.