Let A,B,C be 3× 3 matrices such that A is symmetric and B and C are skew-symmetric. Consider the statements (S1)A^13 B^26-B^26 A^13 is symmetric…

Given,

Matrix $A$A is symmetric, $B$B is skew symmetric and $C$C is skew symmetric,

So, ${A}^{T}=A,{B}^{T}=-B,{C}^{T}=-C$AT=A,BT=−B,CT=−C

Now let $M={A}^{13}{B}^{26}-{B}^{26}{A}^{13}$M=A13B26−B26A13

Then, ${M}^{T}={({A}^{13}{B}^{26}-{B}^{26}{A}^{13})}^{T}$MT=(A13B26−B26A13)T

$$={({A}^{13}{B}^{26})}^{T}-{({B}^{26}{A}^{13})}^{T}$$=(A13B26)T−(B26A13)T

$$={({B}^{T})}^{26}{({A}^{T})}^{13}-{({A}^{T})}^{13}{({B}^{T})}^{26}$$=(BT)26(AT)13−(AT)13(BT)26

$$={B}^{26}{A}^{13}-{A}^{13}{B}^{26}=-M$$=B26A13−A13B26=−M

Hence, $M$M is skew symmetric

Now let, $N={A}^{26}{C}^{13}-{C}^{13}{A}^{26}$N=A26C13−C13A26

Then, ${N}^{T}={({A}^{26}{C}^{13})}^{T}-{({C}^{13}{A}^{26})}^{T}$NT=(A26C13)T−(C13A26)T

$$=-{(C)}^{13}{(A)}^{26}+{A}^{26}{C}^{13}=N$$=−(C)13(A)26+A26C13=N

Hence, $N$N is symmetric.

So, only $S2$S2 is true.