Let A, B, C be 3 times 3 matrices such that A | vedclass

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(A) Given,$A^T = A$,$B^T = -B$,$C^T = -C$.For $(S1)$,let $M = A^{13} B^{26} - B^{26} A^{13}$.Then,$M^T = (A^{13} B^{26} - B^{26} A^{13})^T = (B^{26})^

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Only $S2$ is true

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(A) Given,$A^T = A$,$B^T = -B$,$C^T = -C$.For $(S1)$,let $M = A^{13} B^{26} - B^{26} A^{13}$.Then,$M^T = (A^{13} B^{26} - B^{26} A^{13})^T = (B^{26})^T (A^{13})^T - (A^{13})^T (B^{26})^T$.Since $B^T = -B$,$(B^T)^{26} = (-B)^{26} = B^{26}$.So,$M^T = B^{26} A^{13} - A^{13} B^{26} = -(A^{13} B^{26} - B^{26} A^{13}) = -M$.Thus,$M$ is skew-symmetric. $(S1)$ is false.For $(S2)$,let $N = A^{26} C^{13} - C^{13} A^{26}$.Then,$N^T = (A^{26} C^{13} - C^{13} A^{26})^T = (C^{13})^T (A^{26})^T - (A^{26})^T (C^{13})^T$.Since $C^T = -C$,$(C^T)^{13} = (-C)^{13} = -C^{13}$.So,$N^T = (-C^{13}) A^{26} - A^{26} (-C^{13}) = -C^{13} A^{26} + A^{26} C^{13} = N$.Thus,$N$ is symmetric. $(S2)$ is true.Therefore,only $S2$ is true.

Let $A, B, C$ be $3 \times 3$ 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
$(S2): A^{26} C^{13} - C^{13} A^{26}$ is symmetric
Then,

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Let $A, B, C$ be $3 \times 3$ 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$(S2): A^{26} C^{13} - C^{13} A^{26}$ is symmetricThen,