Let A and B be any two 3 times 3 symmetric an | vedclass

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(C) Given that $A^{T} = A$ and $B^{T} = -B$.For option $A$: Let $C = A^{4} - B^{4}$.$C^{T} = (A^{4} - B^{4})^{T} = (A^{T})^{4} - (B^{T})^{4} = A^{4} -

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$B^{5} - A^{5}$ is a skew-symmetric matrix

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(C) Given that $A^{T} = A$ and $B^{T} = -B$.For option $A$: Let $C = A^{4} - B^{4}$.$C^{T} = (A^{4} - B^{4})^{T} = (A^{T})^{4} - (B^{T})^{4} = A^{4} - (-B)^{4} = A^{4} - B^{4} = C$. Thus,$A^{4} - B^{4}$ is symmetric.For option $B$: Let $C = AB - BA$.$C^{T} = (AB - BA)^{T} = (AB)^{T} - (BA)^{T} = B^{T}A^{T} - A^{T}B^{T} = (-B)(A) - (A)(-B) = -BA + AB = AB - BA = C$. Thus,$AB - BA$ is symmetric.For option $C$: Let $C = B^{5} - A^{5}$.$C^{T} = (B^{5} - A^{5})^{T} = (B^{T})^{5} - (A^{T})^{5} = (-B)^{5} - A^{5} = -B^{5} - A^{5} = -(B^{5} + A^{5})$. This is not necessarily equal to $-C = -(B^{5} - A^{5})$. Thus,this statement is not true.For option $D$: Let $C = AB + BA$.$C^{T} = (AB + BA)^{T} = (AB)^{T} + (BA)^{T} = B^{T}A^{T} + A^{T}B^{T} = (-B)(A) + (A)(-B) = -BA - AB = -(AB + BA) = -C$. Thus,$AB + BA$ is skew-symmetric.Therefore,option $C$ is not true.

Let $A$ and $B$ be any two $3 \times 3$ symmetric and skew-symmetric matrices respectively. Then which of the following is $NOT$ true?

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