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If $\mathrm{A}$ and $\mathrm{B}$ are symmetric matrices of same order such that $\mathrm{AB}+\mathrm{BA}=\mathrm{X}$ and $\mathrm{AB}-\mathrm{BA}=\mathrm{Y}$, then $(\mathrm{XY})^{\mathrm{T}}=$
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The correct answer is:
$-\mathrm{YX}$
$\begin{aligned} & \text { Since } X Y=(A B+B A)(A B-B A) \\ & =(\mathrm{AB}) \mathrm{AB}+(\mathrm{BA})(\mathrm{AB})-(\mathrm{AB})(\mathrm{BA})-(\mathrm{BA})(\mathrm{BA}) \\ & \text { Now }(x y)^{\mathrm{T}}=((A B) \cdot(A B))^{\mathrm{T}}+(B A \cdot A B)^{\mathrm{T}}-(A B \cdot B A)^{\mathrm{T}} \\ & =(A B)^{\mathrm{T}} \cdot(\mathrm{AB})^{\mathrm{T}}+(\mathrm{AB})^{\mathrm{T}} \cdot(\mathrm{BA})-(\mathrm{BD})^{\mathrm{T}}(\mathrm{AB}) \mathrm{T} \\ & -(\mathrm{BA})^{\mathrm{T}}(\mathrm{BA})^{\mathrm{T}} \\ & =\left(B^T \cdot A^T\right)\left(B^T \cdot A^T\right)+\left(B^T \cdot A^T\right) \cdot\left(A^T B^T\right) \\ & -\left(A^T B^T\right)\left(B^T A^T\right)-\left(A^T B^T\right)\left(A^T B^T\right) \\ & =(\mathrm{BA})(\mathrm{BA})+(\mathrm{BA})(\mathrm{BA})-(\mathrm{AB})(\mathrm{BA})-(\mathrm{AB})(\mathrm{AB}) \\ & =\mathrm{BA}(\mathrm{BA}+\mathrm{AB})-\mathrm{AB}(\mathrm{BA}+\mathrm{AB}) \\ & =(\mathrm{BA}-\mathrm{AB})(\mathrm{BA}+\mathrm{AB}) \\ & =-\mathrm{yx} \\ & \end{aligned}$
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