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Question Number 19589 by gourav~ last updated on 13/Aug/17
$${Let}\:{A}\:{and}\:{B}\:{is}\:\mathrm{3}×\mathrm{3}\:{matrix}\:{of}\:{equal}\:{number} \\ $$$${where}\:{A}={symmetric}\:{matrix}\: \\ $$$$....{B}={skew}\:{symmetric}\:{matrix} \\ $$$${and}\:{the}\:{relation}...\:\left({A}+{B}\right)\left({A}−{B}\right)=\left({A}−{B}\right)\left({A}+{B}\right) \\ $$$${then}..{the}\:{value}\:{of}..\:...\:{k} \\ $$$$\:\:\:\:\left({AB}\right)^{{T}} =\left(−\mathrm{1}\right)^{{k}} \left({AB}\right) \\ $$$$\left({a}\right)\:−\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({c}\right)\:\mathrm{2} \\ $$$$ \\ $$$$\left({b}\right)\:\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left({d}\right)\:\mathrm{3} \\ $$
Answered by dioph last updated on 15/Aug/17
$${A}^{\mathrm{2}} −{AB}+{BA}−{B}^{\mathrm{2}} ={A}^{\mathrm{2}} +{AB}−{BA}+{B}^{\mathrm{2}} \\ $$$$\mathrm{2}{AB}\:=\:\mathrm{2}{BA} \\ $$$${AB}\:=\:{BA} \\ $$$$\left({AB}\right)^{{T}} \:=\:\left({BA}\right)^{{T}} \\ $$$$\left({AB}\right)^{{T}} \:=\:{A}^{{T}} {B}^{{T}} \\ $$$$\left({AB}\right)^{{T}} \:=\:{A}\left(−{B}\right) \\ $$$$\left({AB}\right)^{{T}} \:=\:\left(−\mathrm{1}\right)\left({AB}\right) \\ $$