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Question Number 216525 by Mingma last updated on 10/Feb/25

Answered by Rasheed.Sindhi last updated on 10/Feb/25

11.1 A^2 =A , B^2 =B , AB=BA AB=BA⇒(AB)^2 =A^2 B^2 (AB)^2 =A^2 B^2 =AB 11.2 A^2 =A , B^2 =B ⇒^(?) (A+B)^2 =A+B (A+B)^2 =A^2 +AB+BA+B^2 (A+B)^2 =A+AB+BA+B≠A+B in general (A+B)^2 =A+B only when AB+BA=O

$$\mathrm{11}.\mathrm{1}\:\mathrm{A}^{\mathrm{2}} =\mathrm{A}\:,\:\mathrm{B}^{\mathrm{2}} =\mathrm{B}\:,\:\mathrm{AB}=\mathrm{BA} \\ $$$$\mathrm{AB}=\mathrm{BA}\Rightarrow\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{A}^{\mathrm{2}} \mathrm{B}^{\mathrm{2}} \\ $$$$\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{A}^{\mathrm{2}} \mathrm{B}^{\mathrm{2}} =\mathrm{AB} \\ $$$$\mathrm{11}.\mathrm{2}\:\:\:\mathrm{A}^{\mathrm{2}} =\mathrm{A}\:,\:\mathrm{B}^{\mathrm{2}} =\mathrm{B}\:\overset{?} {\Rightarrow}\:\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}+\mathrm{B} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}^{\mathrm{2}} +\mathrm{AB}+\mathrm{BA}+\mathrm{B}^{\mathrm{2}} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}+\mathrm{AB}+\mathrm{BA}+\mathrm{B}\neq\mathrm{A}+\mathrm{B}\:{in}\:{general} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}+\mathrm{B}\:{only}\:{when} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{AB}+\mathrm{BA}=\mathrm{O} \\ $$

Commented by Mingma last updated on 10/Feb/25

Perfect sir! thank you!

Answered by Wuji last updated on 10/Feb/25

A^2 =A−−1 B^2 =B−−−2 AB=BA−−−3 for (AB)^2 =AB (AB)^2 =AB•AB=A(BA)B from (3) (AB)^2 =A(AB)B=A^2 B^2 (AB)^2 =AB {from(1) and (2)} check if (A+B)^2 =(A+B) (A+B)^2 =A^2 +AB+BA+B^2 =B+AB+BA+B (A+B)^2 =A+AB+BA+B=A+B+2AB (A+B)^2 =A+B+2AB ∴ A+B≠idempotent since AB=BA=0

$$\mathrm{A}^{\mathrm{2}} =\mathrm{A}−−\mathrm{1} \\ $$$$\mathrm{B}^{\mathrm{2}} =\mathrm{B}−−−\mathrm{2} \\ $$$$\mathrm{AB}=\mathrm{BA}−−−\mathrm{3} \\ $$$$\mathrm{for}\:\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{AB} \\ $$$$\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{AB}\bullet\mathrm{AB}=\mathrm{A}\left(\mathrm{BA}\right)\mathrm{B} \\ $$$$\mathrm{from}\:\left(\mathrm{3}\right) \\ $$$$\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{A}\left(\mathrm{AB}\right)\mathrm{B}=\mathrm{A}^{\mathrm{2}} \mathrm{B}^{\mathrm{2}} \\ $$$$\left(\mathrm{AB}\right)^{\mathrm{2}} =\mathrm{AB}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left\{\mathrm{from}\left(\mathrm{1}\right)\:\mathrm{and}\:\left(\mathrm{2}\right)\right\} \\ $$$$\: \\ $$$$\: \\ $$$$\mathrm{check}\:\mathrm{if} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\left(\mathrm{A}+\mathrm{B}\right) \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}^{\mathrm{2}} +\mathrm{AB}+\mathrm{BA}+\mathrm{B}^{\mathrm{2}} =\mathrm{B}+\mathrm{AB}+\mathrm{BA}+\mathrm{B} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}+\mathrm{AB}+\mathrm{BA}+\mathrm{B}=\mathrm{A}+\mathrm{B}+\mathrm{2AB} \\ $$$$\left(\mathrm{A}+\mathrm{B}\right)^{\mathrm{2}} =\mathrm{A}+\mathrm{B}+\mathrm{2AB} \\ $$$$\therefore\:\mathrm{A}+\mathrm{B}\neq\boldsymbol{\mathrm{idempotent}} \\ $$$$\mathrm{since}\:\boldsymbol{\mathrm{AB}}=\boldsymbol{\mathrm{BA}}=\mathrm{0} \\ $$

Commented by Rasheed.Sindhi last updated on 10/Feb/25

AB=BA is not given for Q 11.2 Hence AB+BA≠2AB

$$\mathrm{AB}=\mathrm{BA}\:{is}\:{not}\:{given}\:{for}\:{Q}\:\mathrm{11}.\mathrm{2} \\ $$$${Hence}\:\mathrm{AB}+\mathrm{BA}\neq\mathrm{2AB} \\ $$

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