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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

	$11.1 A^{2} = A, B^{2} = B, AB = BA$ $AB = BA \Rightarrow {(AB)}^{2} = A^{2} B^{2}$ ${(AB)}^{2} = A^{2} B^{2} = AB$ $11.2 A^{2} = A, B^{2} = B \overset{?}{\Rightarrow} {(A + B)}^{2} = A + B$ ${(A + B)}^{2} = A^{2} + AB + BA + B^{2}$ ${(A + B)}^{2} = A + AB + BA + B \neq A + B i n g e n e r a l$ ${(A + B)}^{2} = A + B o n l y w h e n$ $AB + BA = O$
	Commented by Mingma last updated on 10/Feb/25
	Perfect sir! thank you!
	Answered by Wuji last updated on 10/Feb/25
	$A^2 =A−−1 B^2 =B−−−2 AB=BA−−−3 for (AB)^2 =AB (AB)^2 =AB•AB=A(BA)B from (3) (AB)^2 =A(AB)B=A^2 B^2 (AB)^2 =AB {from(1) and (2)} check if (A+B)^2 =(A+B) (A+B)^2 =A^2 +AB+BA+B^2 =B+AB+BA+B (A+B)^2 =A+AB+BA+B=A+B+2AB (A+B)^2 =A+B+2AB ∴ A+B≠idempotent since AB=BA=0$
	$A^{2} = A - - 1$ $B^{2} = B - - - 2$ $AB = BA - - - 3$ $for {(AB)}^{2} = AB$ ${(AB)}^{2} = AB ∙ AB = A (BA) B$ $from (3)$ ${(AB)}^{2} = A (AB) B = A^{2} B^{2}$ ${(AB)}^{2} = AB {from (1) and (2)}$ $check if$ ${(A + B)}^{2} = (A + B)$ ${(A + B)}^{2} = A^{2} + AB + BA + B^{2} = B + AB + BA + B$ ${(A + B)}^{2} = A + AB + BA + B = A + B + 2 AB$ ${(A + B)}^{2} = A + B + 2 AB$ $∴ A + B \neq idempotent$ $since AB = BA = 0$
	Commented by Rasheed.Sindhi last updated on 10/Feb/25

	$AB = BA i s n o t g i v e n f o r Q 11.2$ $H e n c e AB + BA \neq 2 AB$
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