Question Number 98369 by nesreen last updated on 13/Jun/20
$$\mathrm{If}\:{A},\:{B}\:\mathrm{are}\:\mathrm{two}\:\mathrm{square}\:\mathrm{matrices}\:\mathrm{such} \\ $$$$\mathrm{that}\:{AB}\:=\:{A}\:\mathrm{and}\:{BA}\:=\:{B},\:\mathrm{then} \\ $$
Answered by bobhans last updated on 13/Jun/20
$$\mathrm{A}^{−\mathrm{1}} \mathrm{AB}=\mathrm{A}^{−\mathrm{1}} \mathrm{A}\:\Rightarrow\:\mathrm{B}\:=\:\mathrm{I} \\ $$$$\mathrm{B}^{−\mathrm{1}} \mathrm{BA}=\mathrm{B}^{−\mathrm{1}} \mathrm{B}\:\Rightarrow\:\mathrm{A}=\mathrm{I}\: \\ $$$$\mathrm{then}\:\mathrm{A}=\mathrm{B}=\mathrm{I} \\ $$
Answered by Rio Michael last updated on 13/Jun/20
$$\mathrm{Also}\:\:{AB}\:=\:{A}\:\mathrm{and}\:{BA}\:=\:{B} \\ $$$$\Rightarrow\:{A}\left({BA}\right)\:=\:{A} \\ $$$$\Rightarrow\:\left({AB}\right)\:{A}\:=\:{A} \\ $$$$\Rightarrow\:\left({A}\right)\left({A}\right)\:=\:{A} \\ $$$$\Rightarrow\:{A}^{\mathrm{2}} \:=\:{A}\:\mathrm{similarly},\:{B}^{\mathrm{2}} \:=\:{B} \\ $$