If-A-and-B-are-invertible-matrices-then-AB-1-B-1-A-1-A-1-B-1-proove-

Question Number 156610 by jlewis last updated on 13/Oct/21

$$\mathrm{If}\:\mathrm{A}\:\mathrm{and}\:\mathrm{B}\:\mathrm{are}\:\mathrm{invertible}\:\mathrm{matrices},\mathrm{then}: \\ $$$$\left(\mathrm{AB}\right)^{−\mathrm{1}} =\mathrm{B}^{−\mathrm{1}} \mathrm{A}^{−\mathrm{1}} \neq\:\mathrm{A}^{−\mathrm{1}} \mathrm{B}^{−\mathrm{1}} \\ $$$$\boldsymbol{\mathrm{proove}}. \\ $$

Answered by physicstutes last updated on 14/Oct/21

$$\mathrm{Consider}\:\mathrm{the}\:\mathrm{theorems}\:{AA}^{−\mathrm{1}} \:=\:{I} \\ $$$$\mathrm{and}\:{BB}^{−\mathrm{1}} \:=\:{I} \\ $$$$\mathrm{now}\:\mathrm{assume}:\:\left({AB}\right)^{−\mathrm{1}} \:=\:{B}^{−\mathrm{1}} {A}^{−\mathrm{1}} \\ $$$$\mathrm{pre}−\mathrm{multiply}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{by}\:{AB} \\ $$$$\Rightarrow\:{AB}\left({AB}\right)^{−\mathrm{1}} \:=\:{ABB}^{−\mathrm{1}} {A}^{−\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:=\:{AIA}^{−\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:=\:{AA}^{−\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:=\:{I}\:\:\mathrm{which}\:\mathrm{is}\:\mathrm{true}! \\ $$$$\:\mathrm{Also}\:\mathrm{consider}: \\ $$$$\left({AB}\right)^{−\mathrm{1}} \:={A}^{−\mathrm{1}} {B}^{−\mathrm{1}} \\ $$$$\mathrm{pre}−\mathrm{multiply}\:\mathrm{both}\:\mathrm{sides}\:\mathrm{by}\:{AB} \\ $$$$\Rightarrow\:\left({AB}\right)\left({AB}\right)^{−\mathrm{1}} \:=\:{ABA}^{−\mathrm{1}} {B}^{−\mathrm{1}} \\ $$$$\Rightarrow\:{I}\:\neq\:{ABA}^{−\mathrm{1}} {B}^{−\mathrm{1}} \: \\ $$

Answered by TheSupreme last updated on 13/Oct/21

$${B}^{−\mathrm{1}} {A}^{−\mathrm{1}} {AB}={B}^{−\mathrm{1}} \left({A}^{−\mathrm{1}} {A}\right){B}={B}^{−\mathrm{1}} {B}={I} \\ $$$${the}\:{inequality}\:{can}\:{be}\:{demonstrate}\:{with}\:{non}\:{commutativity}\:{of}\:{matrix}\:{moltiplication} \\ $$

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