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

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

Answered by physicstutes last updated on 14/Oct/21

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

Answered by TheSupreme last updated on 13/Oct/21

$$B^{-1}{A}^{-1}AB=B^{-1}\left(A^{-1}A\right)B=B^{-1}B=I$$$$theinequalitycanbedemonstratewithnoncommutativityofmatrixmoltiplication$$

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