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Question Number 206868 by depressiveshrek last updated on 28/Apr/24

$$\mathrm{If}A,B\mathrm{and}A+B\mathrm{are}\mathrm{non}-\mathrm{singular}$$$$\mathrm{square}\mathrm{matrices},\mathrm{prove}\mathrm{that}{A}^{-1}+B^{-1}$$$$\mathrm{is}\mathrm{also}\mathrm{non}-\mathrm{singular}.$$

Answered by aleks041103 last updated on 28/Apr/24

$$A^{-1}+B^{-1}=A^{-1}(E+{AB}^{-1})=$$$$=A^{-1}(B+A)B^{-1}$$$$\Rightarrow det(A^{-1}+B^{-1})=det\left(A^{-1}{B}^{-1}(A+B)\right)=$$$$=det\left(A^{-1}\right)det\left(B^{-1}\right)det(A+B)$$$$$$$$Since\mathrm{\exists }{A}^{-1},B^{-1}\Rightarrow$$$$\Rightarrow det(A^{-1}+B^{-1})=\frac{det(A+B)}{det\left(A\right)det\left(B\right)}$$$$SinceA,BandA+Barenonsingular$$$$\Rightarrow det\left(A\right),det\left(B\right),det(A+B)\ne 0$$$$\Rightarrow det(A^{-1}+B^{-1})=\frac{det(A+B)}{det\left(A\right)det\left(B\right)}\ne 0$$$$\Rightarrow det(A^{-1}+B^{-1})\ne 0$$$$\Rightarrow A^{-1}+B^{-1}isalsononsingular.$$

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