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If-A-B-and-A-B-are-non-singular-square-matrices-prove-that-A-1-B-1-is-also-non-singular-

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Question Number 206868 by depressiveshrek last updated on 28/Apr/24
If A, B and A+B are non−singular square matrices, prove that A^(−1) +B^(−1) is also non−singular.
IfA,BandA+Barenon−singularsquarematrices,provethatA−1+B−1isalsonon−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) ⇒det(A^(−1) +B^(−1) )=det(A^(−1) B^(−1) (A+B))= =det(A^(−1) )det(B^(−1) )det(A+B) Since ∃A^(−1) ,B^(−1) ⇒ ⇒det(A^(−1) +B^(−1) )=((det(A+B))/(det(A)det(B))) Since A, B and A+B are nonsingular ⇒det(A), det(B), det(A+B) ≠ 0 ⇒det(A^(−1) +B^(−1) )=((det(A+B))/(det(A)det(B)))≠0 ⇒det(A^(−1) +B^(−1) )≠0 ⇒A^(−1) +B^(−1) is also nonsingular.
A−1+B−1=A−1(E+AB−1)==A−1(B+A)B−1⇒det(A−1+B−1)=det(A−1B−1(A+B))==det(A−1)det(B−1)det(A+B)Since∃A−1,B−1⇒⇒det(A−1+B−1)=det(A+B)det(A)det(B)SinceA,BandA+Barenonsingular⇒det(A),det(B),det(A+B)≠0⇒det(A−1+B−1)=det(A+B)det(A)det(B)≠0⇒det(A−1+B−1)≠0⇒A−1+B−1isalsononsingular.

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