If $ (A + B)^{-1} = A ^{-1} + B^{-1} $ then $\det A = \det B $.

Asked Feb 09 '19 at 13:12

Active Feb 09 '19 at 13:42

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If $A$ and $B$ are $n \times n $ $(n \geq 2)$ nonsingular matrices with real entries such that $(A +B)^{-1} = A^{-1} +B^{-1} $ then show that $ \det A = \det B $. Does this also true for complex entries.

my attempt : $(A+B)(A^{-1}+ B^{-1}) = I$ then $I + AB^{-1} +BA^{-1} = 0$ any hint from here how to proceed?

edited Feb 09 '19 at 13:42

Bernard

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asked Feb 09 '19 at 13:12

CHOUDHARY bhim sen

1

sorry, i have edited. now it is ok from given assumption. – CHOUDHARY bhim sen Feb 09 '19 at 13:16
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Now it is a duplicate. See above ^ – Yanko Feb 09 '19 at 13:23
This question is screaming Sylvester's determinant theorem at me, maybe that will help. – jpmacmanus Feb 09 '19 at 13:26

If $ (A + B)^{-1} = A ^{-1} + B^{-1} $ then $\det A = \det B $.

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