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Let $A$, $B$ be $n\times n(n\ge 2)$ nonsingular matrices with real entries.

a)If $A^{-1}+B^{-1}=(A+B)^{-1}$,then prove that $\det A=\det B$

b)Find the examples of matrices $A$,$B$ satisfying $A^{-1}+B^{-1}=(A+B)^{-1}$.

c)Find the examples of matrices $A,B$ with complex entries such that $A^{-1}+B^{-1}=(A+B)^{-1}$, but $\det A\ne \det B$


I tried the first part I think it may be done by some multiplication right and left side but i failed.And for example i can't derive any example.Is their any process for thinking this type of problem??

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1 Answer 1

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We have:

$$I = (A+B)(A+B)^{-1} = (A+B)(A^{-1} + B^{-1}) = 2I + AB^{-1} + BA^{-1}$$ $$I = (A+B)^{-1}(A+B) = (A^{-1} + B^{-1})(A+B) = 2I + A^{-1}B + B^{-1}A$$

Therefore $$-BA^{-1} = I + AB^{-1}$$ $$-A^{-1}B = I + B^{-1}A$$

Sylvester's determinant identity yields:

$$(-1)^n\frac{\det B}{\det A} = \det(-BA^{-1}) = \det(I + AB^{-1}) = \det(I + B^{-1}A) = \det (-A^{-1}B) = (-1)^n\frac{\det A}{\det B}$$

Therefore

$$(\det A)^2 = (\det B)^2 \implies \det A = \pm\det B$$

On the other hand, we have

$$A + B = (A^{-1} + B^{-1})^{-1} = A(A + B)^{-1}B$$

so

$$\det(A + B) = \det A(A + B)^{-1}B = \frac{\det A \det B}{\det (A + B)} \implies \det A \det B = \det(A + B)^2 > 0$$

Hence, $\det A$ and $\det B$ must be of the same sign so we conclude $\det A = \det B$.

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  • $\begingroup$ How you do this part -->$(A^{-1}+B^{-1})^{-1}=A(A+B)^{-1}B$ ?? $\endgroup$
    – RAM_3R
    Jan 7, 2018 at 18:10
  • $\begingroup$ @RAM_3R $$A^{-1} + B^{-1} = B^{-1} + A^{-1} = B^{-1}(A + B)A^{-1}$$ and now invert the entire equality. $\endgroup$ Jan 7, 2018 at 18:21

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