If A, B and $$(adj (A^{-1})+adj(B^{-1}))$$ are non-singular matrices of same order, then the inverse of $$A(adj(A^{-1}+adj(B^{-1}))^{-1}B$$, is equal to

Property: For any non-singular matrix $$X$$, $$adj(X) = |X|X^{-1}$$.

Simplify $$adj(A^{-1}) = |A^{-1}|(A^{-1})^{-1} = \frac{1}{|A|}A$$.

Let $$K = adj(A^{-1}) + adj(B^{-1}) = \frac{A}{|A|} + \frac{B}{|B|}$$.

The expression is $$M = A \cdot K^{-1} \cdot B$$. We need $$M^{-1}$$.

$$M^{-1} = (A \cdot K^{-1} \cdot B)^{-1} = B^{-1} \cdot K \cdot A^{-1}$$

 Substitute $$K$$ back:

$$M^{-1} = B^{-1} \left( \frac{A}{|A|} + \frac{B}{|B|} \right) A^{-1} = \frac{B^{-1}AA^{-1}}{|A|} + \frac{B^{-1}BA^{-1}}{|B|} = \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$$

 Multiply and divide by $$|A||B|$$ to match options:

$$M^{-1} = \frac{|B|B^{-1} + |A|A^{-1}}{|A||B|} = \frac{adj(B) + adj(A)}{|AB|}$$

Option D is correct.