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(C) We need to find the inverse of the expression $X = A(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))^{-1}B$.Using the property $(XYZ)^{-1}

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$\frac{1}{|AB|}(\operatorname{adj}(B)+\operatorname{adj}(A))$

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(C) We need to find the inverse of the expression $X = A(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))^{-1}B$.Using the property $(XYZ)^{-1} = Z^{-1}Y^{-1}X^{-1}$,we get:$X^{-1} = B^{-1}(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1}))A^{-1}$$X^{-1} = B^{-1}\operatorname{adj}(A^{-1})A^{-1} + B^{-1}\operatorname{adj}(B^{-1})A^{-1}$Using the property $\operatorname{adj}(M^{-1}) = |M^{-1}|M = \frac{1}{|M|}M$,we have $\operatorname{adj}(A^{-1}) = |A^{-1}|A = \frac{1}{|A|}A$ and $\operatorname{adj}(B^{-1}) = |B^{-1}|B = \frac{1}{|B|}B$.Substituting these values:$X^{-1} = B^{-1}(\frac{1}{|A|}A)A^{-1} + B^{-1}(\frac{1}{|B|}B)A^{-1}$$X^{-1} = \frac{1}{|A|}B^{-1}(AA^{-1}) + \frac{1}{|B|}(B^{-1}B)A^{-1}$$X^{-1} = \frac{1}{|A|}B^{-1}I + \frac{1}{|B|}IA^{-1}$$X^{-1} = \frac{B^{-1}}{|A|} + \frac{A^{-1}}{|B|}$Since $B^{-1} = \frac{\operatorname{adj}(B)}{|B|}$ and $A^{-1} = \frac{\operatorname{adj}(A)}{|A|}$,we get:$X^{-1} = \frac{\operatorname{adj}(B)}{|B||A|} + \frac{\operatorname{adj}(A)}{|A||B|} = \frac{1}{|AB|}(\operatorname{adj}(B) + \operatorname{adj}(A))$.

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

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