JEE Mathematics 2025 Question with Solution

Given: $A$, $B$, and $\left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right)$ are non-singular matrices of the same order.

Find: The inverse of

$$A \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right) B$$

Using the relation stated in the solution,

$$\operatorname{adj}(M^{-1}) = |M| M$$

so the worked solution concludes that the required inverse is

$$\frac{1}{|A||B|} \left( \operatorname{adj}(B) + \operatorname{adj}(A) \right).$$

Therefore, the correct option is C.

Given:

$$C = A \left( \operatorname{adj}(A^{-1}) + \operatorname{adj}(B^{-1}) \right) B$$

Find: $C^{-1}$

the solution uses the identity

$$\operatorname{adj}(A^{-1}) = \frac{1}{|A|} A, \qquad \operatorname{adj}(B^{-1}) = \frac{1}{|B|} B$$

because it states that

$$|A^{-1}| = \frac{1}{|A|}, \qquad |B^{-1}| = \frac{1}{|B|}.$$

Substituting into the expression,

$$C = A \left( \frac{1}{|A|} A + \frac{1}{|B|} B \right) B$$

which is written in the solution as

$$C = \frac{1}{|A||B|} A (A + B) B.$$

Then, using the inverse property

$$(XY)^{-1} = Y^{-1} X^{-1},$$

the provided solution concludes

$$C^{-1} = \frac{1}{|A||B|} \left( \operatorname{adj}(B) + \operatorname{adj}(A) \right).$$

Hence, the inverse is $\frac{1}{|A||B|} \left( \operatorname{adj}(B) + \operatorname{adj}(A) \right)$, so the correct option is C.

Note: The intermediate algebra shown in the source is not fully rigorous, but the solution explicitly marks Option C as correct and concludes with this expression.