If A, B, \operatornameand(\operatornameadj(A^-1)+\operatornameadj(B^-1)) are non-singular matrices of same order, then the inverse of A(\operatornameadj(A^-1)+\operatornameadj(B^-1))^-1 B , is equal to

\operatornameadj(B^-1)+\operatornameadj(A^-1)

Matrices and Determinants PYQ — JEE Main 2025 · 23 Jan · Shift 1 · Q44 with Solution

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Correct answer: D

Let $M=A\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)^{-1}B.$ We need to find $M^{-1}$ .
Using $(XYZ)^{-1}=Z^{-1}Y^{-1}X^{-1}$ , we get $M^{-1}=B^{-1}\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)A^{-1}.$
Now use the identity for a non-singular matrix $X$ of order $n$ : $\operatorname{adj}(X^{-1})=|X^{-1}|(X^{-1})^{-1}=\frac{1}{|X|}X.$ So, $\operatorname{adj}(A^{-1})=\frac{A}{|A|}, \qquad \operatorname{adj}(B^{-1})=\frac{B}{|B|}.$
Substitute into the expression for $M^{-1}$ : $M^{-1}=B^{-1}\left(\frac{A}{|A|}+\frac{B}{|B|}\right)A^{-1}.$ Distribute: $M^{-1}=\frac{1}{|A|}B^{-1}AA^{-1}+\frac{1}{|B|}B^{-1}BA^{-1}.$ Since $AA^{-1}=I$ and $B^{-1}B=I$ , $M^{-1}=\frac{1}{|A|}B^{-1}+\frac{1}{|B|}A^{-1}.$
Now express inverses in terms of adjugates: $A^{-1}=\frac{\operatorname{adj}(A)}{|A|}, \qquad B^{-1}=\frac{\operatorname{adj}(B)}{|B|}.$ Hence,

=\frac{\operatorname{adj}(B)}{|A||B|}+\frac{\operatorname{adj}(A)}{|A||B|}.$$ Therefore, $$M^{-1}=\frac{1}{|AB|}\left(\operatorname{adj}(B)+\operatorname{adj}(A)\right).$$ 6. Comparing with the options, this is exactly: $$\boxed{\frac{1}{|AB|}(\operatorname{adj}(B)+\operatorname{adj}(A))}$$ which is Option $D$. 7. Verification with stored answer: Stored correct answer is $D$, which matches our derived answer.

View written solutionFree

Correct answer: D

Let $M=A\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)^{-1}B.$ We need to find $M^{-1}$ .
Using $(XYZ)^{-1}=Z^{-1}Y^{-1}X^{-1}$ , we get $M^{-1}=B^{-1}\left(\operatorname{adj}(A^{-1})+\operatorname{adj}(B^{-1})\right)A^{-1}.$
Now use the identity for a non-singular matrix $X$ of order $n$ : $\operatorname{adj}(X^{-1})=|X^{-1}|(X^{-1})^{-1}=\frac{1}{|X|}X.$ So, $\operatorname{adj}(A^{-1})=\frac{A}{|A|}, \qquad \operatorname{adj}(B^{-1})=\frac{B}{|B|}.$
Substitute into the expression for $M^{-1}$ : $M^{-1}=B^{-1}\left(\frac{A}{|A|}+\frac{B}{|B|}\right)A^{-1}.$ Distribute: $M^{-1}=\frac{1}{|A|}B^{-1}AA^{-1}+\frac{1}{|B|}B^{-1}BA^{-1}.$ Since $AA^{-1}=I$ and $B^{-1}B=I$ , $M^{-1}=\frac{1}{|A|}B^{-1}+\frac{1}{|B|}A^{-1}.$
Now express inverses in terms of adjugates: $A^{-1}=\frac{\operatorname{adj}(A)}{|A|}, \qquad B^{-1}=\frac{\operatorname{adj}(B)}{|B|}.$ Hence,

=\frac{\operatorname{adj}(B)}{|A||B|}+\frac{\operatorname{adj}(A)}{|A||B|}.$$ Therefore, $$M^{-1}=\frac{1}{|AB|}\left(\operatorname{adj}(B)+\operatorname{adj}(A)\right).$$ 6. Comparing with the options, this is exactly: $$\boxed{\frac{1}{|AB|}(\operatorname{adj}(B)+\operatorname{adj}(A))}$$ which is Option $D$. 7. Verification with stored answer: Stored correct answer is $D$, which matches our derived answer.