Verified Solution • 2025–26 Session moderate

QJEEM25Ma632d: JEE Main 2025 Mathematics - Class 12 JEEM25Ma632d Solution: If are non-singular matrices of same order, then the inverse of …

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

Multiple Choice Options

  • A AB−1∣A∣+BA−1∣B∣\frac{A B^{-1}}{|A|}+\frac{B A^{-1}}{|B|}
  • B adj⁡(B−1)+adj⁡(A−1)\operatorname{adj}\left(\mathrm{B}^{-1}\right)+\operatorname{adj}\left(\mathrm{A}^{-1}\right)
  • C AB−1+A−1 B\mathrm{AB}^{-1}+\mathrm{A}^{-1} \mathrm{~B}
  • D 1∣AB∣(adj⁡(B)+adj⁡(A))\frac{1}{|A B|}(\operatorname{adj}(B)+\operatorname{adj}(A))

Step-by-Step Verified Solution

amp;[A(adj⁡( A−1)+adj⁡(B−1))−1⋅ B]−1amp; B−1⋅(adj⁡( A−1)+adj⁡(B−1))⋅A−1amp; B−1adj⁡( A−1)A−1+B−1(adj⁡( B−1))⋅A−1amp; B−1∣ A−1∣I+∣B−1∣IA−1amp; B−1∣ A∣+A−1∣ B∣amp;⇒adjB⁡∣B∣∣A∣+adj⁡∣A∣∣B∣amp;=1∣ A∣∣B∣(adjB⁡+adjA⁡)\begin{aligned}& {\left[\mathrm{A}\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \cdot \mathrm{~B}\right]^{-1}} \\& \mathrm{~B}^{-1} \cdot\left(\operatorname{adj}\left(\mathrm{~A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\& \mathrm{~B}^{-1} \operatorname{adj}\left(\mathrm{~A}^{-1}\right) \mathrm{A}^{-1}+\mathrm{B}^{-1}\left(\operatorname{adj}\left(\mathrm{~B}^{-1}\right)\right) \cdot \mathrm{A}^{-1} \\& \mathrm{~B}^{-1}\left|\mathrm{~A}^{-1}\right| \mathrm{I}+\left|\mathrm{B}^{-1}\right| \mathrm{IA}^{-1} \\& \frac{\mathrm{~B}^{-1}}{|\mathrm{~A}|}+\frac{\mathrm{A}^{-1}}{|\mathrm{~B}|} \\& \Rightarrow \frac{\operatorname{adjB}}{|\mathrm{B}||\mathrm{A}|}+\frac{\operatorname{adj}}{|\mathrm{A}||\mathrm{B}|} \\& =\frac{1}{|\mathrm{~A}||\mathrm{B}|}(\operatorname{adjB}+\operatorname{adjA})\end{aligned}

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