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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2025 · 23 January · Morning Shift · Mathematics · Matrices

If A, B , and (adj (A^ -1 )+adj(B^ -1 )) are non-singular matrices...

If A,B\mathrm{A}, \mathrm{B}, 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))−1 B\mathrm{A}\left(\operatorname{adj}\left(\mathrm{A}^{-1}\right)+\operatorname{adj}\left(\mathrm{B}^{-1}\right)\right)^{-1} \mathrm{~B}, is equal to

Options

  1. AAB−1+A−1 B\mathrm{AB}^{-1}+\mathrm{A}^{-1} \mathrm{~B}
  2. Badj⁡(B−1)+adj⁡(A−1)\operatorname{adj}\left(\mathrm{B}^{-1}\right)+\operatorname{adj}\left(\mathrm{A}^{-1}\right)
  3. CAB−1∣A∣+BA−1∣B∣\frac{A B^{-1}}{|A|}+\frac{B A^{-1}}{|B|}
  4. D1∣AB∣(adj⁡(B)+adj⁡(A))\frac{1}{|A B|}(\operatorname{adj}(B)+\operatorname{adj}(A))

Correct answer

D. 1∣AB∣(adj⁡(B)+adj⁡(A))\frac{1}{|A B|}(\operatorname{adj}(B)+\operatorname{adj}(A))

Step-by-step solution

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