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Mathematics · Matrices

JEE Main 2026 — 4 April, Morning Shift — Question 29

Let A=[112−201135]\mathrm{A} = \left[ \begin{array}{ccc}1 & 1 & 2\\ -2 & 0 & 1\\ 1 & 3 & 5 \end{array} \right] . Then the sum of all elements of the matrix adj⁡(adj⁡(2(adj⁡A)−1))\operatorname{adj}(\operatorname{adj}(2(\operatorname{adj}A)^{-1})) is equal to :

  1. Option A:

    3

  2. Option B:

    4

  3. Option C:

    -4

  4. Option D:

    -3

    Correct

Answer: D

Step-by-step solution

∣A∣=−4|\mathrm{A}|=-4 As (adjA⁡)−1=A∣A∣=−A4(\operatorname{adjA})^{-1}=\frac{\mathrm{A}}{|\mathrm{A}|}=\frac{-\mathrm{A}}{4} B=2(adj⁡ A)−1=−A2\mathrm{B}=2(\operatorname{adj} \mathrm{~A})^{-1}=\frac{-\mathrm{A}}{2} ∣B∣=∣−A2∣|\mathrm{B}|=\left|\frac{-\mathrm{A}}{2}\right| =−18∣ A∣=12=\frac{-1}{8}|\mathrm{~A}|=\frac{1}{2} adj⁡(adj⁡B)=∣B∣n−2 B=∣B∣B\operatorname{adj}(\operatorname{adj} \mathrm{B})=|\mathrm{B}|^{\mathrm{n}-2} \mathrm{~B}=|\mathrm{B}| \mathrm{B} =B2=−A4=\frac{\mathrm{B}}{2}=\frac{-\mathrm{A}}{4} Sum of element =−124=−3=\frac{-12}{4}=-3

Answer key and solution verified before publishing.

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Exam
JEE Main 2026
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Matrices
Topic
Adjoint of a Square Matrix

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