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JEE Main 2026 · 4 Apr · Shift 1, Mathematics question 6

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

Choose one option
Answer and worked solution

Official answer: D

Determinant. det⁡A=1(0−3)−1(−10−1)+2(−6)=−4\det A=1(0-3)-1(-10-1)+2(-6)=-4.

Simplify. (adj⁡A)−1=Adet⁡A(\operatorname{adj}A)^{-1}=\dfrac A{\det A}, so M=2(adj⁡A)−1=−A2M=2(\operatorname{adj}A)^{-1}=-\dfrac A2, with det⁡M=−18(−4)=12\det M=-\frac18(-4)=\frac12.

Double adjoint (3 × 3): adj⁡(adj⁡M)=(det⁡M)M=−A4\operatorname{adj}(\operatorname{adj}M)=(\det M)M=-\dfrac A4.

Sum of elements. The entries of AA sum to 12, so the answer is −3-3.

(D)

Topics:Adjoint and inverse
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