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JEE Main 202622 JanuaryMorning Shift
Exam
JEE Main
Year
2026
Date
22 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2026 · 22 January · Morning Shift · Mathematics · Matrices

Let A be a 3 × 3 matrix such that A+A^ T =O . If A [ array c 1 -1 0...

Let A be a 3×33 \times 3 matrix such that A+AT=O\mathrm{A}+\mathrm{A}^{\mathrm{T}}=\mathrm{O}. If A\mathrm{A}\left[1−10\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[332\begin{array}{l}3 \\ 3 \\ 2\end{array}\right], \mathrm{A}^{2}\left[1−10\begin{array}{c}1 \\ -1 \\ 0\end{array}\right]=\left[−319−24\begin{array}{c}-3 \\ 19 \\ -24\end{array}\right] and det⁡(adj⁡(2adj⁡( A+\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}(\mathrm{~A}+\mathrm{I})))=(2)α⋅(3)β⋅(11)γ,α,β,γ)))=(2)^{\alpha} \cdot(3)^{\beta} \cdot(11)^{\gamma}, \alpha, \beta, \gamma are non-negative integers, then α+β+γ\alpha+\beta+\gamma is equal to ____\_\_\_\_

Correct answer

0

Step-by-step solution

AA is skew-symmetric: A = \begin{bmatrix}0 & -c & b\c & 0 & -a\\-b & a & 0\end{bmatrix}.
From A[1−10]=[332]A\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}3\\3\\2\end{bmatrix}: c=3c = 3, a+b=−2a+b = -2.
From A2[1−10]=[−319−24]A^2\begin{bmatrix}1\\-1\\0\end{bmatrix} = \begin{bmatrix}-3\\19\\-24\end{bmatrix}: b=3b = 3, a=−5a = -5.
A+I=[1−33315−3−51]A+I = \begin{bmatrix}1&-3&3\\3&1&5\\-3&-5&1\end{bmatrix}, det⁡(A+I)=44\det(A+I) = 44.
For 3×33 \times 3 matrix: det⁡(adj(M))=(det⁡M)2\det(\text{adj}(M)) = (\det M)^2.
det⁡(2 adj(A+I))=8⋅442=8⋅1936=15488=27⋅112\det(2\,\text{adj}(A+I)) = 8 \cdot 44^2 = 8 \cdot 1936 = 15488 = 2^7 \cdot 11^2.
det⁡(adj(2 adj(A+I)))=154882=214⋅114\det(\text{adj}(2\,\text{adj}(A+I))) = 15488^2 = 2^{14} \cdot 11^4.
α=14, β=0, γ=4\alpha = 14,\ \beta = 0,\ \gamma = 4. α+β+γ=18\alpha+\beta+\gamma = 18.
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