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← JEE Main 2022 · 27 Jul · Shift 2
MathematicsInteger+4 / -1Question 21 of 90

JEE Main 2022 · 27 Jul · Shift 2, Mathematics question 21

Consider a matrix A=[αβγα2β2γ2β+γγ+αα+β]A=\begin{bmatrix}\alpha&\beta&\gamma\\\alpha^2&\beta^2&\gamma^2\\\beta+\gamma&\gamma+\alpha&\alpha+\beta\end{bmatrix}, where α,β,γ\alpha,\beta,\gamma are three distinct natural numbers.

If det⁡(adj⁡(adj⁡(adj⁡(adj⁡A))))(α−β)16(β−γ)16(γ−α)16=232×316\dfrac{\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32}\times3^{16}, then the number of such 3-tuples (α,β,γ)(\alpha,\beta,\gamma) is ______.

Answer and worked solution

Official answer: 42

Determinant. Adding row 1 to row 3 gives det⁡A=±(α+β+γ)(α−β)(β−γ)(γ−α)\det A=\pm(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha).

Adjugates. Four adjugates raise it to the 16th power, so (α+β+γ)16=232316(\alpha+\beta+\gamma)^{16}=2^{32}3^{16} and α+β+γ=12\alpha+\beta+\gamma=12.

Count. Ordered distinct triples: (112)−3⋅4−1=42\binom{11}2-3\cdot4-1=42.

Answer: 42

Topics:Adjoint and inverseDeterminants and propertiesDistribution and arrangements
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