Consider a matrix $A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \\ \alpha^{2} & \beta^{2} & \gamma^{2} \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}\right]$, where $\alpha, \beta, \gamma$ are three distinct natural numbers. If $\frac{\det(\text{adj}(\text{adj}(\text{adj}(\text{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32} \times 3^{16}$, then the number of such 3-tuples $(\alpha, \beta, \gamma)$ is ______.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Determinants

Consider a matrix

A = α α^{2} β + γ β β^{2} γ + α γ γ^{2} α + β

, where

α, β, γ

are three distinct natural numbers. If

\frac{d e t ( adj ( adj ( adj ( adj A ))))}{( α - β ) ^{16} ( β - γ ) ^{16} ( γ - α ) ^{16}} = 2^{32} \times 3^{16}

, then the number of such 3-tuples

(α, β, γ)

is ______.

Answer: 42

Computing

∣ A ∣

by row operations:

∣ A ∣ = (α + β + γ) (α - β) (β - γ) (γ - α)

. Using the property

∣ adj (adj (adj (adj A))) ∣ = ∣ A ∣^{16}

for a

3 \times 3

matrix, we get

(α + β + γ)^{16} = 2^{32} \cdot 3^{16} = (12)^{16}

, so

α + β + γ = 12

. The number of positive integer solutions is

(2 11) = 55

. Subtracting 1 case where all are equal and 12 cases where two are equal gives

55 - 13 = 42

distinct tuples.

JEE Main · 2022 · 27 Jul 2022 · Shift 2 · Integer answer