JEE Main 2022 — Mathematics Algebra

Given $A=[\begin{matrix}\alpha & \beta & \gamma \\ {\alpha }^{2} & {\beta }^{2} & {\gamma }^{2} \\ \beta +\gamma & \gamma +\alpha & \alpha +\beta \end{matrix}]$A=[αα2β+γ​ββ2γ+α​γγ2α+β​]

$${R}_{3}\rightarrow {R}_{3}+{R}_{1}$$R3​→R3​+R1​

$$A=[\begin{matrix}\alpha & \beta & \gamma \\ {\alpha }^{2} & {\beta }^{2} & {\gamma }^{2} \\ \alpha +\beta +\gamma & \alpha +\beta +\gamma & \alpha +\beta +\gamma \end{matrix}]$$A=[αα2α+β+γ​ββ2α+β+γ​γγ2α+β+γ​]

$$\Rightarrow |A|=|\alpha +\beta +\gamma ||\begin{matrix}\alpha & \beta & \gamma \\ {\alpha }^{2} & {\beta }^{2} & {\gamma }^{2} \\ 1 & 1 & 1\end{matrix}|$$⇒∣A∣=∣α+β+γ∣∣αα21​ββ21​γγ21​∣

$$\Rightarrow |A|=(\alpha +\beta +\gamma )(\alpha -\beta )(\beta -\gamma )(\gamma -\alpha )$$⇒∣A∣=(α+β+γ)(α−β)(β−γ)(γ−α)

We know $|adjA|={|A|}^{n-1}$∣adjA∣=∣A∣n−1

$$\Rightarrow |adj(adjA)|={|A|}^{{(n-1)}^{2}}$$⇒∣adj(adjA)∣=∣A∣(n−1)2

$$\Rightarrow |adj(adj(adj(adjA)))|={|A|}^{{(n-1)}^{4}}$$⇒∣adj(adj(adj(adjA)))∣=∣A∣(n−1)4

Here $|adj(adj(adj(adjA)))|={|A|}^{{2}^{4}}={|A|}^{16}$∣adj(adj(adj(adjA)))∣=∣A∣24=∣A∣16

Given $\frac{|(adj(adj(\mathrm{adj}(adjA)))|}{{(\alpha -\beta )}^{16}{(\beta -\gamma )}^{16}{(\gamma -\alpha )}^{16}}={2}^{32}\times {3}^{16}$(α−β)16(β−γ)16(γ−α)16∣(adj(adj(adj(adjA)))∣​=232×316

$$\Rightarrow \frac{{(\alpha +\beta +\gamma )}^{16}{(\alpha -\beta )}^{16}{(\beta -\gamma )}^{16}{(\gamma -\alpha )}^{16}}{{(\alpha -\beta )}^{16}{(\beta -\gamma )}^{16}{(\gamma -\alpha )}^{16}}={2}^{32}\times {3}^{16}$$⇒(α−β)16(β−γ)16(γ−α)16(α+β+γ)16(α−β)16(β−γ)16(γ−α)16​=232×316

$$\therefore {(\alpha +\beta +\gamma )}^{16}={2}^{32}\cdot {3}^{16}$$∴(α+β+γ)16=232⋅316

$$\Rightarrow {(\alpha +\beta +\gamma )}^{16}={(12)}^{16}$$⇒(α+β+γ)16=(12)16

$$\Rightarrow \alpha +\beta +\gamma =12$$⇒α+β+γ=12

$$\because \alpha ,\beta ,\gamma \in N$$∵α,β,γ∈N

$$(\alpha -1)+(\beta -1)+(\gamma -1)=9$$(α−1)+(β−1)+(γ−1)=9

Possible number all tuples $(\alpha ,\beta ,\gamma )$(α,β,γ) will be $C211=55$C211=55

$1$1 case for $\alpha =\beta =\gamma$α=β=γ and $12$12 case when any two of these are equal are also included here but $\alpha \neq \beta \neq \gamma$α=β=γ

Hence, number of distinct tuples $(\alpha ,\beta ,\gamma )$(α,β,γ)

$$=55-13=42$$=55−13=42