Consider a matrix $A=\left[\begin{array}{ccc}\alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \ga

Question

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

Professor PC Thomas Classes · Answer

(42) : We have given\ $A=\left[\begin{array}{lll}\alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ \beta+\gamma & \gamma+\alpha & \alpha+\beta\end{array}ight]$\ Applying $R_{3} ightarrow R_{3}+R_{1}$, we get\ $|A|=\left|\begin{array}{ccc}\alpha & \beta & \gamma \ \alpha^{2} & \beta^{2} & \gamma^{2} \ \alpha+\beta+\gamma & \alpha+\beta+\gamma & \alpha+\beta+\gamma\end{array}ight|$\ $\Rightarrow \quad|A|=(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\alpha-\alpha)$\ $|\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A)))|=|A|^{(n-1)^{4}}=|A|^{16}$\ $	herefore \frac{\operatorname{det}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj} A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}$\ $=\frac{(\alpha+\beta+\gamma)^{16}(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}$\ $\Rightarrow \quad(\alpha+\beta+\gamma)^{16}=(12)^{16}$\ $\Rightarrow \alpha+\beta+\gamma=12$\ Here, $\alpha, \beta, \gamma \in N$\ $	herefore \quad(\alpha-1)+(\beta-1)+(\gamma-1)=9$\ Number of solution is ${ }^{9+3-1} C_{3-1}=55$ ways\ Case I: When $\alpha=\beta=\gamma$. There is only one way.\ Case II : When two of them are equal then there are total 12 cases ways.\ $	herefore \quad$ Required $3-$ tuples $=55-1-12=42$\