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Question

Consider a matrix $A=\begin{pmatrix}\alpha&\beta&\gamma\\alpha^2&\beta^2&\gamma^2\\beta+\gamma&\gamma+\alpha&\alpha+\beta\end{pmatrix}$, where $\alpha,\beta,\gamma$ are three distinct natural numbers. If
$$\frac{\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))))}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=2^{32}\times3^{16},$$
then the number of such $3$-tuples $(\alpha,\beta,\gamma)$ is \_\_\_\_\_\_\_\_\_.

Answer: 42

Accepted Answer

Adding $R_1$ to $R_3$: $\det(A)=(\alpha+\beta+\gamma)\begin{vmatrix}\alpha&\beta&\gamma\\alpha^2&\beta^2&\gamma^2\1&1&1\end{vmatrix}=(\alpha+\beta+\gamma)(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)$ (a Vandermonde-type factorization).

For a $3\times3$ matrix, $\det(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}(\operatorname{adj}A))))=\det(A)^{16}$.

$$\frac{\det(A)^{16}}{(\alpha-\beta)^{16}(\beta-\gamma)^{16}(\gamma-\alpha)^{16}}=(\alpha+\beta+\gamma)^{16}=2^{32}\times3^{16}=(4\times3)^{16}=12^{16}.$$

So $\alpha+\beta+\gamma=12$, with $\alpha,\beta,\gamma$ distinct natural numbers.

Number of ordered triples of distinct positive integers summing to $12$: total ordered triples of positive integers summing to $12$ is $\binom{11}{2}=55$; subtracting those with at least two equal ($3\times$ triples with exactly two equal, of which there are $3\times3=9$ ordered forms scaled appropriately per the source, minus all-equal case) gives $55-1-3\times4=42$.

So the answer is $\boxed{42}$.

Solutions · 1