Question

\(\begin{array}{l} \text{Consider a matrix}~A=\begin{bmatrix}\alpha & \beta & \gamma \\\alpha^2 & \beta^2 & \gamma^2 \\\beta+\gamma & \gamma+\alpha & \alpha+\beta \\\end{bmatrix}\end{array}\)

where \(α, β, γ\) are three distinct natural numbers.
If \(\begin{array}{l}\frac{\text{det(adj(adj(adj(adj A))))}}{\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}}=2^{32}\times3^{16},\end{array}\)
then the number of such 3-tuples \((α, β, γ)\) is _________.

Answer 1

Correct Answer: 42

\(\begin{array}{l} \text{det}\left(A\right)=\begin{vmatrix}\alpha & \beta & \gamma \\\alpha^2 & \beta^2 & \gamma^2 \\\beta+\gamma & \gamma+\alpha & \alpha+\beta \\\end{vmatrix}\end{array}\)

\(\begin{array}{l} R_3 \rightarrow R_3 + R_1\end{array}\)

\(\begin{array}{l} \Rightarrow \left(\alpha+\beta+\gamma\right)\begin{vmatrix}\alpha & \beta & \gamma \\\alpha^2 & \beta^2 & \gamma^2 \\1 & 1 & 1 \\\end{vmatrix}\end{array}\)

\(\begin{array}{l}\therefore det \left(A\right) = \left(\alpha + \beta + \gamma\right) \left(\alpha – \beta\right) \left(\beta – \gamma\right) \left(\gamma – \alpha\right)\end{array}\)
Also, det (adj (adj (adj (adj (A)))))
\(\begin{array}{l} =\left(\text{det}\left(A\right)\right)^{2^4}=\left(\text{det}\left(A\right)\right)^{16}\end{array}\)
\(\begin{array}{l} \therefore\ \frac{\left(\alpha+\beta+\gamma\right)^{16}\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}}{\left(\alpha-\beta\right)^{16}\left(\beta-\gamma\right)^{16}\left(\gamma-\alpha\right)^{16}}=\left(4.3\right)^{16}\end{array}\)
\(\begin{array}{l}\Rightarrow \alpha + \beta + \gamma = 12 \end{array}\)
\(\begin{array}{l}\Rightarrow \left(\alpha, \beta, \gamma\right) \text{distinct natural triplets}\end{array}\)
\(\begin{array}{l}= ^{11}C_2 – 1 – ^3C_2 \left(4\right) = 55 – 1 – 12\\ = 42 \end{array}\)