Consider a matrix A=[α, β, γ; α², β², γ²; β+γ, γ+α, α+β], where α, β, γ are three distinct natural numbers.
If det(adj(adj(adj(adj A))))/(α-β)¹⁶(β-γ)¹⁶(γ-α)¹⁶=2³² × 3¹⁶, then the number of such 3 - tuples (α, β, γ) is ____________.

Solution

Maths sir · 5 steps

Add row 1 to row 3 to pull out α+β+γ. Then each adj just squares the determinant again.

1. Simplify det A. Do R₃ → R₃ + R₁. Every entry of row 3 becomes α+β+γ, so take it out as a common factor and row 3 becomes 1, 1, 1. What remains is a Vandermonde determinant, which equals (α-β)(β-γ)(γ-α).

   det A = (α+β+γ)(α-β)(β-γ)(γ-α)

2. Repeated adjoint. For a 3 × 3 matrix, det(adj M) = (det M)². Taking adj four times squares the determinant four times, so the numerator is (det A)^(2⁴) = (det A)¹⁶.
3. Find the sum. Dividing by the three difference factors leaves (α+β+γ)¹⁶ = 2³² × 3¹⁶ = 12¹⁶. The numbers are natural, so their sum is positive, which gives α+β+γ = 12.
4. Count tuples. The number of ordered triples of natural numbers with sum 12 is C(11, 2) = 55. Remove (4,4,4). Also remove the triples with exactly two numbers equal: 2a + c = 12 with a ∈ {1, 2, 3, 5} gives 4 choices, and each fits 3 positions, so 12 triples.
5. So. The number of triples is 55 - 1 - 12 = 42.

The common slip: 7. Counting unordered sets of three distinct numbers, such as {1, 2, 9}, gives 7. The question asks for ordered 3-tuples, so each set counts 3! = 6 times.

The idea to keep. For an n × n matrix, applying adj k times raises det A to the power (n-1)ᵏ. A row like β+γ is a hint to add it to the row α, β, γ and get the full sum.

Answer from the NTA final answer key. Solution by PrepWiser.