Let α ∈(0, ∞) and A=[1, 2, α; 1, 0, 1; 0, 1, 2]. If det(adj(2 A-Aᵀ) · adj(A-2 Aᵀ))=2⁸, then (det(A))² is equal to:

Solution

Maths sir · 4 steps

Spot that (A-2Aᵀ)ᵀ = -(2A-Aᵀ), so both determinants are tied to one number.

1. Name the matrices. Let B = 2A - Aᵀ and C = A - 2Aᵀ. For a 3 × 3 matrix, det(adj M) = (det M)², so the given condition becomes (det B)²(det C)² = 2⁸.
2. Link C to B. Cᵀ = Aᵀ - 2A = -B. A 3 × 3 matrix flips the sign of its determinant when negated, so det C = det Cᵀ = -det B. That gives (det B)⁴ = 2⁸, so |det B| = 4.

   B = (1, 3, 2α; 0, 0, 1; -α, 1, 2), det B = 1(0-1) - 3(0+α) + 0 = -1-3α

3. Find alpha. Since α > 0, |det B| = 1 + 3α = 4, so α = 1.
4. So. det A = 1(0-1) - 2(2-0) + α(1-0) = α - 5 = -4. Hence (det A)² = 16, option (a).

The idea to keep. For an n × n matrix, det(adj M) = (det M)ⁿ⁻¹. When two matrices are built from A and Aᵀ, check whether one is the transpose of the other, up to a sign.

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