Let α ∈ (0, ∞) and A = <table><tr><td>1</td><td>2</td><td>α</td></tr><tr><td>1</td><td>0</td><td>1</td></tr><tr><td>0</td><td>1</td><td>2</td></tr></table>. If det(adj(2A - Aᵀ).adj(A - 2Aᵀ)) = 2⁸, then (det(A))² is equal to:
Step-by-Step Solution
Key Concept: Use properties of determinants and adjoints: det(adj(M)) = (det(M))^(n-1) where n=3, so det(adj(M)) = (det(M))^2. Also det(AB) = det(A)det(B).
Let M = 2A - Aᵀ and N = A - 2Aᵀ. Note that N = -(2Aᵀ - A) = -(2A - Aᵀ)ᵀ = -Mᵀ. So det(N) = det(-Mᵀ) = (-1)^3 det(M) = -det(M). The given equation is det(adj(M)adj(N)) = 2^8. This is det(adj(M))det(adj(N)) = 2^8. Since det(adj(M)) = (det(M))^2, we have (det(M))^2(det(N))^2 = 2^8. Substituting det(N) = -det(M), we get (det(M))^2(-det(M))^2 = 2^8, so (det(M))^4 = 2^8, which means (det(M))^2 = 2^4 = 16. Now, A = [[1, 2, \alpha], [1, 0, 1], [0, 1, 2]]. det(A) = 1(0-1) - 2(2-0) + \alpha(1-0) = -1 - 4 + \alpha = \alpha - 5. Aᵀ = [[1, 1, 0], [2, 0, 1], [\alpha, 1, 2]]. 2A - Aᵀ = [[2-1, 4-1, 2\alpha-0], [2-2, 0-0, 2-1], [0-\alpha, 2-1, 4-2]] = [[1, 3, 2\alpha], [0, 0, 1], [-\alpha, 1, 2]]. det(2A - Aᵀ) = 1(0-1) - 3(0+\alpha) + 2\alpha(0-0) = -1 - 3\alpha. We have (det(2A - Aᵀ))^2 = 16, so (-1 - 3\alpha)^2 = 16. Since \alpha > 0, 1 + 3\alpha = 4, so 3\alpha = 3, \alpha = 1. Then det(A) = \alpha - 5 = 1 - 5 = -4. Thus (det(A))^2 = (-4)^2 = 16.
Correct Answer: 3