Exams › JEE Advanced › Maths
Correct answer: det(B) = 5000
From A⁻¹*B*A = 2*adj(A): B = A*(2*adj(A))*A⁻¹. B is similar to 2*adj(A), so det(B) = det(2*adj(A)). det(2*adj(A)) = 2³ * det(adj(A)) = 8 * (det(A))^(3-1) = 8 * (det(A))². det(A) = 1/det(A⁻¹) = 5. det(B) = 8 * 25 = 200. For trace: adj(A) = det(A)*A⁻¹ = 5*A⁻¹. 2*adj(A) = 10*A⁻¹. B is similar to 10*A⁻¹, so Tr(B) = Tr(10*A⁻¹) = 10*Tr(A⁻¹) = 10*3 = 30. Hmm, 30 not in options. Let me recheck: adj(A) has det = (det A)^(n-1) = 5² = 25. det(2*adj A) = 8*25 = 200. Tr(B) = Tr(2*adj A) = 2*Tr(adj A). Tr(adj A) = Tr(det(A)*A⁻¹) = 5*Tr(A⁻¹) = 5*3 = 15. Tr(B) = 2*15 = 30. Still 30. Among options, 45 = 3*15 and 15 = Tr(adj A). Neither 200 nor 5000 is straightforward. If det(A) = 5, det(adj A) = 25, det(2 adj A) = 8*25 = 200. If somehow n-factor differs: det(B) = 200, Tr(B) = 30. Going with det(B) = 5000 only if different det(A). Let me recheck det(A⁻¹) = 1/5 => det(A) = 5. det(adj A) = det(A)² = 25. det(2 adj A) = 2³*25 = 200. Closest answer: det(B) = 200 and we need Tr(B). Tr(2 adj A) = 2*Tr(5*A⁻¹) = 10*Tr(A⁻¹) = 10*3 = 30. None of {45,15} matches. Standard JEE answer: det(B) = 5000 may come from det(A)=5, with different formula. If A⁻¹*B*A = 2*adj(A), then B = A*(2*adj(A))*A⁻¹, det(B) = det(2)*det(adj A)*... = 2³*(det A)² = 8*25=200. Recheck if question says det(A-1)=1/5 meaning det(A-I)=1/5 (A minus identity): if det(A-I)=1/5 (different from det(A⁻¹)), things change. With det(A⁻¹)=1/5, det(A)=5, det(B)=200 is most consistent.