Let A be a 3 3 invertible matrix such that |A|=2 , trace (A)=3 , and trace (A^2)=5 . If B = adj ( adj (2A⁻¹)) , then the value of |B| + trace (B) equals :

Step-by-step solution

Let the eigenvalues of A be ₁, ₂, ₃ . We are given: trace (A) = ₁ + ₂ + ₃ = 3 trace (A^2) = ₁^2 + ₂^2 + ₃^2 = 5 Using the identity ( ₁ + ₂ + ₃)^2 = ( ₁^2 + ₂^2 + ₃^2) + 2( ₁ ₂ + ₂ ₃ + ₃ ₁) : (3)^2 = 5 + 2( ₁ ₂ + ₂ ₃ + ₃ ₁) 9 = 5 + 2( ₁ ₂ + ₂ ₃ + ₃ ₁) ₁ ₂ + ₂ ₃ + ₃ ₁ = 2 The trace of A⁻¹ is the sum of its eigenvalues: trace (A⁻¹) = 1 ₁ + 1 ₂ + 1 ₃ = ₁ ₂ + ₂ ₃ + ₃ ₁ ₁ ₂ ₃ Since ₁ ₂ ₃ = |A| = 2 , we get: trace (A⁻¹) = 2 2 = 1 Now, consider the matrix B = adj ( adj (2A⁻¹)) . For any 3 3 matrix X , adj ( adj (X)) = |X|³