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Let A be a 3x3 real matrix with |A| = 2 and A² * adj(A) = [[2, 2, 0], [0, 2, 2], [k, 0, 2]]. Which of the following statements is/are correct?
- A³ = 2I
- |kA| = 16
- A² = 2A
- trace of A³ = 6
Correct answer: |kA| = 16
Solution
Since A*adj(A) = |A|*I = 2I, we get adj(A) = 2*A^(-1). So A²*adj(A) = A²*(2A^(-1)) = 2A. Thus 2A = [[2,2,0],[0,2,2],[k,0,2]], giving A = [[1,1,0],[0,1,1],[k/2,0,1]]. For |A| = 2: expanding, det(A) = 1*(1-0) - 1*(0-k/2) + 0 = 1 + k/2 = 2 => k = 2. So A = [[1,1,0],[0,1,1],[1,0,1]]. Computing A³ gives [[2,3,3],[3,2,3],[3,3,2]], so A³ ≠ 2I (option A false). trace(A³) = 2+2+2 = 6 (option D true). A² = [[1,2,1],[1,1,2],[2,1,1]] ≠ 2A (option C false). |kA| = |2A| = 2³*|A| = 8*2 = 16 (option B true).
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