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If A * adj(A²) = [[1, 0, 2], [2, 1, 0], [1, 0, 1]], find the absolute value of the sum of all elements of adj(A), where adj(X) denotes the adjoint (classical adjugate) of matrix X.

1. 6
2. 7
3. 8
4. 9

Correct answer: 8

Solution

Let M = [[1,0,2],[2,1,0],[1,0,1]]. Using adj(A²) = (adj A)² and adj(A) = det(A) * A^(-1): A*(adj A)² = A * det(A)² * (A^(-1))² = det(A)² * A^(-1) = det(A) * (det(A) * A^(-1)) = det(A) * adj(A). So det(A) * adj(A) = M. Take determinants: det(A) * det(adj A) = det(M). For 3x3: det(adj A) = det(A)², so det(A)³ = det(M). Also det(A) * adj(M)... wait: det(det(A) * adj(A)) = det(A)³ * det(adj A)/det(A)... let me redo: det(M) = det(det(A) * adj(A)) = det(A)³ * det(adj A). Since det(adj A) = (det A)² for 3x3: det(M) = det(A)³ * det(A)² = det(A)⁵. Compute det(M): expanding along row 1: 1*(1*1 - 0*0) - 0 + 2*(2*0 - 1*1) = 1 - 2 = -1. So det(A)⁵ = -1, giving det(A) = -1. Then adj(A) = M / det(A) = M / (-1) = -M = [[-1,0,-2],[-2,-1,0],[-1,0,-1]]. Sum of elements = -1+0-2-2-1+0-1+0-1 = -8. Absolute value = 8.

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