If the system of equations x + ay + z = 0, ax + y + z = 0, x + y + az = 0 has a non-trivial solution, then the value of a^3 + 3a is:
Correct: C
For a homogeneous system to have a non-trivial solution, the determinant of the coefficient matrix must be zero. The matrix is [[1,a,1],[a,1,1],[1,1,a]]. Computing the determinant: |A| = 1(1·a - 1·1) - a(a·a - 1·1) + 1(a·1 - 1·1) = (a - 1) - a(a^2 - 1) + (a - 1) = 2(a - 1) - a^3 + a = -a^3 + 3a - 2. Setting |A| = 0: -a^3 + 3a - 2 = 0 => a^3 - 3a + 2 = 0 => (a-1)^2(a+2) = 0. So a = 1 or a = -2. If a = 1, the matrix becomes all ones, rank 1, non-trivial solutions exist. If a = -2, rank is 2, still non-trivial solutions. For a = 1: a^3 + 3a = 1 + 3 = 4. For a = -2: a^3 + 3a = -8 - 6 = -14. Since the problem asks for 'the value', it likely refers to a = 1 (the repeated root), giving 4. So the answer is 4.