If ω is a non-real cube root of unity, then what is a root of the following equation? |x+1 ω ω²; ω x+ω² 1; ω² 1 x+ω| = 0
A.x = 0✓ Correct
B.x = 1
C.x = ω
D.x = ω²
Explanation
Apply C1 → C1 + C2 + C3. The first column becomes: (x+1+ω+ω², ω+x+ω²+1, ω²+1+x+ω) = (x, x, x) since 1+ω+ω² = 0. Factor out x: the determinant = x·|1 ω ω²; 1 x+ω² 1; 1 1 x+ω|. So x = 0 is a root.
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