Matrices & Determinants
Matrices and Determinants
Allen Star Batch
Grade 12
Question:
If $1, \omega, \omega^2$ are cube roots of unity then system of equations: $x + 2\omega y + 3\omega^2 z = 1 - \omega^2$, $2x + 3\omega y + \omega^2 z = \omega^2 - \omega$, $3x + \omega y + 2\omega^2 z = \omega - 1$ has
unique solution
infinitely many solutions
no solution
exactly 3 different solutions
Step-by-Step Solution
Key Concept: For a system of linear equations AX = B, existence and uniqueness of solutions depend on the determinant of coefficient matrix A. If det(A) ≠ 0, the system has a unique solution by Cramer's rule. Here, properties of cube roots of unity (1 + ω + ω² = 0 and ω³ = 1) must be used to evaluate the 3×3 determinant.
Evaluate the determinant $\begin{vmatrix} 1 & 2\omega & 3\omega^2 \\ 2 & 3\omega & \omega^2 \\ 3 & \omega & 2\omega^2 \end{vmatrix}$ by expanding along the first row. Taking $R_1 - 2R_2 + 3R_3$ gives the computation $5 - 2(4-3) + 3(2-9) = 5 - 2 - 21 = -18$, using properties of cube roots of unity where $1 + \omega + \omega^2 = 0$.
Correct Answer: 1