If ω is a cube root of unity, then a root of the equation is $| \begin{matrix} x + 1 & ω & ω^{2} \\ ω & x + ω^{2} & 1 \\ ω^{2} & 1 & x + ω \end{matrix} | = 0$

Question

Question

This question was previously asked in

BPSC Asstt. Prof. ME Held on Nov 2015 (Advt. 22/2014)

Answer 1

Concept:

If ω is the cube root of unity, i.e. ω³ = 1.

then 1 + ω + ω² = 0.

ω⁴ = ω³ω = ω [∵ ω³ =1]

Calculation:

Given:

$| \begin{matrix} x + 1 & ω & ω^{2} \\ ω & x + ω^{2} & 1 \\ ω^{2} & 1 & x + ω \end{matrix} | = 0$

C'₁ = C₁ + C₂

$| \begin{matrix} x + 1 + ω + ω^{2} & ω & ω^{2} \\ x + 1 + ω + ω^{2} & x + ω^{2} & 1 \\ x + 1 + ω + ω^{2} & 1 & x + ω \end{matrix} | = 0$

$| \begin{matrix} x & ω & ω^{2} \\ x & x + ω^{2} & 1 \\ x & 1 & x + ω \end{matrix} | = 0 (∵ 1 + ω + ω^{2} = 0)$

R'₂ = R₂ - R₁ and R'₃ = R₃ - R₁

$| \begin{matrix} x & ω & ω^{2} \\ 0 & x + ω^{2} - ω & 1 - ω^{2} \\ 0 & 1 - ω & x + ω - ω^{2} \end{matrix} | = 0$

Expanding along first column:

∴ x[(x + ω² - ω)(x + ω - ω²) - (1 - ω)(1 - ω²)]

∴ x[x² + ωx - ω²x + ω²x + ω³ - ω⁴ - ωx - ω² + ω³ - 1 + ω² + ω - ω³]

∴ x³ = 0 (∵ ω³ = 1 and ω⁴ = ω³ω ⇒ ω)

∴ x = 0.