Question

If ∆₁ and ∆₂ be the determinant.

$\left| \begin{matrix} 1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \end{matrix} \right|$and $\left| \begin{matrix} 1 & 1 & \omega \\ 1 & 1 & \omega^{2} \\ \omega^{2} & \omega & 1 \end{matrix} \right|$.

ω, ω² are imaginary cube root of unity. Then $\frac{\Delta_{1}}{\Delta_{2}}$=

• A. 3$\sqrt{3}$I
• B. $\sqrt{3}$I
• C. –$\sqrt{3}$I
• D. None on these

Answer 1

Answer: (B)

$\sqrt{3}$I

Answer 2

Applying R₁ + R₂­ + R₃ Putting 1 + ω + ω² = 0

∆₁ = 3(ω² – ω) = 3$\left\lbrack \frac{- 1 - i\sqrt{3}}{2} - \frac{- 1 + i\sqrt{3}}{2} \right\rbrack$= $3\sqrt{3}$i

Applying R₁ – R₂ making two zero

∆₂ = (ω – ω²)² = ω² + ω⁴ – 2ω³ = ω² + ω + 1 – 3 = –3 ⇒ $\frac{\Delta_{1}}{\Delta_{2}}$ = $\sqrt{3}$i