Question

If w is cube root of unity, then

D= $\left| \begin{matrix} x + 1 & \omega & \omega^{2} \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix} \right|$ = ........

• A. x³ + 1
• B. x³ + w
• C. x³ + w²
• D. x³

Answer 1

Answer: (D)

x³

Answer 2

R₁ ® R₁ + R₂ + R₃

D =$\left| \mspace{6mu}\begin{matrix} x & x & x \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix}\mspace{6mu} \right|$Q (1 + w + w² = 0)

D = x $\left| \mspace{6mu}\begin{matrix} 1 & 1 & 1 \\ \omega & x + \omega^{2} & 1 \\ \omega^{2} & 1 & x + \omega \end{matrix}\mspace{6mu} \right|$

D = x$\left| \mspace{6mu}\begin{matrix} 1 & 0 & 0 \\ \omega & x + \omega^{2}–\omega & 1 - x - \omega^{2} \\ \omega^{2} & 1 - \omega^{2} & x + \omega - 1 \end{matrix}\mspace{6mu} \right|$

D = x[(x+w² – w) (x +w –1) – (1–w²)(1– x– w²)]

D = x [x²] = x³