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Question: At what value of \(x,\) will\(\left| \begin{matrix} x + \omega^{2} & \omega & 1 \\ \omega & \omega^{...

At what value of x,x, will∣x+ω2ω1ωω21+x1x+ωω2∣=0\left| \begin{matrix} x + \omega^{2} & \omega & 1 \\ \omega & \omega^{2} & 1 + x \\ 1 & x + \omega & \omega^{2} \end{matrix} \right| = 0.

A

x=0x = 0

B

x=1x = 1

C

x=−1x = - 1

D

None of these

Answer

x=0x = 0

Explanation

Solution

∣x+ω2ω1ωω21+x1x+ωω2∣\left| \begin{matrix} x + \omega^{2} & \omega & 1 \\ \omega & \omega^{2} & 1 + x \\ 1 & x + \omega & \omega^{2} \end{matrix} \right| = 0

Check at x=0,x = 0, we get ∣ω2ω1ωω211ωω2∣\left| \begin{matrix} \omega^{2} & \omega & 1 \\ \omega & \omega^{2} & 1 \\ 1 & \omega & \omega^{2} \end{matrix} \right|

= ω2(ω4−ω)−ω(ω3−1)+1(ω2−ω2)\omega^{2}(\omega^{4} - \omega) - \omega(\omega^{3} - 1) + 1(\omega^{2} - \omega^{2})

= ω2(ω−ω)−ω(1−1)+0=0\omega^{2}(\omega - \omega) - \omega(1 - 1) + 0 = 0 Or

1 + \omega + \omega^{2} + x & \omega & 1 \\ 1 + \omega + \omega^{2} + x & \omega^{2} & 1 + x \\ 1 + \omega + \omega^{2} + x & x + \omega & \omega^{2} \end{matrix} \right|$$ by $C_{1} \rightarrow C_{1} + C_{2} + C_{3}$ $= \left| \begin{matrix} x & \omega & 1 \\ x & \omega^{2} & 1 + x \\ x & x + \omega & \omega^{2} \end{matrix} \right|$, ($\because 1 + \omega + \omega^{2} = 0$) = 0, if $x = 0$.