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KCET · Maths · Complex Number

If \(\omega\) is an imaginary cube root of unity, then the value of \(\left[\begin{array}{ccc}1 & \omega^{2} & 1-\omega^{4} \\ \omega & 1 & 1+\omega^{5} \\ 1 & \omega & \omega^{2}\end{array}\right]\) is

  1. A \(-4\)
  2. B \(\omega^{2}-4\)
  3. C \(\omega^{2}\)
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(B) \(\omega^{2}-4\)

Step-by-step Solution

Detailed explanation

Given, \(\omega^{3}=1\) and \(1+\omega+\omega^{2}=0\)
\[
=\left|\begin{array}{ccc}
1 & \omega^{2} & 1-\omega^{4} \\
\omega & 1 & 1+\omega^{5} \\
1 & \omega & \omega^{2}
\end{array}\right|=\left|\begin{array}{ccc}
1 & \omega^{2} & 1-\omega \\
\omega & 1 & 1+\omega^{2} \\
1 & \omega & \omega^{2}
\end{array}\right|
\]
Expand with respect to \(R_{1}\)
\[
\begin{aligned}
&=\left(\omega^{2}-\omega-\omega^{3}\right)-\omega^{2}\left(\omega^{3}-1-\omega^{2}\right) \\
&\quad+(1-\omega)\left(\omega^{2}-1\right) \\
&=\left(\omega^{2}-\omega-1\right)-\omega^{2}\left(1-1-\omega^{2}\right) \\
&\quad+(1-\omega)\left(\omega^{2}-1\right) \\
&=\omega^{2}-\omega-1+\omega^{4}+\omega^{2}-1-\omega^{3}+\omega \\
&=2 \omega^{2}-2+\omega-1=2 \omega^{2}+\omega-3 \\
&=\omega^{2}+\left(\omega+\omega^{2}\right)-3 \\
&=\omega^{2}-1-3=\omega^{2}-4
\end{aligned}
\]