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NDA 2024
Exam
NDA
Year
2024
Subject
Mathematics
Chapter
Complex Number

NDA 2024 · Mathematics · Complex Number

If ω=-(1)/(2)+i 3 2 then what is | array ccc 1+ω & 1+ω^2 & ω+ω^2 1...

If ω=−12+i32\omega=-\frac{1}{2}+i \frac{\sqrt{3}}{2} then what is \left|1+ω1+ω2ω+ω21ωω21ω1ω21\begin{array}{ccc} 1+\omega & 1+\omega^2 & \omega+\omega^2 \\ 1 & \omega & \omega^2 \\ \frac{1}{\omega} & \frac{1}{\omega^2} & 1 \end{array}\right| \text{ equal to? }

Options

  1. A00
  2. Bω\omega
  3. Cω2\omega^2
  4. D1−ω21-\omega^2

Correct answer

A. 00

Step-by-step solution

Suppose \mathrm{A}=\left|1+ω1+ω2ω+ω21ωω21ω1ω21\begin{array}{ccc}1+\omega & 1+\omega^2 & \omega+\omega^2 \\ 1 & \omega & \omega^2 \\ \frac{1}{\omega} & \frac{1}{\omega^2} & 1\end{array}\right| Now C1→C1+C2+C3\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2+\mathrm{C}_3, we get ⇒A=∣2(1+ω+ω2)1+ω2ω+ω2(1+ω+ω2)ωω2(1+ω+ω2)ω21ω21∣⇒A=∣2(0)1+ω2ω+ω20ωω201ω21∣=0\begin{aligned} & \Rightarrow \mathrm{A}=\left|\begin{array}{ccc} 2\left(1+\omega+\omega^2\right) & 1+\omega^2 & \omega+\omega^2 \\ \left(1+\omega+\omega^2\right) & \omega & \omega^2 \\ \frac{\left(1+\omega+\omega^2\right)}{\omega^2} & \frac{1}{\omega^2} & 1 \end{array}\right| \\ & \Rightarrow \mathrm{A}=\left|\begin{array}{ccc} 2(0) & 1+\omega^2 & \omega+\omega^2 \\ 0 & \omega & \omega^2 \\ 0 & \frac{1}{\omega^2} & 1 \end{array}\right|=0 \end{aligned}\left\{\because\right.\text{ all }\text{ element }\text{ of } \mathrm{C}_1\text{ is }\text{ zero } \}$.
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