SATHEE: Complex Numbers 5 Question 7

Complex Numbers 5 Question 7

7. Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ , then value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 1 & - 1 - ω^{2} & ω^{2} 1 & ω^{2} & ω \end{array} |$ is

(2002, 1M)

(a) $3 ω$

(b) $3 ω (ω - 1)$

(d) $3 ω (1 - ω)$

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Answer:

Correct Answer: 7. (b)

Solution:

$L e t Δ = | \begin{array}{ccc} 1 & 1 & 1 1 & - 1 - ω^{2} & ω^{2} 1 & ω^{2} & ω \end{array} |$

Applying $R_{2} \to R_{2} - R_{1}; R_{3} \to R_{3} - R_{1}$

$\begin{aligned} \begin{array}{llll} 1 & 1 & 1 \end{array} \\ = 0 - 2 - ω^{2} ω^{2} - 1 \\ 0 ω^{2} - 1 ω - 1 \\ = (- 2 - ω^{2}) (ω - 1) - {(ω^{2} - 1)}^{2} \\ = - 2 ω + 2 - ω^{3} + ω^{2} - (ω^{4} - 2 ω^{2} + 1) \\ = 3 ω^{2} - 3 ω = 3 ω (ω - 1) [∵ ω^{4} = ω] \end{aligned}$

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Complex Numbers 5 Question 7

7. Let ω=−12+i32, then value of the determinant |111 1−1−ω2ω2 1ω2ω| is

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7. Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2}$ , then value of the determinant $| \begin{array}{ccc} 1 & 1 & 1 1 & - 1 - ω^{2} & ω^{2} 1 & ω^{2} & ω \end{array} |$ is