Let = - 1 2 + i 3 2 . Then the value of the determinant | 1 1 1 1 - 1 - 2 2 1 2 4 | is , Determinant [2002]

Question

Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2} .$ Then the value of the determinant $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ is [2002]

Smart Achievers · Accepted Answer

(2)

Given that $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2},$ then $ω^{2} = - \frac{1}{2} - i \frac{\sqrt{3}}{2} .$

Also, $1 + ω + ω^{2} = 0$ and $ω^{3} = 1$

Now, $Δ =$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ $=$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $(∵ ω = - 1 - ω^{2} and ω^{3} = 1)$ $[C_{1} \to C_{1} + C_{2} + C_{3}]$

$Δ =$ $| \begin{matrix} 3 & 1 & 1 \\ 0 & ω & ω^{2} \\ 0 & ω^{2} & ω \end{matrix} |$ $(∵ 1 + ω + ω^{2} = 0)$

On expanding along $C_{1}$ , we get

$Δ = 3 (ω^{2} - ω^{4}) = 3 (ω^{2} - ω) = 3 ω (ω - 1)$

Solution

Q.
Let $ω = - \frac{1}{2} + i \frac{\sqrt{3}}{2} .$ Then the value of the determinant $| \begin{matrix} 1 & 1 & 1 \\ 1 & - 1 - ω^{2} & ω^{2} \\ 1 & ω^{2} & ω^{4} \end{matrix} |$ is [2002]

1 $3 ω$

2 $3 ω (ω - 1)$

3 $3 ω^{2}$

4 $3 ω (1 - ω)$

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Solution

Q. Let ω=-12+i32. Then the value of the determinant |1111-1-ω2ω21ω2ω4| is [2002]

1 3ω

2 3ω(ω-1)

3 3ω2

4 3ω(1-ω)