Let be a complex number such that where If then is equal to

Question

Let $ω$ be a complex number such that $2 ω + 1 = z$ , where $z = \sqrt{- 3}$ . If $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$ , then $3 k$ is equal to

Smart Achievers · Accepted Answer

(1)

$Given, 2 ω + 1 = z$ $2 ω + 1 = \sqrt{- 3}$ $[∴ z = \sqrt{- 3}] \Rightarrow ω = \frac{- 1 + \sqrt{3} i}{2}$

$Since, ω is cube root of unity.$ $∴$ $ω^{2} = \frac{- 1 - \sqrt{3} i}{2} and ω^{3 n} = 1$

$Now,$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$

$\Rightarrow$ $| \begin{matrix} 1 & 1 & 1 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$[∴ 1 + ω + ω^{2} = 0 and ω^{7} = {(ω^{3})}^{2} \cdot ω = ω]$

$On applying R_{1} \to R_{1} + R_{2} + R_{3}, we get$ $| \begin{matrix} 3 & 1 + ω + ω^{2} & 1 + ω + ω^{2} \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$=$ $| \begin{matrix} 3 & 0 & 0 \\ 1 & ω & ω^{2} \\ 1 & ω^{2} & ω \end{matrix} |$ $= 3 k$

$\Rightarrow 3 (ω^{2} - ω^{4}) = 3 k$

$\Rightarrow (ω^{2} - ω) = k$

$∴$ $k = (\frac{- 1 - \sqrt{3} i}{2}) - (\frac{- 1 + \sqrt{3} i}{2}) = - \sqrt{3} i = - z$

Solution

Q.
Let $ω$ be a complex number such that $2 ω + 1 = z$ , where $z = \sqrt{- 3}$ . If $| \begin{matrix} 1 & 1 & 1 \\ 1 & - ω^{2} - 1 & ω^{2} \\ 1 & ω^{2} & ω^{7} \end{matrix} |$ $= 3 k$ , then $3 k$ is equal to

1 $- 3 z$

2 $3 z$

3 $- 3$

4 $+ 3$

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Solution

Q. Let ω be a complex number such that 2ω+1=z, where z=-3. If |1111-ω2-1ω21ω2ω7|=3k, then 3k is equal to

1 -3z

2 3z

3 -3