Question

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

• A. z
• B. -1
• C. 1
• D. -z

Answer 1

Answer: (D)

-z

Answer 2

$2\omega + 1 = z,\, z = \sqrt{3}i$ $\omega = \frac{-1+\sqrt{3}i}{2}\to$ Cube root of unity. $C_{1} \to C_{1}+ C_{2}+ C_{3}$ $\begin{vmatrix}1&1&1\\\ 1&-1-\omega^{2}&\omega^{2}\\\ 1&\omega^{2}&\omega^{7}\end{vmatrix} = \begin{vmatrix}1&1&1\\\ 1&\omega&\omega^{2}\\\ 1&\omega^{2}&\omega\end{vmatrix} = \begin{vmatrix}3&1&1\\\ 0&\omega&\omega^{2}\\\ 0&\omega^{2}&\omega\end{vmatrix}$ $= 3\left(\omega^{2}-\omega^{4}\right)$ $= 3\left[\left(\frac{-1-\sqrt{3}i}{2}\right)-\left(\frac{-1+\sqrt{3}i}{2}\right)\right]$ $= -3\sqrt{3}i$ $= -3z$ $\therefore k = -z$