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EEJ MAIN Mathematics QUESTION #6038
Question 1

Let $\omega$ be a complex number with $2\omega+1=z$ where $z=\sqrt{-3}$. If $\begin{vmatrix}1&1&1\\1&-\omega^2-1&\omega^2\\1&\omega^2&\omega^7\end{vmatrix}=3k$, find $k$.

  • $-z$
  • $z$✔️
  • $-1$
  • $1$
Correct Answer Explanation

Since $z=\sqrt{-3}=i\sqrt{3}$ and $2\omega+1=i\sqrt{3}$, so $\omega=\frac{-1+i\sqrt{3}}{2}=e^{2\pi i/3}$ (a primitive cube root of unity).

Properties: $\omega^3=1$, $1+\omega+\omega^2=0$, so $-\omega^2-1=\omega$ and $\omega^7=\omega$.

Matrix becomes: $\begin{vmatrix}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{vmatrix}$

This determinant $= 3(\omega-\omega^2)\cdot... $ After evaluation $= 3i\sqrt{3} = 3z$... so $k=z$.