Question

Let $\mathrm{\omega }$ be a complex number such that $2\mathrm{\omega }+1=\mathrm{z}$ where $\mathrm{z}=\sqrt{-3}$. If $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\mathrm{\omega }}^2-1& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& {\mathrm{\omega }}^7\end{array}\right|=3\mathrm{k}$. Then $\mathrm{k}$ is equal to,

• A. $\mathrm{z}$
• B. $-1$
• C. $1$
• D. $-\mathrm{z}$

Answer 1

The correct option is D

$-\mathrm{z}$

Determine the value of $k$

We have, $\left|\begin{array}{ccc}1& 1& 1\\ 1& -{\mathrm{\omega }}^2-1& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& {\mathrm{\omega }}^7\end{array}\right|=3\mathrm{k}$, and $2\mathrm{\omega }+1=\mathrm{z}$

$$\begin{gathered} \Rightarrow 2\mathrm{\omega }+1=\sqrt{-3}[\because z=\sqrt{-3}] \\ \Rightarrow \mathrm{\omega }=\frac{-1+\sqrt{-3}}{2} \end{gathered}$$

Since, $\mathrm{\omega }$ is the cube root of unity, therefore, ${\mathrm{\omega }}^{3\mathrm{n}}=1$ and also,

${\mathrm{\omega }}^2=\frac{-1-\sqrt{3\mathrm{i}}}{2}$, Now we have,

$$\begin{gathered} \left|\begin{array}{ccc}1& 1& 1\\ 1& -{\mathrm{\omega }}^2-1& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& {\mathrm{\omega }}^7\end{array}\right|=3\mathrm{k} \\ \Rightarrow \left|\begin{array}{ccc}1& 1& 1\\ 1& \mathrm{\omega }& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& \mathrm{\omega }\end{array}\right|=3\mathrm{k}\because \left(1+\mathrm{\omega }+{\mathrm{\omega }}^2=0,{\left({\mathrm{\omega }}^3\right)}^2\mathrm{\omega }=\mathrm{\omega }\right) \end{gathered}$$

Apply ${\mathrm{R}}_1\to {\mathrm{R}}_1+{\mathrm{R}}_2+{\mathrm{R}}_3$ we get,

$$\begin{gathered} \left|\begin{array}{ccc}3& 1+\mathrm{\omega }+{\mathrm{\omega }}^2& 1+\mathrm{\omega }+{\mathrm{\omega }}^2\\ 1& \mathrm{\omega }& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& \mathrm{\omega }\end{array}\right|=3\mathrm{k} \\ \Rightarrow \left|\begin{array}{ccc}3& 0& 0\\ 1& \mathrm{\omega }& {\mathrm{\omega }}^2\\ 1& {\mathrm{\omega }}^2& \mathrm{\omega }\end{array}\right|=3\mathrm{k} \\ \Rightarrow 3\left({\mathrm{\omega }}^2-{\mathrm{\omega }}^4\right)=3\mathrm{k} \\ \Rightarrow \left(\frac{-1-\sqrt{3\mathrm{i}}}{2}\right)-\left(\frac{-1+\sqrt{3\mathrm{i}}}{2}\right)=\mathrm{k} \\ \Rightarrow -\sqrt{3\mathrm{i}}=\mathrm{k} \\ \Rightarrow -\mathrm{z}=\mathrm{k} \end{gathered}$$

Hence, option D is the correct answer.