Question

tomorrow is my jee
someone help​

Answer 1

The complex number $\omega$ is given as,

$\longrightarrow 2\omega+1=z\quad\quad\dots(i)$

where $z=\sqrt{-3}.$

$\longrightarrow 2\omega+1=\sqrt{-3}$

Squaring both sides,

$\longrightarrow (2\omega+1)^2=(\sqrt{-3})^2$

$\longrightarrow 4\omega^2+4\omega+1=-3$

$\longrightarrow 4\omega^2+4\omega+4=0$

$\longrightarrow \omega^2+\omega+1=0\quad\quad\dots(1)$

Multiplying $\omega-1$ to both sides,

$\longrightarrow(\omega^2+\omega+1)(\omega-1)=0$

$\longrightarrow\omega^3-1=0\quad\quad[\because\,a^3-b^3=(a-b)(a^2+ab+b)]$

$\longrightarrow\omega^3=1\quad\quad\dots(2)$

Then,

$\longrightarrow\left|\begin{array}{ccc}1&1&1\\1&-\omega^2-1&\omega^2\\1&\omega^2&\omega^7\end{array}\right|=3k$

$\longrightarrow\left|\begin{array}{ccc}1&1&1\\1&-(\omega^2+1)&\omega^2\\1&\omega^2&(\omega^3)^2\cdot\omega\end{array}\right|=3k$

From (1) and (2),

$\longrightarrow\left|\begin{array}{ccc}1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega\end{array}\right|=3k$

Performing the operations $R_2\to R_1-R_2$ and $R_3\to R_1-R_3,$

$\longrightarrow\left|\begin{array}{ccc}1&1&1\\0&1-\omega&1-\omega^2\\0&1-\omega^2&1-\omega\end{array}\right|=3k$

Evaluating the determinant,

$\longrightarrow(1-\omega)^2-(1-\omega^2)^2=3k$

$\longrightarrow(1-\omega+1-\omega^2)(1-\omega-1+\omega^2)=3k$

$\longrightarrow(2-(\omega^2+\omega))(\omega^2-\omega)=3k$

From (1),

$\longrightarrow(2-(-1))(\omega^2-\omega)=3k$

$\longrightarrow3(\omega^2-\omega)=3k$

$\longrightarrow\omega^2-\omega=k$

$\longrightarrow\omega^2+\omega+1-2\omega-1=k$

From (1),

$\longrightarrow0-(2\omega+1)=k$

From (i), we get,

$\longrightarrow\underline{\underline{k=-z}}$

Hence (4) is the answer.