Matrices and Determinants 2 Question 11
13. 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) $z$
(c) -1
(d) 1
(2017 Main)
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Answer:
Correct Answer: 13. (a)
Solution:
- Given, $2 \omega+1=z$
$ \begin{aligned} \Rightarrow & 2 \omega+1 =\sqrt{-3} & {[\because z=\sqrt{-3}] } \\ \Rightarrow & \omega =\frac{-1+\sqrt{3} i}{2} & \end{aligned} $
Since, $\omega$ is cube root of unity.
$\therefore \quad \omega^{2}=\frac{-1-\sqrt{3} i}{2}$ and $\omega^{3 n}=1$
Now, $\begin{vmatrix}1 & 1 & 1 \\ 1 & -\omega^{2}-1 & \omega^{2} \\ 1 & \omega^{2} & \omega^{7}\end{vmatrix}=3 k$
$ \Rightarrow \quad\begin{vmatrix} 1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \end{vmatrix}=3 k $
$ \left[\because 1+\omega+\omega^{2}=0 \text { and } \omega^{7}=\left(\omega^{3}\right)^{2} \cdot \omega=\omega\right] $
On applying $R_{1} \rightarrow R_{1}+R_{2}+R_{3}$, we get
$ \begin{vmatrix} 3 & 1+\omega+\omega^{2} & 1+\omega+\omega^{2} \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \end{vmatrix}=3 k $
$\Rightarrow $ $ \begin{vmatrix} 3 & 0 & 0 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega \end{vmatrix}=3 k $
$\Rightarrow 3\left(\omega^{2}-\omega^{4}\right)=3 k $
$\Rightarrow \left(\omega^{2}-\omega\right)=k $
$\therefore k=\frac{-1-\sqrt{3} i}{2}-\frac{-1+\sqrt{3} i}{2}=-\sqrt{3} i=-z $
