# If ω is a non-real cube root of unity and n is not a multiple of 3, then

Question:

If $\omega$ is a non-real cube root of unity and $n$ is not a multiple of 3, then

$\Delta=\left|\begin{array}{ccc}1 & \omega^{n} & \omega^{2 n} \\ \omega^{2 n} & 1 & \omega^{n} \\ \omega^{n} & \omega^{2 n} & 1\end{array}\right|$ is equal to

(a) 0

(b) $\omega$

(c) $\omega^{2}$

(d) 1

Solution:

(a) 0

$\Delta=\mid 1 \quad w^{n} \quad w^{2 n}$

$w^{2 n} \quad 1 \quad w^{n}$

$w^{n} \quad w^{2 n} \quad 1 \mid$

$=\mid 1+w^{n}+w^{2 n} \quad w^{n} \quad w^{2 n}$

$w^{2 n}+1+w^{n} \quad 1 \quad w^{n}$

$w^{n}+w^{2 n}+1 \quad w^{2 n} \quad 1 \mid$     [Appplying $C_{1} \rightarrow C_{1}+C_{2}+C_{3}$ ]

Now,

$1+w+w^{2}=0$        $[\because w$ is a complex cube root of unity $]$

$\Rightarrow 1+w^{n}+w^{2 n}=0$           $[\because n$ is not a multiple of 3$]$

$\Rightarrow \Delta=\mid \begin{array}{lll}0 & w^{n} & w^{2 n}\end{array}$

$\begin{array}{lll}0 & 1 & w^{n}\end{array}$

$0 \quad w^{2 n} 1 \mid=0$