Question

Let $\omega =-\frac{1}{2}+i\frac{\sqrt{3}}{2}$, then the value of the determinant$\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$ is

• A. $3\omega$
• B. $3\omega (\omega -1)$
• C. $3{\omega }^2$
• D. $3\omega (1-\omega )$

Answer 1

The correct option is B

$3\omega (\omega -1)$

Explanation for correct option:

Finding the value of the determinant:

Given, $\left|\begin{array}{ccc}1& 1& 1\\ 1& -1-{\omega }^2& {\omega }^2\\ 1& {\omega }^2& \omega \end{array}\right|$

We know that $1+\omega +{\omega }^2=0$

Apply the operation on the first row $R_1$ as follows

$R_1\to R_1+R_2+R_3$ we get,

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

Now evaluating the value of the determinant we get,

$$\begin{gathered} =3\left({\omega }^2-{\omega }^4\right) \\ =3\left({\omega }^2-\omega \right)\because {\omega }^4=\omega \\ =3\omega \left(\omega -1\right) \end{gathered}$$

Therefore the value of the determinant is equal to $3\omega \left(\omega -1\right)$

Hence, the correct answer is option (B).