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

Answer 1

The correct option is A $-1$

Given,

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

$=1\cdot det\left(\begin{array}{cc}1+{\omega }^2& {\omega }^2\\ {\omega }^2& \omega \end{array}\right)-1\cdot det\left(\begin{array}{cc}1& {\omega }^2\\ 1& \omega \end{array}\right)+1\cdot det\left(\begin{array}{cc}1& 1+{\omega }^2\\ 1& {\omega }^2\end{array}\right)$

$=1\cdot (\omega (1+{\omega }^2)-{\omega }^4)-1\cdot (\omega -{\omega }^2)+1\cdot (-1)$

$=-{\omega }^4+{\omega }^3+{\omega }^2-1$

$=-\omega +1+{\omega }^2-1$

$={\omega }^2-\omega$

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

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

$=-1$