Question

If $\omega$ is a complex cube root of unity then $\left|\begin{array}{ccc}1& 1+\omega & 1+{\omega }^2\\ 1+\omega & 1+{\omega }^2& 1\\ 1+{\omega }^2& 1& 1+\omega \end{array}\right|=$

• A. $-2$
• B. $4$
• C. $0$
• D. $2$

Answer 1

Answer: (B)

$4$

$1+w+w^2=0,w^3=1$

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

$=1(w^3-1)+w^2(w^4+w)-w(-w^2-w^2)$

$=0+2w^3+w\left(2w^2\right)$

$=2w^3+2w^3$

$=4w^3$

$=4$