Question

If $(\omega \ne 1)$ is a cube root of unity, then
$\left|\begin{array}{ccc}1& 1+i+{\omega }^2& {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|$

• A. $0$
• B. $1$
• C. $i$
• D. $\omega$

Answer 1

The correct option is A $0$

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

$R_1\to R_1+R_3$

$\left|\begin{array}{ccc}1-i& {\omega }^2+\omega & {\omega }^2-1\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|=\left|\begin{array}{ccc}1-i& -1& {\omega }^2-1\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|$ (as $1+\omega +{\omega }^2=0$)

As $R_1$ and $R_2$ are same. Hence determinant is zero.