Question

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

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

Answer 1

The correct option is A $0$

$\left|\begin{array}{ccc}1& 1+i+{\omega }^2& {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|$
$\left|\begin{array}{ccc}1& i-\omega & {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -i+\omega -1& -1\end{array}\right|(\because 1+\omega +{\omega }^2=0)$
$R_3\to R_3-R_2$
$=\left|\begin{array}{ccc}1& i-\omega & {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -1& -i+\omega & -{\omega }^2\end{array}\right|$
$=-\left|\begin{array}{ccc}1& i-\omega & {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ 1& i-\omega & {\omega }^2\end{array}\right|$

On expanding the above determinant we get,
$=0$