Question

If one of the cube roots of $1$ be $\omega$, then $\left|\begin{array}{ccc}1& 1+{\omega }^2& {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -1+\omega & -1\end{array}\right|$

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

Answer 1

The correct option is D $0$

Let $\Delta =\omega$, then $\left|\begin{array}{ccc}1& 1+{\omega }^2& {\omega }^2\\ 1-i& -1& {\omega }^2-1\\ -i& -1+\omega & -1\end{array}\right|$

Applying $C_1\to C_1-C_2+C_3$
$=\Delta \left|\begin{array}{ccc}0& 1+{\omega }^2& {\omega }^2\\ 1+{\omega }^2-i& -1& {\omega }^2-1\\ -\omega -i& \omega -1& -1\end{array}\right|$

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

$=(-\omega -1)[-1\{(-1-{\omega }^2)-{\omega }^2(\omega -1\left)\right\}+1({\omega }^4-1+{\omega }^2\left)\right]$

$=(-\omega -1)[+1+{\omega }^2+{\omega }^3-{\omega }^2+1(\omega -1+{\omega }^2\left)\right]$

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

$=\omega (-\omega -1)(1+\omega +{\omega }^2)=0(\because 1+\omega +{\omega }^2=0)$