Question

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

• A. $\omega$
• B. $1$
• C. $i$
• D. None of these
  

Answer 1

Answer: (D)

None of these

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

$R_3\to R_3+R_1$

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

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

Expanded along $R_2$