Complex Numbers And Quadratic Equations question 259
Question: If $ \omega (\ne 1) $ is a cube root of unity, then $ \begin{vmatrix} 1 & 1+i+{{\omega }^{2}} & {{\omega }^{2}} \\ 1-i & -1 & {{\omega }^{2}}-1 \\ -i & -i+\omega -1 & -1 \\ \end{vmatrix} $ is equal to [IIT 1995]
Options:
A) 0
B) 1
C) $ \omega $
D) $ i $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \Delta = \begin{vmatrix} 1 & 1+i+{{\omega }^{2}} & {{\omega }^{2}} \\ 1-i & -1 & {{\omega }^{2}}-1 \\ -i & -i+\omega -1 & -1 \\ \end{vmatrix} $ $ = \begin{vmatrix} 1 & i-\omega & {{\omega }^{2}} \\ 1-i & -1 & {{\omega }^{2}}-1 \\ -i & -i+\omega -1 & -1 \\ \end{vmatrix} $ = 0 ( $ \because $ Two rows are identical)
BETA
