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Home  >>  JEEMAIN and NEET  >>  Mathematics  >>  Class12  >>  Matrices

If $\omega(\neq 1)$ is a cube root of unity,then $\begin{vmatrix}1 &1+i+\omega^2&\omega^2\\1-i&-1&\omega^2-1\\-i&-i+\omega-1&-1\end{vmatrix}$=

(A) 0

(B) 1

(C) i

(D) $\omega$

1 Answer

$\begin{vmatrix}1 &1+i+\omega^2&\omega^2\\1-i&-1&\omega^2-1\\-i&-i+\omega-1&-1\end{vmatrix}$
Apply $R_1\rightarrow R_1-R_2+R_3$
$\begin{vmatrix}0 &0&0\\1-i&-1&\omega^2-1\\-i&-i+\omega-1&-1\end{vmatrix}$
$\Rightarrow 0$
$(1+\omega+\omega^2)=0$
Hence (a) is the correct answer.
answered Nov 20, 2013 by sreemathi.v
 

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