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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class12  >>  Determinants
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Let $\omega=-\large\frac{1}{2}$$+i\large\frac{\sqrt 3}{2}$.Then the value of the determinant $\begin{vmatrix}1&1&1\\1&-1-\omega^2&\omega^2\\1&\omega^2&\omega^4\end{vmatrix}$ is

$(a)\;3\omega\qquad(b)\;3\omega(\omega-1)\qquad(c)\;3\omega^2\qquad(d)\;3\omega(1-\omega)$

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1 Answer

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Given that
$\omega=-\large\frac{1}{2}$$+i\large\frac{\sqrt 3}{2}$
$\omega^2=-\large\frac{1}{2}$$-i\large\frac{\sqrt 3}{2}$
Also $1+\omega+\omega^2=0,\omega^3=1$
Now given determinant is
$\Delta=\begin{vmatrix}1 &1&1\\1&-1-\omega^2&\omega^2\\1 &\omega^2&\omega^4\end{vmatrix}$
$\;\;\;=\begin{vmatrix}1 &1&1\\1&\omega&\omega^2\\1 &\omega^2&\omega\end{vmatrix}$
$1+\omega+\omega^2=0$
Expanding along $C_1$ we get
$3(\omega^2-\omega^4)=3(\omega^2-\omega)=3\omega(\omega-1)$
Hence (b) is the correct answer.
answered Nov 20, 2013 by sreemathi.v
 

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