Question

Let $\omega=-\frac{1}{2}+i\frac{\sqrt 3}{2},$ then value of the determinant $\begin {vmatrix} 1 & 1& 1 \\ 1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega \\ \end {vmatrix} is $

• A. $3 \omega$
• B. $3 \omega(\omega-1)$
• C. $3 \omega^2$
• D. $3 \omega(1- \omega)$

Answer 1

Answer: (B)

Let $\Delta=\begin {vmatrix}1 & 1& 1 \\1 & -1-\omega^2 & \omega^2 \\ 1 & \omega^2 & \omega \\\end {vmatrix} is$
Applying $R_2\rightarrow R_2-R_1;R_3\rightarrow R_3-R_1$
=$\begin {vmatrix}1 & 1& 1 \\0 & -2-\omega^2 & \omega^2-1 \\0 & \omega^2-1 & \omega-1 \\\end {vmatrix} $
$=(-2-\omega^2)(\omega-1)-(\omega^2-1)^2$
$=-2\omega+2-\omega^3+\omega^2-(\omega^4-2\omega^2+1)$
$=3\omega^2-3\omega=3\omega(\omega-1) [\because \omega^4=\omega]$