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# Evaluate $\left|\begin{array}{ccc} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$

Toolbox:
• (i) If two rows or columns are identical then the value of the determinant is zero
• (ii)Elementary transformation can be made by
• (a) interchanging two rows or columns
• (b) By adding or subtracting two or more rows or columns.
Let $\Delta=\left|\begin{array}{ccc} 1 & \omega^n & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$
We know $\omega^3=1\qquad Therefore\; \omega^{3n}=1$
Hence substitute for 1 in $R_1$ in the determinant we get
$\Delta=\left|\begin{array}{ccc} \omega^{3n} & \omega^{n} & \omega^{2n} \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$
Now let us take $\omega^n$ as the common factor from $R_1$
$\Delta=\omega^{n}\left|\begin{array}{ccc} \omega^{2n} & 1 & \omega^n \\ \omega^{2n} & 1 & \omega^{n} \\ \omega^n & \omega^{2n} & 1 \end{array}\right|$
Here two rows $R_1$ and $R_2$ are identical .
Hence the value of the determinant is zero
Therefore $\Delta=0$
Solution : option A is correct

edited Dec 24, 2013