A determinant!

[Tolaso J Kos]

Let $a=e^{2\pi i /9}$. Prove that:

$$\begin{vmatrix}
1 &a^2 &a^7 \\
a^4& a^6 &a^8 \\
a^5& a &a^3
\end{vmatrix}=-3$$

[Tsakanikas Nickos]

\( \begin{aligned}
\begin{vmatrix} 1 & a^2 & a^7 \\ a^4 & a^6 & a^8 \\ a^5 & a & a^3 \end{vmatrix} \; \; &\stackrel{R_{2} \to R_{2} - a^4R_{1} \; , \; R_{3} \to R_{3} - a^5R_{1}}{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=}
\begin{vmatrix} 1 & a^2 & a^7 \\ 0 & 0 & a^8-a^{11} \\ 0 & a - a^7 & a^3 - a^{12} \end{vmatrix} \stackrel{R_{2} \leftrightarrow R_{3}}{=\!=\!=\!=\!=} \\ \\
&=- \begin{vmatrix} 1 & a^2 & a^7 \\ 0 & a - a^7 & a^3 - a^12 \\ 0 & 0 & a^8-a^{11} \end{vmatrix} \\ \\
&=- (a-a^7)(a^8-a^{11}) \\ \\
&=-a^9(1-a^6)(1-a^3) \stackrel{a^9 = 1 \; , \;a^3 + a^6 = -1 }{=\!=\!=\!=\!=\!=\!=\!=\!=\!=\!=} \\ \\
&=-3
\end{aligned} \)