If $$\omega $$ is the cube root of unity, then what is one root of the equation \[\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x}&{ - 2{\omega ^2}}\\ 2&\omega &{ - \omega }\\ 0&\omega &1 \end{array}} \right| = 0\,?\]

Solution :
Given matrix is :
\[\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x}&{ - 2{\omega ^2}}\\ 2&\omega &{ - \omega }\\ 0&\omega &1 \end{array}} \right| = 0\]
By $${C_2} \to {C_2} + {C_3},$$   we get
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}&{ - 2{\omega ^2}}\\ 2&0&{ - \omega }\\ 0&{1 + \omega }&1 \end{array}} \right| = 0\]
\[ \Rightarrow \left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}&{ - 2{\omega ^2}} \\ 2&0&{ - \omega } \\ 0&{ - {\omega ^2}}&1 \end{array}} \right| = 0\left[ {\because 1 + \omega = - {\omega ^2}} \right]\]
\[ \Rightarrow {\omega ^2}\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2{\omega ^2}}\\ 2&{ - \omega } \end{array}} \right| + 1\left| {\begin{array}{*{20}{c}} {{x^2}}&{ - 2x - 2{\omega ^2}}\\ 2&{ - 0} \end{array}} \right| = 0\]
$$\eqalign{ & \Rightarrow {\omega ^2}\left( { - \omega {x^2} + 4{\omega ^2}} \right) - \left( { - 4x - 4{\omega ^2}} \right) = 0 \cr & \Rightarrow - {x^2} + 4\omega + 4x + 4{\omega ^2} = 0 \cr & \Rightarrow - {x^2} + 4\omega - 4x - 4 - 4\omega = 0 \cr & \Rightarrow - {x^2} - 4x - 4 = 0 \cr & \Rightarrow {\left( {x + 2} \right)^2} = 0 \cr & \Rightarrow x = - 2. \cr} $$