Question

Using properties of determinants, prove the following :

\[\begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix} = \left( 1 - a^3 \right)^2\].

Answer 1

Let \[∆ = \begin{vmatrix}1 & a & a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix}\]

Applying R₁ → R₁ + R₂ + R₃, we get

\[∆ = \begin{vmatrix}1 + a + a^2 & 1 + a + a^2 & 1 + a + a^2 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix}\]

\[ = \left( 1 + a + a^2 \right) \begin{vmatrix}1 & 1 & 1 \\ a^2 & 1 & a \\ a & a^2 & 1\end{vmatrix}\]

Applying C₂ → C₂ − C₁ and C₃ → C₃ − C₁, we get

\[∆ = \left( 1 + a + a^2 \right) \begin{vmatrix}1 & 0 & 0 \\ a^2 & 1 - a^2 & a - a^2 \\ a & a^2 - a & 1 - a\end{vmatrix}\]

\[ = \left( 1 + a + a^2 \right) \left( 1 - a \right) \left( 1 - a \right) \begin{vmatrix}1 & 0 & 0 \\ a^2 & 1 + a & a \\ a & - a & 1\end{vmatrix}\]

\[ = \left( 1 - a^3 \right) \left( 1 - a \right) \begin{vmatrix}1 & 0 & 0 \\ a^2 & 1 + a & a \\ a & - a & 1\end{vmatrix}\]

Expanding along R₁, we get
∆ = (1 − a³) (1 − a) {[(1 + a) + a²] − 0 + 0}
   = (1 − a³) (1 − a) (1 + a + a²)
   = (1 − a³) (1 − a³)
   = (1 − a³)²