Prove the following identities – |(1,a,a^2)(a^2,1,a)(a,a^2,1)|= (a^3 - 1)^2 ​

Question

Prove the following identities –

\(\begin{vmatrix} 1 & a & a^2 \\[0.3em] a^2& 1 & a \\[0.3em] a & a^2 &1 \end{vmatrix}\) = (a³ - 1)²

Answer 1

Let Δ = \(\begin{vmatrix} 1 & a & a^2 \\[0.3em] a^2& 1 & a \\[0.3em] a & a^2 &1 \end{vmatrix}\)

Recall that the value of a determinant remains same if we apply the operation Rⁱ→ Rⁱ + kR_j or C_i→ C_i + kC_j.

Applying R₁→ R₁ + R₂, we get

⇒ Δ = (a² + a + 1) = \(\begin{vmatrix} 1 & 0 &0 \\[0.3em] a^2& 1-a^2 & a-a^2 \\[0.3em] a & a^2-a &1-a \end{vmatrix}\)

Expanding the determinant along R₁, we have

Δ = (a² + a + 1)(1)[(1 – a²)(1 – a) – (a² – a)(a – a²)] 

⇒ Δ = (a² + a + 1)(1 – a – a² + a³ – a³ + a⁴ + a² – a³) 

⇒ Δ = (a² + a + 1)(1 – a – a³ + a⁴) 

⇒ Δ = (a² + a + 1)(a⁴ – a³ – a + 1) 

⇒ Δ = (a² + a + 1)[a³(a – 1) – (a – 1)] 

⇒ Δ = (a² + a + 1)(a – 1)(a³ – 1) 

⇒ Δ = (a – 1)(a² + a + 1)(a³ – 1) 

⇒ Δ = (a³ – 1)(a³ – 1) 

∴ Δ = (a³ – 1)²

Thus,

\(\begin{vmatrix} 1 & a & a^2 \\[0.3em] a^2& 1 & a \\[0.3em] a & a^2 &1 \end{vmatrix}\) = (a³ - 1)²