Question

Using properties of determinants, show that `|(x, p, q),(p, x, q),(q, q, x)| = (x - p)(x^2 + px - 2q)^2`

Answer 1

To prove:

`|(x, p, q),(p, x, q),(q, q, x)| = (x - p)(x^2 + px - 2q^2)`

Let Δ = LHS = `|(x, p, q),(p, x, q),(q, q, x)|`

Applying C₁ → C₁ – C₂, we get

`Δ = |(x - p, p, q),(p - x, x, q),(q - q, q, x)|`

Taking (x – p) as common from C₁, we get

`Δ = (x - p) |(1, p, q),(-1, x, q),(0, q, x)|`

Now, expanding along R₁, we get

Δ = (x – p) [1(x² – q²) – p(–x – 0) + q(–q – 0)]

= (x – p) (x² – q² + px – q²)

= (x – p) (x² + px – 2q²)

= RHS.

Hence Proved.