Question

Show that Δ = `|(x, "p", "q"),("p", x, "q"),("q", "q", x)| = (x - "p")(x^2 + "p"x - 2"q"^2)` 

Answer 1

Applying C₁ → C₁ – C₂, we have

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

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

= `(x - "p")|(0, "p" + x, 2"p"),(-1, x, "q"),(0, "q", x)|`  Applying R₁ → R₁ – R₂ 

Expanding along C₁, we have

Δ = `(x - "p")("p"x + x^2 - 2"q"^2)`

= `(x - "p")(x^2 + "p"x - 2"q"^2)`