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