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Question
If `|("a", "b", "a"alpha + "b"),("b", "c", "b"alpha + "c"),("a"alpha + "b", "b"alpha + "c", 0)|` = 0, prove that a, b, c are in G. P or α is a root of ax2 + 2bx + c = 0
Solution
Let Δ = `|("a", "b", "a"alpha + "b"),("b", "c", "b"alpha + "c"),("a"alpha + "b", "b"alpha + "c", 0)|`
= `|("a", "b", "a"alpha),("b", "c", "b"alpha),("a"alpha + "b", "b"alpha + "c", -("b"alpha + c))| ("C"_3 -> "C"_3 - "C"_2)`
= `|("a", "b", 0),("b", "c", 0),("a"alpha + "b", "b"alpha + "c", -("b"alpha + c)),(, , -("a"alpha^2 + "b"alpha))| ("C"_3 -> "C"_3 - alpha"C"_1)`
= `|("a", "b", 0),("b", "c", 0),("a"alpha + "b", "b"alpha + "c", -("a"alpha^2 + 2"b"alpha + c))|` expanding along C3
We get – (aα2 + 2bα + c)[ac – b2]
So Δ = 0
⇒ (aα2 + 2bα + c)(ac – b2)
= – 0
= 0
⇒ aα2 + 2bα + c = 0
or
ac – b2 = 0
(i.e.) a is a root of ax2 + 2bx + c = 0
or
ac = b2
⇒ a, b, c are in G.P.
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