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Question
Prove tha `|("bc" - "a"^2, "ca" - "b"^2, "ab" - "c"^2),("ca" - "b"^2, "ab" - "c"^2, "bc" - "a"^2),("ab" - "c"^2, "bc" - "a"^2, "ca" - "b"^2)|` is divisible by a + b + c and find the quotient.
Solution
Δ = `|("bc" - "a"^2, "ca" - "b"^2, "ab" - "c"^2),("ca" - "b"^2, "ab" - "c"^2, "bc" - "a"^2),("ab" - "c"^2, "bc" - "a"^2, "ca" - "b"^2)|`
[Applying C1 → C1 – C2 and C2 → C2 – C3]
Δ = `|("bc" - "a"^2 - "ca" + "b"^2,"ca" - "b"^2 - "ab" + "c"^2, "ab" - "c"^2),("ca" - "b"^2 - "ab" + "c"^2, "ab" - "c"^2 - "bc" + "a"^2, "bc" - "a"^2),("ab" - "c"^2 - "bc" + "a"^2, "bc" - "a"^2 - "ca" + "b"^2, "ca" - "b"^2)|`
= `|(("b" - "a")("a" + "b" + "c"), ("c" - "b")("a" + "b" + "c"), "ab" - "c"^2),(("c" - "b")("a" + "b" + "c"), ("a" - "c")("a" + "b" + "c"), "bc" - "a"^2),(("a" - "c")("a" + "b" + "c"), ("b" - "a")("a" + "b" + "c"), "ca" - "b"^2)|`
[Taking (a + b + c) common from C1 and C2 each]
Δ = `("a" + "b" + "c")^2 |("b" - "a", "c" - "b", "ab" - "c"^2),("c" - "b", "a" - "c", "bc" - "a"^2),("a" - "c", "b" - "a", "ca" - "b"^2)|`
[Applying R1 → R1 + R2 + R3]
Δ = `("a" + "b" + "c")^2 |(0, 0, "ab" + "bc" + "ca" - ("a"^2 + "b"^2 + "c"^2)),("c" - "b", "a" - "c", "bc" - "a"^2),("a" - "c", "b" - "a", "ca" - "b"^2)|`
[Expanding along R1]
Δ = `("a" + "b" + "c")^2 ["ab" + "bc" + "ca" - ("a"^2 + "b"^2 + "c"^2)][("c" - "b")("b" - "a") - ("a" - "c")^2]`
= `("a" + "b" + "c")^2 ("ab" + "bc" + "ca" - "a"^2 - "b"^2 - "c"^2) xx ("bc" - "ac" - "b"^2 + "ab" - "a"^2 - "c"^2 + 2"ac")`
= (a + b + c)[(a + b + c)(a2 + b2 + c2 – ab – bc – ca)2]
Hence, given determinant is divisible by (a + b + c) and quotient is (a + b + c)(a2 + b2 + c2 – ab – bc – ca)2
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