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Question
Prove that `|("a"^2, "bc", "ac" + "c"^2),("a"^2 + "ab", "b"^2, "ac"),("ab", "b"^2 + "bc", "c"^2)| = 4"a"^2"b"^2"c"^2`
Solution
Let Δ = `|("a"^2, "bc", "ac" + "c"^2),("a"^2 + "ab", "b"^2, "ac"),("ab", "b"^2 + "bc", "c"^2)|`
Δ = `|(2"a"^2 + 2"ab", 2"b"^2 + 2"bc", 2"c"^2 + 2"ac"),("a"^2 + 2"ab", 2"b"^2 + "bc", "c"^2 + "ac"),("ab", "b"^2 + "bc", "c"^2)| {:("R"_1 -> "R"_1 + "R"_2 + "R"_3),("R"_2 -> "R"_2 + "R"_3):}`
= `2 |("a"^2 + "ab", "b"^2 + 2"bc", "c"^2 + "ac"),("a"^2 + 2"ab", 2"b"^2 + "bc", "c"^2 + "ac"),("ab", "b"^2 + "bc", "c"^2)|`
= `2 |("a"^2 + "ab", "b"^2 + "bc", "c"^2 + "ac"),("a"^2 + "ab", "b"^2, "ac"),("ab", "b"^2 + "bc", "c"^2)| "R"_1 -> "R"_1 - "R"_2`
= `2|(0, "bc", "c"^2),("a"^2, -"bc", "ac" - "c"^2),("ab", "b"^2 + "bc", "c"^2)| {:("R"_1 -> "R"_1 - "R"_2),("R"_2 -> "R"_2 - "R"_3):}`
= `2|(0, "bc", "c"^2),("a"^2, 0, "ac"),("ab", "b"^2, 0)| {:("R"_2 -> "R"_2 + "R"_1),("R"_3 -> "R"_3 - "R"_11):}`
= `2"abc" |(0, "b", "c"),("a", 0, "c"),("a", "b", 0)|`
= 2abc [0 – b(0 – ac) + c(ab – 0)]
= 2abc [abc + abc]
= 2abc × 2abc
Δ = 4a2b2c2
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