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Question
Show that `|("b" + "c", "bc", "b"^2"C"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|` = 0
Solution
`|("b" + "c", "bc", "b"^2"C"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)| = "abc"/"abc" |("b" + "c", "bc", "b"^2"C"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|`
= `1/"abc" |("ab" + "ac", "abc", "ab"^2"c"^2),("bc" + "ab", "abc", "bc"^2"a"^2),("ca" + "bc", "abc", "ca"^2"b"^2)| {:("R"_1 -> "aR"_1),("R"_2 -> "bR"_2),("R"_3 -. "cR"_3):}`
= `(("abc")("abc"))/"abc" |("ab" + "ac", 1, "bc"),("bc" + "ab", 1, "ca"),("ca" + "bc", 1, "ab")|`
Taking out abc from column c2 and c3
`"C"_1 -> "C"_1 + "C"_3`
= `("abc") |("ab" + "bc" + "ca", 1, "bc"),("ab" + "bc" + "ca", 1, "ca"),("ab" + "bc" + "ca", 1, "ab")|`
= `("abc")("ab" + "bc" + "ca") |(1, 1, "bc"),(1, 1, "ca"),(1, 1, "ab")|`
= (abc)(ab + bc + ca) × 0
= 0
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