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

Answer 1

`|("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