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Question
Prove that `|(1 + "a", 1, 1),(1, 1 + "b", 1),(1, 1, 1 + "c")| = "abc"(1 + 1/"a" + 1/"b" + 1/"c")`
Solution
Let Δ = `|(1 + "a", 1, 1),(1, 1 + "b", 1),(1, 1, 1 + "c")|`
Δ = `|("a", 0, -"c"),(0, "b", -"c"),(1, 1, 1 + "c")| {:("R"_1 -> "R"_1 - "R"_3),("R"_2 -> "R"_2 - "R"_3):}`
= a[b(1 + c) + c(1)] – 0 – c[0 – b]
= a[b + bc + c] + bc
= ab + abc + ac + bc
= abc + ab + bc + ac
= abc
`["abc"/"abc" + "ab"/"abc" + "bc"/"abc" + "ac"/"abc"] = "abc" [1 + 1/"c" + 1/"a" + 1/"b"]`
Δ = `"abc" [1 + 1/"a" + 1/"b" + 1/"c"]``
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