Question

Prove that `|(1 + "a", 1, 1),(1, 1 + "b", 1),(1, 1, 1 + "c")| = "abc"(1 + 1/"a" + 1/"b" + 1/"c")`

Answer 1

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"]``