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Question
Prove that `|(1, "a", "a"^2 - "bc"),(1, "b", "b"^2 - "ca"),(1, "c", "c"^2 - "ab")|` = 0
Solution
L.H.S = `|(1, "a", "a"^2 - "bc"),(1, "b", "b"^2 - "ca"),(1, "c", "c"^2 - "ab")|`
= `|(0, "a" - "b", "a"^2 - "bc" - "b"^2 + "ac"),(0, "b" - "c", "b"^2 - "ac" - "c"^2 + "ab"),(1, "c", "c"^2 - "ab")| {:("R"_1 -> "R"_1 - "R"_2),("R"_2 -> "R"_2 - "R"_3):}`
= `|(0, "a" - "b", ("a"^2 - "b"^2) + ("ac" - "bc")),(0, "b" - "c", ("b"^2 - "c"^2) + ("ab" - "ac")),(1, "c", "c"^2 - "ab")|`
= `|(0, "a" - "b", ("a" + "b")("a" - "b") + "c"("a" - "b")),(0, "b" - "c", ("b" + "c")("b" - "c") + "a"("b" - "c")),(1, "c", "c"^2 - "ab")|`
= `|(0, "a" - "b", ("a" - "b")("a" + "b" + "c")),(0, "b" - "c", ("b" - "c")("a" + "b" + "c")),(1, "c", "c"^2 - "ab")|`
Taking (a – b)(b – c) from R1 and R2 respectively
We get (a – b)(b – c) `|(0, 1, "a" + "b" + "c"),(0, 1, "a" + "b" + "c"),(1, "c", "c"^2 - "ab")|` expanding along C1
(a – b)(b – c) {0 – 0 + 1[(a + b + c) – (a + b + c)]} = 0
=R.H.S
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