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Question
Answer the following question:
Without expanding determinant 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
L.H.S. = `|("b" + "c", "bc", "b"^2"c"^2),("c" + "a", "ca", "c"^2"a"^2),("a" + "b", "ab", "a"^2"b"^2)|`
Taking bc, ca, ab common from R1, R2, R3 respectively, we get
L.H.S. = `("bc")("ca")("ab")|(("b" + "c")/"bc", 1, "bc"),(("c" + "a")/"ca", 1, "ca"),(("a" + "b")/"ab", 1, "ab")|`
Taking abc common from C3, we get
L.H.S. = `("a"^2"b"^2"c"^2)("abc")|(1/"c" + 1/"b", 1, 1/"a"),(1/"a" + 1/"c", 1, 1/"b"),(1/"b" + 1/"a", 1, 1/"c")|`
Applying C1 → C1 + C3, we get
L.H.S. = `"a"^3"b"^3"c"^3|(1/"a" + 1/"b" + 1/"c", 1, 1/"a"),(1/"a" + 1/"b" + 1/"c", 1, 1/"b"),(1/"a" + 1/"b" + 1/"c", 1, 1/"c")|`
Taking `(1/"a" + 1/"b" + 1/"c")` common from C1, we get
L.H.S. = `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")|(1, 1, 1/"a"),(1, 1, 1/"b"),(1, 1, 1/"c")|`
= `"a"^3"b"^3"c"^3 (1/"a" + 1/"b" + 1/"c")(0)` ...[∵ C1 and C2 are identical]
= 0
= R.H.S.
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