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Question
Using properties of determinant show that
`|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc
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Solution
L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`
By C1 + (C2 + C3), we get,
L.H.S. = `|(2("a" + "b"), "a", "b"),(2("a" + "c"), "a" + "c", "c"),(2("b" + "c"), "c", "b" + "c")|`
By taking 2 common from C1, we get,
L.H.S. = `2|("a" + "b", "a", "b"),("a" + "c", "a" + "c", "c"),("b" + "c", "c", "b" + "c")|`
By C1 – (C2 + C3), we get,
L.H.S. = `2|(0, "a", "b"),(-"c", "a" + "c", "c"),(-"c", "c", "b" + "c")|`
By C2 + C1 and C3 + C1, we get
L.H.S. = `2|(0, "a", "b"),(-"c", "a", 0),(-"c", 0, "b")|`
By taking c, a, b common from C1, C2, C3 respectively, we get,
L.H.S. = `2("c" "a" "b")|(0, 1, 1),(-1, 1, 0),(-1, 0, 1)|`
= 2abc [0 (1 – 0) – 1 ( – 1 + 0) + 1 (0 + 1)]
= 2abc (0 + 1 + 1)
= 4abc
= RHS.
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