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Question
Answer the following question:
Without expanding determinant show that
`|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0
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Solution
Let D = `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|`
By taking ( –1) common from each of R1, R2, R3, we get,
D = `(-1)^3|(0, -"a", -"b"),("a", 0, -"c"),("b", "c", 0)|`
By interchanging rows and columns, we get,
D = `(-1)|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|`
∴ D = – 1(D)
∴ 2D = 0
∴ D = 0
∴ `|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0
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