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Question
Without expanding the determinants, 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)|`
Taking (– 1) common from R1, R2, R3, we get
D = `(- 1)^3 |(0, -"a", -"b"),("a", 0, -"c"),("b", "c", 0)|`
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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