Question

Answer the following question:

Without expanding determinant show that

`|(0, "a", "b"),(-"a", 0, "c"),(-"b", -"c", 0)|` = 0

Answer 1

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