Question

Without expanding the determinants, 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)|`

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