Question

Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0

Answer 1

Let A = `[(0, "a", "b"),(-"a", 0, -"c"),(-"b", "c", 0)]`

Here A^T = `[(0, -"a", -"b"),("a", 0, "c"),("b", -"c", 0)]`

= – A

⇒ A is a skew symmetric matrix

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

= 0(0 + c²) – a[0 – bc] + b[– ac + 0]

= abc – abc

= 0