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Question
Write the general form of a 3 × 3 skew-symmetric matrix and prove that its determinant is 0
Solution
Let A = `[(0, "a", "b"),(-"a", 0, -"c"),(-"b", "c", 0)]`
Here AT = `[(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
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