Question

Using properties of determinants, show that `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|` = 4abc.

Answer 1

L.H.S. = `|("a" + "b", "a", "b"),("a", "a" + "c", "c"),("b", "c", "b" + "c")|`

Applying C₁ → C₁ – (C₂ + C₃), we get

L.H.S. = `|(0, "a", "b"),(-2"c", "a" + "c", "c"),(-2"c", "c", "b" + "c")|`

Taking (– 2) common from C_1, we get

L.H.S. = ` - 2|(0, "a", "b"),("c", "a" + "c", "c"),("c", "c", "b" + "c")|`

Applying C₂ → C₂ – C₁ and C₃ → C₃ – C₁, we get

L.H.S. = `-2|(0, "a", "b"),("c", "a", 0),("c", 0, "b")|`

= - 2[0(ab – 0) – a(bc – 0) + b(0 – ac)]
= – 2(0 – abc – abc)
= – 2(– 2abc)
= 4abc
= R.H.S.