Question

Using properties of determinant 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")|`

By C₁ + (C₂ + C₃), we get,

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

By taking 2 common from C₁, we get,

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

By C₁ – (C₂ + C₃), we get,

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

By C₂ + C₁ and C₃ + C₁, we get

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

By taking c, a, b common from C₁, C₂, C₃ respectively, we get,

L.H.S. = `2("c"  "a"  "b")|(0, 1, 1),(-1, 1, 0),(-1, 0, 1)|`

= 2abc [0 (1 – 0) – 1 ( – 1 + 0) + 1 (0 + 1)]

= 2abc (0 + 1 + 1)

= 4abc

= RHS.