Question

Show that `|("b" + "c", "a" - "c", "a" - "b"),("b" - "c", "c" + "a", "b" - "a"),("c" - "b", "c" - "a", "a" + b")|` = 8abc

Answer 1

Let |A| = `|("b" + "c", "a" - "c", "a" - "b"),("b" - "c", "c" + "a", "b" - "a"),("c" - "b", "c" - "a", "a" + b")|`

Put a = 0

|A| = `|("b" + "c", - "c", - "b"),("b" - "c", "c", "b"),("c" - "b", "c", "b")|`

= `"bc"|("b" + "c", -1, -1),("b" - "c", 1, 1),("c" - "b", 1, 1)|`

Since two columns identical

= bc × 0 = 0

∴ a – 0 is a factor.

That is, a is a factor.

Put b = 0 in |A|

|A| = `|("b", "a", "a" - "b"),("b", "a", "b" - "a"),(-"b", -"a", "a" + "b")|`

= `"ab" |(1, 1, "a" - "b"),(1, 1, "b" - "a"),(-1, -1, "a" + "b")|`

Since two columns identical

= ab × 0 = 0

∴ c – 0 is a factor.

That is, c is a factor.

The degree of the product of the factors abc is 3.

The degree of the product of leading diagonal elements (b + c)(c + a)(a + b) is 3.

∴ The other factor is the constant factor k.

`|("b" + "c", "a" - "c", "a" - "b"),("b" - "c", "c" + "a", "b" - "a"),("c" - "b", "c" - "a", "a" + "b")|` = kabc

Put a = 1

b = 1

c = 1

`|(1 + 1, 1 - 1, 1 - 1),(1 - 1, 1 + 1, 1 - 1),(1 - 1, 1 - 1, 1 + 1)|` = k × 1 × 1 × 1

`|(2, 0, 0),(0, 2, 0),(0, 0, 2)|` = k

2 × 2 × 2 = 8

⇒ k = 8

∴ `|("b" + "c", "a" - "c", "a" - "b"),("b" - "c", "c" + "a", "b" - "a"),("c" - "b", "c" - "a", "a" + b")|` = 8abc