Question

Show that `|("b" + "C", "a", "a"^2),("c" + "a", "b", "b"^2),("a" + "b", "c", "c"^2)|` = (a + b + c)(a – b)(b – c)(c – a)

Answer 1

Let |A| = `|("b" + "C", "a", "a"^2),("c" + "a", "b", "b"^2),("a" + "b", "c", "c"^2)|` 

Put a = b in |A|

|A| = `|("b" + "c", "b", "b"^2),("c" + "b", "b", "b"^2),("b" + "b", "c", "c"^2)|`

|A| = `|("b" + "c", "b", "b"^2),("b" + "c", "b", "b"^2),("b" + "b", "c", "c"^2)|`

Since two rows are idenctical

|A| = 0

Since two rows are idenctical

|A| = 0

∴ a – b is a factor of |A|.

The given determinant is in cyclic symmetric form in a, b and c.

Therefore, b – c and c – a are also factors.

The degree of the product of the factors (a – b)(b – c)(c – a) is 3 and the degree of the product of the leading diagonal elements (b + c) . b . c² is 4.

Therefore, the other factor is k(a + b + c).

 `|("b" + "C", "a", "a"^2),("c" + "a", "b", "b"^2),("a" + "b", "c", "c"^2)|` = k(a + b + c)(a – b) × (b – c)(c – a)

Put a = 1, b = 2, c = 3 we get

`|(2 +3, 1, 1^2),(3 + 1, 2, 2^2),(1 + 2, 3, 3^2)|` = k(1 + 2 + 3)(1 – 2) × (2 – 3)(3 – 1)

`|(5, 1, 1),(4, 2, 4),(3, 3, 9)|` = k × 6 ×  –1 × –1 × 2

5(18 – 12) – 1(36 – 12) + 1(12 – 6) = 12k

5 × 6 – 24 + 6 = 12k

30 – 24 + 6 = 12k

12 = 12

⇒ k = 1

∴ `|("b" + "C", "a", "a"^2),("c" + "a", "b", "b"^2),("a" + "b", "c", "c"^2)|` = (a + b + c)(a – b)(b – c)(c – a)