Question

Show that `|(1, 1, 1),(x, y, z),(x^2, y^2, z^2)|` = (x – y)(y – z)(z – x)

Answer 1

Let |A| = `|(1, 1, 1),(x, y, z),(x^2, y^2, z^2)|`

Put x = y

|A| = `|(1, 1, 1),(x, x, z),(x^2, x^2, z^2)|`

|A| = 0 since two columns identical

∴ x – y is a factor of A.

The given determinant is in the cyclic symmetric form in x, y, and z.

Therefore, y – z and z – x are also factors of |A|.

The degree of the product of the factors (x – y)(y – z)(z – x) is 3 and the degree of the product of the leading diagonal elements 1, y, z² is 3.

Therefore, the other factor is the constant factor k.

`|(1, 1, 1),(x, y, z),(x^2, y^2, z^2)|` = k(x – y)(y – z)(z – x)

Put x = 0, y = 1, z = -1 we get

`|(1, 1, 1),(0, 1,-1),(0^2, 1^2, (-1)^2)|` = k(0 – 1)(1 + 1)(– 1 – 0)

`|(1, 1, 1),(0, 1, -1),(0, 1, 1)|` = 2k

Expanding along the first column

1(1 + 1) = 2k

2 = 2k

⇒ k = 1

∴ `|(1, 1, 1),(x, y, z),(x^2, y^2, z^2)|` = (x – y)(y – z)(z – x)