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