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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)
Solution
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)
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