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Question
Solve the following problems by using Factor Theorem:
Show that `|(x, "a", "a"),("a", x, "a"),("a", "a", x)|` = (x – a)2 (x + 2a)
Solution
Let |A| = `|(x, "a", "a"),("a", x, "a"),("a", "a", x)|`
Pt x = a in |A|, we get
|A| = `|("a", "a", "a"),("a", "a", "a"),("a", "a", "a")|`
By putting x = a , we have three rows of |A| are identical.
Therefore (x – a)2 is a factor of |A|
Put x = – 2a in |A|
|A| = `|(-2"a" + "a", "a"),("a", -2"a", "a"),("a", "a", -2"a")|`
= `|(-2"a" + "a" + "a", "a", "a"),("a" - 2"a" + "a", -2"a", "a"),("a" + "a" - 2"a","a", -2"a")|`
= `|(0, "a", "a"),(0, -2"a", "a"),(0, "a", -2"a")|`
∴ x + 2a is a factor of |A|. The degree of the product of the factors (x – a)2 (x + 2a) is 3.
The degree of tfie product of the leading diagonal elements x . x . x is 3.
∴ The other factor is the contant factor k.
∴ `|(x, "a", "a"),("a", x, "a"),("a", "a", x)| "k"(x - "a")^2 (x + 2"a")`
Put x = – a
`|(-"a", "a", "a"),("a", , "a"),("a", "a", - "a")| = "k"(- "a" - "a")^2 (- "a" + 2a")`
`"a"^3 |(-1, 1, 1),(1, -1, 1),(1, 1, -1)| = "k" xx 4"a"^2 xx "a"`
a3 [– 1(1 – 1) – 1( – 1 – 1) + 1(1 + 1)] = k . 4a3
a3 [0 + 2 + 2 ] = 4 ka3
4a3 = 4ka3
k = 1
∴ `|(x, "a", "a"),("a", x, "a"),("a", "a", x)|` = (x – a)2 (x + 2a)
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