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)

Answer 1

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)