VITEEE 2017 Paper

Given |

x+α	β	γ
γ	x+β	α
α	β	x+γ

| = 0
Operate C1 → C1+C2+C3
= 0
= (x + α + β + γ) |

1	β	γ
1	x+β	α
1	β	x+γ

| = 0
⇒ x + α + β + γ = 0 ⇒ x = - (α + β + γ)
Again if
|

1	β	γ
1	x+β	α
1	β	x+γ

| = 0 ⇒ |

1	β	γ
0	x	α−γ
0	0	x

| = 0
⇒ x2 = 0 ⇒ x = 0
∴ Solutions of the equation are x = 0,
- (α + β + γ)