Question

Show that

`|(x-3,x-4,x-alpha),(x-2,x-3,x-beta),(x-1,x-2,x-gamma)|=0`, where α, β, γ are in A.P.

 

Answer 1

Given:α, β, γ are in A.P.

Now,

\[2\beta = \alpha + \gamma\]

 

`Delta=|(x-3,x-4,x-alpha),(x-2,x-3,x-beta),(x-1,x-2,x-gamma)|`

`Delta=1/2|(x-3,x-4,x-alpha),(2x-4,2x-6,2x-2beta),(x-1,x-2,x-gamma)|`    `["Applying"  R_2->2R_2]`

`Delta=1/2|(x-3,x-4,x-alpha),(0,0,-2beta+alpha+gamma),(x-1,x-2,x-gamma)|`     `[because 2beta=alpha+gamma]`      `["Applying"  R_2->R_2-(R_1+R_3)]`

`Delta=1/2|(x-3,x-4,x-alpha),(0,0,0),(x-1,x-2,x-gamma)|`

Δ = 0