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Show that ∣ ∣ ∣ ∣ ∣ X − 3 X − 4 X − α X − 2 X − 3 X − β X − 1 X − 2 X − γ ∣ ∣ ∣ ∣ ∣ = 0 , Where α, β, γ Are in A.P.

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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.

 

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Solution

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
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Chapter 5: Determinants - Exercise 6.2 [Page 61]

APPEARS IN

R.D. Sharma Mathematics Volume 1 and 2 [English] Class 12
Chapter 5 Determinants
Exercise 6.2 | Q 48 | Page 61

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