Matrices and Determinants 2 Question 34

38. Let a>0,d>0. Find the value of the determinant

|1a1a(a+d)1(a+d)(a+2d)1(a+d)1(a+d)(a+2d)1(a+2d)(a+3d)1(a+2d)1(a+2d)(a+3d)1(a+3d)(a+4d)|

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Answer:

Correct Answer: 38.

4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)

Solution:

  1. Given, a>0,d>0 and let

Δ=|1a1a(a+d)1(a+d)(a+2d)1(a+d)1(a+d)(a+2d)1(a+2d)(a+3d)1(a+2d)1(a+2d)(a+3d)1(a+3d)(a+4d)|

Taking 1a(a+d)(a+2d) common from R1,

1(a+d)(a+2d)(a+3d) from R2,1(a+2d)(a+3d)(a+4d) from R3Δ=1a(a+d)2(a+2d)3(a+3d)2(a+4d)(a+d)(a+2d)(a+2d)a(a+2d)(a+3d)(a+3d)(a+d)(a+3d)(a+4d)(a+4d)(a+2d)

Δ=1a(a+d)2(a+2d)3(a+3d)2(a+4d)Δ

where, Δ=|(a+d)(a+2d)(a+2d)a(a+2d)(a+3d)(a+3d)(a+d)(a+3d)(a+4d)(a+4d)(a+2d)|

Applying R2R2R1,R3R3R2

Δ=|(a+d)(a+2d)(a+2d)a(a+2d)(2d)dd(a+3d)(2d)dd|

Applying R3R3R2

Δ=|(a+d)(a+2d)(a+2d)a(a+2d)2ddd2d200|

Expanding along R3, we get

Δ=2d2|a+2dadd|Δ=(2d2)(d)(a+2da)=4d4Δ=4d4a(a+d)2(a+2d)3(a+3d)2(a+4d)



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