Matrices and Determinants 2 Question 8

9. Let a1,a2,a3.,a10 be in GP with ai>0 for i=1,2,,10 and S be the set of pairs (r,k),r,kN (the set of natural numbers) for which

|logea1ra2klogea2ra3klogea3ra4klogea4ra5klogea5ra6klogea6ra7klogea7ra8klogea8ra9klogea9ra10k|=0

Then, the number of elements in S, is (2019 Main, 10 Jan II)

(a) 4

(b) 2

(c) 10

(d) infinitely many

1b2 value of det(A)b is

(2019 Main, 10 Jan II)

(a) 3

(b) 23

(c) 23

(d) 3

Show Answer

Answer:

Correct Answer: 9. (c)

Solution:

  1. Given, |logea4ra5klogea5ra6klogea6ra7k logea7ra8klogea8ra9klogea9ra10k|=0

On applying elementary operations

C2C2C1 and C3C3C1, we get

logea1ra2klogea2ra3klogea1ra2k

logea4ra5klogea5ra6klogea4ra5k

logea7ra8klogea8ra9klogea7ra8k

logea3ra4klogea1ra2klogea6ra7klogea4ra5klogea9ra10klogea7ra8k∣=0

|logea1ra2klogea2ra3ka1ra2klogea3ra4ka1ra2klogea4ra5klogea5ra6ka4ra5klogea6ra7ka4ra5klogea7ra8klogea8ra9ka7ra8klogea9ra10ka7ra8k|=0logemlogen=logemn

[a1,a2,a3,a10 are in GP, therefore put a1=a,a2=aR,a3=aR2,,a10=aR9]

logear+kRklogear+kRr+2kar+kRklogear+kR3r+4klogear+kR4r+5kar+kR3r+4klogear+kR6r+7klogear+kR7r+8kar+kR6r+7klogear+kR2r+3kar+kRklogear+kR5+6kar+kR3r+4k=0logear+kR8r+9kar+kR6r+7k

logeR2r+2k=logeR2(r+k)=2logeRr+k]

So, value of determinant will be zero for any value of (r,k),r,kN.

Set ’ S ’ has infinitely many elements.



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