Matrices and Determinants 1 Question 3

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3. Let A=cosαsinα sinαcosα,(αR) such that A32=01 10. Then, a value of α is

======= ####3. Let A=cosαsinαsinαcosα,(αR) such that A32=0110. Then, a value of α is

3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed

(a) π32

(b) 0

(c) π64

(d) π16

(2019 Main, 8 April I)

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Answer:

Correct Answer: 3. (c)

Solution:

  1. Given, matrix A=[cosαsinαsinαcosα]

A2=[cosαsinαsinαcosα]cosαsinαsinαcosα]=[cos2αsin2αcosαsinαsinαcosαsinαcosα+cosαsinαsin2α+cos2α]=[cos2αsin2αsin2αcos2α]

Similarly,

An=[cos(nα)sin(nα)sin(nα)cos(nα)],nNA32=[cos(32α)sin(32α)sin(32α)cos(32α)]=[0110] (given) 

So, cos(32α)=0 and sin(32α)=1

32α=π2α=π64



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