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Q.

Let  x∈ℝ and let P=111022003,Q=2xx040xx6 and R=PQP−1           Then which of the following options is/are correct

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a

R=det2xx040xx5+8 for all real values of x

b

For x = 1, there exists a unit vector αi^+βj^+γk^  for which

c

For x = 0, if  R1ab=61ab  then a + b = 5

d

There exists a real number x such that PQ = QP

answer is A.

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Detailed Solution

|R|=PQP-1=|Q|=412-x2=410-x2+8=2xx040xx5+8 If x=1⇒|R|=44≠0⇒α=β=γ=0 For x=0⇒R=111022003200040006166-3003-2002=16126402480036R-6I=-41230-243000-4+a+2b3=0,-2a+4b3=0⇒b=3,a=2⇒a+b=5 If PQ=QP⇒ no value of
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