Let x ∈ R and let P = 1 1 1 0 2 2 0 0 3 ,   Q = 2 x x 0 4 0 x x 6 and R = P Q P - 1 . Then which…

Let x∈R and let P=111022003, Q=2xx040xx6 and R=PQP-1.
Then which of the following options is/are correct?
  1. For x=1, there exists a unit vector αi^+βj^+γk^ for which Rαβγ=000
  2. There exists a real number x such that PQ=QP
  3. det⁡R=det⁡2xx040xx5+8, for all x∈R
  4. For x=0, if R1ab=61ab, then a+b=5

Solution

P=111022003Q=2xx040xx6
Option (C)
Now R=PQP-1
det⁡R=det⁡Rdet⁡Qdet⁡P-1
det⁡R=det⁡Qdet⁡P-1=1det⁡P
det⁡R=det⁡2xx040xx6
det⁡R=48-4x2
now det⁡2xx040xx5=40-4x2
det⁡R=det⁡2xx040xx5+8
Option A
Rαβγ=000 must have not trivial solution
So det⁡R=0
48-4x2=0⇒x=±23
Option D
R1ab=66a6b
PQP-11ab=66a6b …(i)
P-1=166-3003-2002
Putting P, Q, P-1 in equation (i)
161264024800361ab=66a6b
⇒12+6a+4b=36
24a+8b=36a
a=2 and b=3
a+b=5
Option B
PQ=QP
PQ=QP⇒PQP-1=Q
R=Q
Not possible for any value of x.

Asked in: JEE Advanced 2019 (Paper 2)

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