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Question

Let xR and let P=111022003, Q=2xx040xx6 and R=PQP1. Then which of the following options is/are correct?

A
There exists a real number x such that PQ=QP
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B
det R=det2xx040xx5+8, for all xR
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C
For x=0, if R1ab=61ab, then a+b=5
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D
For x=1, there exists a unit vector α^i+β^j+γ^k for which Rαβγ=000
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Solution

The correct option is C For x=0, if R1ab=61ab, then a+b=5
Option 2:
L.H.S.
R=PQP1
det R=|R|=|PQP1|=|P||Q||P1|=|P||Q|1|P|=|Q|
|R|=|Q|=484x2
det Q=|Q|=∣ ∣2xx040xx6∣ ∣=484x2
R.H.S.
det 2xx040xx5+8=404x2+8=484x2=det R

Option 1:
If PQ=QPPQP1=QPP1R=Q
P1=16630032002 and
R=PQP1=161110220032xx040xx6630032002

=166x+123x+6410x12x2484x18x0366x
Put x=0
R=⎢ ⎢ ⎢ ⎢ ⎢21230443006⎥ ⎥ ⎥ ⎥ ⎥
For any value of x, RQ
Therefore, PQ=QP doesn't exist for any real number x.

Option 3:
For x=0 R1ab=61ab⎢ ⎢ ⎢ ⎢ ⎢21230443006⎥ ⎥ ⎥ ⎥ ⎥1ab=66a6b⎢ ⎢ ⎢ ⎢ ⎢2+a+2b34a+4b36b⎥ ⎥ ⎥ ⎥ ⎥=66a6b
Comparing the element of both the matrix, we get
2+a+2b3=6 ...(i) and 4a+4b3=6a...(ii)
Solving both the equation
a=2,b=3a+b=5

Option 4:
For x=1,R=⎢ ⎢ ⎢ ⎢33212423305⎥ ⎥ ⎥ ⎥
Rαβγ=000⎢ ⎢ ⎢ ⎢ ⎢3α+32βγ2α+4β+23γ3α+5γ⎥ ⎥ ⎥ ⎥ ⎥=000
After solving, we will get trivial soloution i.e.,
α=β=γ=0
So for x=1, there is no unit vector α^i+β^j+γ^k for which Rαβγ=000

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