Let x∈ℝ and let P=111022003,Q=2xx040xx6 and R=PQP−1 Then which of the following options is/are correct
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
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.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
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