The correct option is C For x=0, if R⎡⎢⎣1ab⎤⎥⎦=6⎡⎢⎣1ab⎤⎥⎦, then a+b=5
Option 2:
L.H.S.
R=PQP−1
det R=|R|=|PQP−1|=|P||Q||P−1|=|P||Q|1|P|=|Q|
⇒|R|=|Q|=48−4x2
⎛⎜⎝∵det Q=|Q|=∣∣
∣∣2xx040xx6∣∣
∣∣=48−4x2⎞⎟⎠
R.H.S.
det ⎡⎢⎣2xx040xx5⎤⎥⎦+8=40−4x2+8=48−4x2=det R
Option 1:
If PQ=QP⇒PQP−1=QPP−1⇒R=Q
∵P−1=16⎡⎢⎣6−3003−2002⎤⎥⎦ and
R=PQP−1=16⎡⎢⎣111022003⎤⎥⎦⎡⎢⎣2xx040xx6⎤⎥⎦⎡⎢⎣6−3003−2002⎤⎥⎦
=16⎡⎢⎣6x+123x+64−10x12x248−4x18x036−6x⎤⎥⎦
Put x=0
R=⎡⎢
⎢
⎢
⎢
⎢⎣21230443006⎤⎥
⎥
⎥
⎥
⎥⎦
For any value of x, R≠Q
Therefore, PQ=QP doesn't exist for any real number x.
Option 3:
For x=0 R⎡⎢⎣1ab⎤⎥⎦=6⎡⎢⎣1ab⎤⎥⎦⇒⎡⎢
⎢
⎢
⎢
⎢⎣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=3⇒a+b=5
Option 4:
For x=1,R=⎡⎢
⎢
⎢
⎢⎣332−12423305⎤⎥
⎥
⎥
⎥⎦
R⎡⎢⎣αβγ⎤⎥⎦=⎡⎢⎣000⎤⎥⎦⇒⎡⎢
⎢
⎢
⎢
⎢⎣3α+32β−γ2α+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 R⎡⎢⎣αβγ⎤⎥⎦=⎡⎢⎣000⎤⎥⎦