Answer:
Option B,C
Explanation:
Here, P=[3−1−220α3−50]
Now, |P|=3(5α)+1(−3α)−2(−10)=12α+20 ...........(i)
∴ adj(P)= [5α2α−10−10612α−(3α+4)2]T
= [5α−10−α2α6−3α−4−10122] .....(ii)
As, PQ=kI
⇒ |P||Q|=|kI|
⇒ |P||Q|=k3
⇒ |P|(k22)=k3 [given,|Q|=k22]
⇒ |P|=2k ..............(iii)
∵ PQ=ki
∵ Q=kp−1I
=k.adjP|P|=k(adjP)2k [ from Eq. (iii)]
=adjP2
=12[5α−10−α2α6−3α−4−10122]
∴ q23=−3α−42[given,q23=−k8]
⇒ −(3α+4)2=−k8
⇒ (3α+4)×4=k
⇒ 12α+16=k .......(iv)
From Eq .(iii) . |P|=2k
⇒ 12α+20=2k
[ from Eq. (i)].......(v) On solving Eqs. (iv) and (v) , we get α=−1 and K=4 ........(vi) ∴ 4α−k+8=−4−4+8=0 ∴ Option (b) is correct.Now, |P adj (Q)| =|P| |adj Q|= 2k(k22)2=k52=2102=29 ∴ Option (c) correct