Let `p=[(3,-1,-2),(2,0,alpha),(3,-5,0)],` where `alpha in RR.` Suppose `Q=[q_(ij)]` is a matrix such that `PQ=kI,` where `k in RR, k != 0 and I`is the identity matrix of order 3. If `q_23=-k/8 and det(Q)=k^2/2,` then

Options

A. `alpha0,k=8`
B. `4alpha-k+8=0`
C. `det(PadjQ)=2^9`
D. `det(Q adjP)2^13`

Answer

B, C

Explanation

We have,
`PQ =kl and det (Q) =k^2/2`
`rArr det(PQ)=det(kI) and det (Q)=k^2/2`
`rArr det (P) det(Q) =k^3and det (Q) =k^2/2`
`rArr k^2/2det(P) =k^3and det(Q) =k^2/2`
`rArr det (P) =2k and det (Q) =k^2/2`
Again, PQ = kI
`rArr Q=P^(-1) (kI) =kP^(-1)`
`rArr Q=k(1/absPadjP)`
`rArr Q=k(1/(2k) adjP)`
`rArr Q=1/2adjP`
`rArr q_23 =1/2(adjP) _23 =1/2" cofactor " P_32 =-1/2{:abs((3,-2),(2,alpha)):}`
`rArr =-k/8=-1/2(3alpha +4)`
`rArr k=12alpha +16 ...(i)`
Now, `det (P) =2K`
`rArr {:abs((3,-1,-2),(2,0,alpha),(3,-5,0))=2k:}`
`rArr 15 alpha-3alpha+20=2k`
`rArr k=6alpha+10`
Solving (i) and (ii), we get : `alpha=-1 k=4`.
`:. 4alpha-k+8=-4-4+8=0`
So, option (b) is correct.
Now,
`det(P(adjQ)) =abs(PadjQ)=absPabs(adjQ)`
`=2kabs(Q)^2=2k(k^2/2)^2=1/2k^5=2^9`
and,
`det (Q(adjP))= abs(QadjP)=absQ abs(adjP)`
`=absQabs(P)^2=k^2/2xx(2k)^2=2k^4=2xx(4)^45=2xx(4)^4=2^9`
So, option (c) is correct.