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Let p=[(3,-1,-2),(2,0,alpha),(3,-5,0)], ...

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=kl,` where `k in RR, k != 0 and l` is the identity matrix of order 3. If `q_23=-k/8 and det(Q)=k^2/2,` then

A

`alpha=0, k=8`

B

`4alpha-k+8=0`

C

det `("P adj (Q)")=2^(9)`

D

det `("Q adj (P)")=2^(13)`

Text Solution

Verified by Experts

The correct Answer is:
B, C

`PQ=KI`
`:. Q=kP^(-1)I=kP^(-1)`
`:. Q=k ("adj (P)")/(|P|)=k/((12alpha+20)) [(-,-,-),(-,-,(-3alpha-4)),(-,-,-)]`
Comparing `q_(23)`, we get
`(-k)/8=(-k(3alpha+4))/((12alpha+20))`
`implies alpha=-1`
Also `PQ=kI`
`implies` det (P) . Det `(Q)=k^(3)`
`:. (12 alpha+20) k^(2)/2=k^(3)`
`:. k=6alpha+10=4`
`:. 4 alpha-k+8=0`
det (P adj (Q))= det (P) det (adj (Q))
`=8xx("det"(Q))^(2)`
`=8xx8^(2)`
`=2^(9)`
det (Q adj (P))= det (Q) det (adj (P))
`=2^(9)`
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