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,` thenA. `alpha = 0, k=8`B. `4alpha -k + 8 =0`C. `det (padj(Q) ) = 2^(9)`D. `det (Qadj(P) ) = 2^(13)`

Let `p=[(3,-1,-2),(2,0,alpha),(3,-5,0)],` where `alpha in RR.` Suppose

1.	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,` thenA. `alpha = 0, k=8`B. `4alpha -k + 8 =0`C. `det (padj(Q) ) = 2^(9)`D. `det (Qadj(P) ) = 2^(13)`
Answer» Correct Answer - B::C `because PQ = kI rArr (P.Q)/k = I rArr P^(-1) = Q/k" " ...(i)` Also `abs(P) = 12 alpha +20 " " (ii)` and `" "`given ` q_(23) = (-k)/8` Comoaring the third element of `2^(nd)` row no both sides, we get `1/((12alpha + 20))(-(3alpha + 4)) = 1/k xx (-k)/8` `rArr 24 alpha + 32= 12 alpha + 20` ` alpha = -1` ...(iii) From (ii), `abs(P)=8` ...(iv) Also `PQ= kI` `rArrabs(PQ) = abs(kI)` `rArr abs(P) abs(Q) = k^(3)` `rArr 8xx (k^(2))/2 = k^(3) " " (because abs(P) = 8, abs(Q) = k^(2)/2)` `therefore k= 4 " " ...(v)` (b) `4 alpha - k + 8 = -4 -4 + 8 = 0` (c) `det(P " adj" (Q)) = abs(P) abs("adj" Q) = abs(P) abs(Q)^(2) = 8xx8^(2) = 2^(9)` (d) `det(Q " adj" (P)) = abs(Q) abs("adj" P) = abs(Q) abs(P)^(2) = 8xx8^(2) = 2^(9)`

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