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

Options

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

Answer

B, C

Explanation

`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)`