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

Discussion

No Comment Found

Related InterviewSolutions

Let `A=" diag "[3,-5,7]" and "B=" diag "[-1,2,4].` ltbr. Find (i) (A+B) (ii) (A-B) (iii) -5A (iv) (2A+3B).`
A2=A THEN FIND VALUE OF (I+A)2 -3A
If `A=[(1,2),(2,1)]` and `f(x)=(1+x)/(1-x)`, then f(A) isA. `[(1,1),(1,1)]`B. `[(2,2),(2,2)]`C. `[(-1,-1),(-1,-1)]`D. none of these
Id `[1//25 0x1//25]=[5 0-a5]^(-2)`, then the value of `x`is`a//125`b. `2a//125`c. `2a//25`d. none of theseA. `a//125`B. `2a//125`C. `2a//25`D. none of these
If `A=[(a+ib,c+id),(-c+id,a-ib)]` and `a^(2)+b^(2)+c^(2)+d^(2)=1`, then `A^(-1)` is equal toA. `[(a-ib,-c-id),(c-id,a+ib)]`B. `[(a+ib,-c+id),(-c+id,a-ib)]`C. `[(a-ib,-c-id),(-c-id,a+ib)]`D. none of these
If the matrix A is such that `({:(1,3),(0,1):})A=({:(1,1),(0,-1):})`, then what is A equal to ?A. `{:[(1,4),(-1,0)]:}`B. `{:[(1,-4),(1,0)]:}`C. `{:[(1,-4),(0,-1)]:}`D. none of these
If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`A.B.C.D.
If `A=[[cos alpha, -sin alpha] , [sin alpha, cos alpha]], B=[[cos2beta, sin 2beta] , [sin 2 beta, -cos2beta]]` where `0 lt beta lt pi/2` then prove that `BAB=A^(-1)` Also find the least positive value of `alpha` for which `BA^4B= A^(-1)`
If `A(alpha, beta)=[("cos" alpha,sin alpha,0),(-sin alpha,cos alpha,0),(0,0,e^(beta))]`, then `A(alpha, beta)^(-1)` is equal toA. `A(-alpha, -beta)`B. `A(-alpha, beta)`C. `A(alpha, -beta)`D. `A(alpha, beta)`
`{:A=[(cos^2 alpha,cosalphasinalpha),(cosalphasinalpha,sin^2alpha)]:}` `{:B=[(cos^2 beta,cosbetasinbeta),(cosbetasinbeta,sin^2beta)]:}` are two matrices such that the product AB is the null matrix, then `(alpha-beta)` is

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

Discussion

No Comment Found

Related InterviewSolutions

Reply to Comment