InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 851. |
If `A=[(1,-1,1),(0,2,-3),(2,1,0)],B=(adjA)` and `C=5A,` then `(|adjB|)/(|C|)=`A. 5B. 25C. `-1`D. 1 |
|
Answer» Correct Answer - d |
|
| 852. |
If A is a square matrix of order `n xx n` and k is a scalar, then `adj (kA)` is equal to (1) `k adj A` (2) `k^n adj A` (3) `k^(n-1) adj A` (4) `k^(n+1) adj A`A. k adj AB. `k^n adj A`C. `k^(n-1)adjA`D. `k^(n+1)adjA` |
| Answer» Correct Answer - C | |
| 853. |
If `A=[(cosx,sinx),(-sinx,cosx)] and A.(adjA)=k[(1,0),(0,1)]` then the value of `k` isA. `sinx cos x`B. 1C. 2D. 3 |
| Answer» Correct Answer - B | |
| 854. |
If `A`is a singular matrix, then adj `A`isa. singularb. non singularc. symmetric d.not definedA. singularB. non-singularC. symmetricD. not defined |
| Answer» Correct Answer - A | |
| 855. |
If `A`is a singular matrix, then adj `A`isa. singularb. non singularc. symmetric d.not definedA. identity matrixB. null matrixC. scalar matrixD. none of these |
| Answer» Correct Answer - B | |
| 856. |
If `A`is a singular matrix, then adj `A`isa. singularb. non singularc. symmetric d.not definedA. non-sigularB. singularC. symmetricD. not defined |
|
Answer» Correct Answer - B We have, `absA=0. Let A be a square matrix of order n. Then, `abs(adjA)abs(A)^(n-1)` `rArr abs(adjA)=0` `rArr " adj A is a singular matrix."` |
|
| 857. |
If `A`is a singular matrix, then adj `A`isa. singularb. non singularc. symmetric d.not definedA. singularB. non-singularC. symmeticD. not defined |
| Answer» Correct Answer - D | |
| 858. |
If `A=[(1,2),(3,4),(5,6)]` and `B=[(-3,-2),(1,-5),(4,3)]`, then find `D=[(p,q),(r,s),(t,u)]` such that `A+B-D=O`. |
|
Answer» Here `A+B-D=O` `:. D=A+B` `implies [(p,q),(r,s),(t,u)]=[(1,2),(3,4),(5,6)]+[(-3,-2),(1,-5),(4,3)]` `= [(1-3,2-2),(3+1,4-5),(5+4,6+3)]=[(-2,0),(4,-1),(9,9)]` `D=[(-2,0),(4,-1),(9,9)]` |
|
| 859. |
Find the values of x for which matrix `[(3,-1+x,2),(3,-1,x+2),(x+3,-1,2)]` is singular. |
|
Answer» Given matrix is singular `:. |(3,x-1,2),(3,-1,x+2),(x+3,-1,2)|=0` `implies |(0,x,-x),(3,-1,x+2),(x+3,-1,2)|=0" "[R_(1) rarr R_(1)-R_(2)]` `implies |(0,x,-x),(-x,0,x),(x+3,-1,2)|=0" "[R_(2) rarr R_(2)-R_(3)]` `implies |(0,x,0),(-x,0,x),(x+3,-1,1)|=0" "[C_(3) rarr C_(3)+C_(2)]` `implies -x[(-x)-x(x+3)]=0` `implies x(x^(2)+4x)=0` `implies x=0, -4` |
|
| 860. |
Show that the matrix `A = [[1 , a,alpha , aalpha],[1, b, beta, b beta ],[1 ,c,gamma ,cgamma ]]` is of renk 3 provided no two of a, b, c are equal and no two of `alpha ,beta,gamma ` are equal.A.B.C.D. |
|
Answer» We have , `A= [[1,a, alpha , aalpha],[1, b ,beta,b beta ],[1 ,c, gamma,cgamma]]` Applying `R_(2) rarr R_(2) - R_(1) and R_(3) rarr R_(3) -R_(1),` we get `A= [[1,a, alpha , aalpha],[0, b-a ,beta-alpha,b beta-aalpha ],[0,c-a, gamma-alpha,cgamma-aalpha]]` Applying `C_(2) rarr C_(2) - aC_(1), C_(3) rarr C_(3)- alphaC_(1) and C_(4) rarr C_(4) - a alpha C_(1), ` we get `A= [[1,0, 0 , 0],[0, b-a ,beta-alpha,b beta-aalpha ],[0,c-a, gamma-alpha,cgamma-aalpha]]` Applying ` C_(4) rarr C_(4) - alpha C_(2) - bC_(3)` we get `A= [[1,0, 0 , 0],[0, b-a ,beta-alpha,0 ],[0,c-a, gamma-alpha,(c-b)(gamma-alpha)]]` For `p(A)= 3` `c- a ne 0 , gamma - alpha ne 0, c-bne 0, b-a ne0 , beta - alpha ne 0 ` i.e.,`ane b, bnec, cnea and alpha ne beta, beta ne gamma , gamma ne alpha ` |
|
| 861. |
For `alpha, beta, gamma in R`, let `A=[(alpha^(2),6,8),(3,beta^(2),9),(4,5,gamma^(2))]` and `B=[(2alpha,3,5),(2,2beta,6),(1,4,2gamma-3)]` |
|
Answer» We have, tr (A)=tr (B), `:. alpha^(2)+beta^(2)+gamma^(2)=2alpha+2beta+2 gamma-3` `implies (alpha^(2)-2alpha+1)+(beta^(2)-2beta+1)+(gamma^(2)-2gamma+1)=0` `implies (alpha-1)^(2)+(beta-1)^(2)+(gamma-1)^(2)=0` `implies alpha=1, beta=1, gamma=1` `:. (alpha^(-1)+beta^(-1)+gamma^(-1))=(1+1+1)=3` |
|
| 862. |
If `A=((alpha,beta),(gamma,-alpha))`. such that `A^2=I`. prove that `1-alpha^2-beta gamma=0` |
|
Answer» Here, `A = [[alpha, beta],[gamma, -alpha]]` We are given, `A^2 = I` `:. [[alpha, beta],[gamma, -alpha]] [[alpha, beta],[gamma, -alpha]] = [[1,0],[0,1]]` `=>[[alpha^2+betagamma, alphabeta -alphabeta],[gammaalpha-gammaalpha, betagamma+alpha^2]] = [[1,0],[0,1]]` `=> alpha^2+betagamma = 1` `=>1-alpha^2-betagamma = 0` |
|
| 863. |
If `alpha, beta, gamma` are three real numbers and `A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),cos (gamma-beta),1)]` then which of following is/are true ?A. A is singularB. A is symmetricC. A is orthogonalD. A is not invertible |
|
Answer» Correct Answer - A::B::D `A=[(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)][(cos alpha,cos beta,cos gamma),(sin alpha,sin beta,sin gamma),(0,0,0)]` Clearly, A is symmetric and `|A|=0`, hence, singular and not invertiable. Also, `A A^(T) ne 1` |
|
| 864. |
Let `A=[a_(ij)]_(mxxn)` be a matrix such that `a_(ij)=1` for all I,j. Then ,A. rank (A) gt 1B. rank (A) = 1C. rank (A) = mD. rank (A) = n |
| Answer» Correct Answer - B | |
| 865. |
If `S=[(0,1,1),(1,0,1),(1,1,0)]` and `A=[(b+c,c-a,b-a),(c-b,c+b,a-b),(b-c,a-c,a+b)]` `(a, b, c ne 0)`, then `SAS^(-1)` isA. symmetric matrixB. diagonal matrixC. invertible matrixD. singular matrix |
|
Answer» `S[(0,1,1),(1,0,1),(1,1,0)]` or `S^(-1)=1/2 [(-1,1,1),(1,-1,1),(1,1,-1)]` We have, `SA=[(0,1,1),(1,0,1),(1,1,0)][(b+c,c-a,b-a),(c-b,c+a,a-b),(b-c,a-c,a+b)]` `=[(0,2a,2a),(2b,0,2b),(2c,2c,0)]` `:. SAS^(-1)=[(0,2a,2a),(2b,0,2b),(2c,2c,0)] 1/2 [(-1,1,1),(1,-1,1),(1,1,-1)]` `=[(0,a,a),(b,0,b),(c,c,0)][(-1,1,1),(1,-1,1),(1,1,-1)]` `=[(2a,0,0),(0,2b,0),(0,0,2c)]` `=` diag `(2a, 2b, 2c)` |
|
| 866. |
If `D_(1)` and `D_(2)` are two `3xx3` diagonal matrices, then which of the following is/are true ?A. `D_(1)D_(2)` is a diagonal matrixB. `D_(1)D_(2)=D_(2)D_(1)`C. `D_(1)^(2)+D_(2)^(2)` is a diagonal matrixD. none of these |
| Answer» All are properties of diagonal matrix. | |
| 867. |
Let A be a squarre matrix of order of order 3 satisfies the matrix equation `A^(3) -6 A ^(2) + 7 A - 8 I = O and B = A- 2 I . ` Also, `det A = 8.` If `adj((B/2)^(-1))= (p/q)B`, where `p,q in N` , the least valuse of `(p+q)` is equal toA. 7B. 9C. 29D. 41 |
|
Answer» Correct Answer - A `because B = A -2I ` `therefore A^(-1) B = I - 2A^(-1)` ...(i) `adj[(B/2)^(-1)]= (B/2)/abs(B/2) = (B/2)/(1/8abs(B)) = (4B)/abs(B) = 4/10 B [because abs(B) = 10 ]` `= 2/5 B = p/q B` [given] `therefore p= 2 and q = 5 ` Hence, p + q = 7 |
|
| 868. |
IF ` A=[(1,1,0),(2,1,5),(1,2,1)],`then `a_(11) A_(21) +a_(12)+a_(12)A_(22)+a_(13)A_(23)` =A. 1B. 0C. `-1`D. 2 |
|
Answer» Correct Answer - b |
|
| 869. |
If matrix `[(1,2,-1),(3,4,5),(2,6,7)]` and its inverse is denoted by `A^(-1)=[(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(31),a_(32),a_(33))]` value of `a_(32)=`A. `(1)/(34)`B. `(-2)/(17)`C. `(1)/(17)`D. `(-1)/(17)` |
|
Answer» Correct Answer - d |
|
| 870. |
Let A be the `2 xx 2` matrix given by `A=[a_("ij")]` where `a_("ij") in {0, 1, 2, 3, 4}` such theta `a_(11)+a_(12)+a_(21)+a_(22)=4` then which of the following statement(s) is/are true ?A. Number of matrices A such that the trace of A equal to 4, is 5B. Number of matrices A, such that A is invertible is 18C. Absolute difference between maximum value and minimum value of det (A) is 8D. Number of matrices A such that A is either symmetric (or) skew symmetric and det (A) is divisible by 2, is 5. |
|
Answer» (1) Possible matrices are `[(1,0),(0,3)], [(3,0),(0,1)], [(2,0),(0,2)], [(4,0),(0,0)], [(0,0),(0,4)]` (2) Using 0, 0, 2, 2, there are two matrices which are invertible which are `[(2,0),(0,2)]` and `[(0,2),(2,0)]` Using 0, 0, 1, 3, there are four matrices which are invertible. Using 0, 1, 1, 2, there are twelve matrices which are invertible. Using 0, 0, 0, 4 and using 1, 1, 1, 1 no matrix is formed, which isinvertiable. `:.` Total number of matrices `= 18` (3) `|4-(-4)|=8` (4) There are five matrices, which are either symmetric or skew symetric and whose determinant is divisible by 2. These are `[(2,0),(0,2)], [(0,2),(2,0)], [(0,0),(0,4)], [(4,0),(0,0)], [(1,1),(1,1)]` |
|
| 871. |
If `A= [a_(ij)] _(nxxn)` and f is a function, we define `f (A) = [f(a_(ij))] _(nxxn)."Let"A = [[pi/2-theta,theta],[-theta,pi/2-theta]]` thenA. sin A is invertibleB. sin A = cos AC. sin A is orthogonalD. sin 2 A=2 sin A cos A |
|
Answer» Correct Answer - A::C `sin A = [[cos theta ,sin theta],[-sin theta , cos theta]] and cos theta = [[sin theta , cos theta ],[cos theta , sin theta ]]` `therefore abs(sin A) cos^(2) theta + sin ^(2) theta = 1 ne 0 ` Hence, sin A is incertible. Also, `(sinA)(sin A)^(T) = [[cos theta ,sin theta],[-sin theta , cos theta]][[cos theta ,-sin theta],[sin theta , cos theta]]` ` = [[1,0],[0,1]] = I` Hence, sin A is orthogonal. Also ,`2sinA" " sin A = 2[[cos theta ,sin theta],[-sin theta , cos theta]][[sin theta,cos theta ],[ cos theta,sin theta]]` `= 2 [[sin 2theta, 1 ],[cos 2 theta, 0]] ne sin 2 A` |
|
| 872. |
Let A be a square matrix of order 3 satisfies the relation `A^(3)-6A^(2)+7A-8I=O` and `B=A-2I`. Also, det. `A=8`, thenA. det. `("adj. "(I-2A^(-1))=25/16`B. adj. `((B/2)^(-1))=B/10`C. det. `("adj. "(I-2A^(-1)))=75/32`D. adj. `((B/2)^(-1))=(2B)/5` |
|
Answer» Given that `B=A-2I` ...(i) And `A^(3)-6A^(2)+7A-8I=O` `:. (A-2I)^(3)=5A` ...(ii) `:. B^(3)=5A` `implies |B^(3)|=|5A|=5^(3)|A|=5^(3)xx8` `implies |B|=10` Now, from (i), we get `:. A^(-1) B=I-2A^(-1)` `:. |"adj. "(I-2A^(-1))|=|I-2A^(-1)|^(2)` `=|A^(-1)B|^(2)=((|B|)/(|A|))^(2)` `=(5/4)^(2)` `=25/16` adj `((B/2)^(-1))=("adj."(B/2))^(-1)=("adj."("adj." B/2))/(|"adj." B/2|)` `=(|B/2|B/2)/(|B/2|^(2))=(B/2)/(|B/2|)=(4B)/(|B|)=(2B)/5` |
|
| 873. |
If `A=[(cos theta,sin theta,0),(-sin theta,cos theta,0),(0,0,1)]`, where `A_(11), A_(12), A_(13)` are co-factores of `a_(11), a_(12), a_(13)` respectively , then the value of `a_(11)A_(11)+a_(12)A_(12)+a_(13)=`A. `-1`B. 1C. 0D. `(1)/(2)` |
|
Answer» Correct Answer - b |
|
| 874. |
If the adjoint of a 3 3 matrix P is1 4 4 2 1 7 1 1 3, then the possible value(s) of thedeterminant of P is (are) (A) 2 (B) 1 (C) 1 (D) 2A. -2B. -1C. 1D. 2 |
|
Answer» Correct Answer - A::D Given, `"adj " P = [[1,4,4],[2,1,7],[1,1,3]]` `rArr abs(adj P )= abs((1,4,4),(2,1,7),(1,1,3))` `= 1(-4) - 4 (-1) + 4 (1) = 4` `rArr abs(P)^(3-1) = 4` `rArr abs(P) = pm 2` |
|
| 875. |
If A is a 3x3 matrix and B is its adjoint matrix the determinant of B is 64 then determinant of A isA. 64B. `ppm64`C. `pm8`D. 18 |
|
Answer» Correct Answer - C We have, `absB=64` `rArr abs(adjA)=64rArr abs(A)^2=64rArr absA=pm8`. |
|
| 876. |
Let `A = [(1,0,0), (2,1,0), (3,2,1)],` and `U_1, U_2 and U_3` are columns of a `3 xx 3` matrix `U`. If column matrices `U_1, U_2 and U_3` satisfy `AU_1 = [(1),(0),(0)], AU_2 = [(2),(3),(0)], AU_3 = [(2),(3),(1)]` then the sum of the elements of the matrix `U^(-1)` isA. -1B. 0C. 1D. 3 |
|
Answer» Correct Answer - B `because " Adj"U = ((-1,-2,0),(-7,-5,-3),(9,-6,3))` `therefore U^(-1) =("Adj"U)/abs(U) = ("Adj"U)/3` `rArr` sum of the elements of `U^(1) = 1/3(-1-2+0-7-5-3+9+6+3)=0` |
|
| 877. |
Let `A = [(1,0,0), (2,1,0), (3,2,1)],` and `U_1, U_2 and U_3` are columns of a `3 xx 3` matrix `U`. If column matrices `U_1, U_2 and U_3` satisfy `AU_1 = [(1),(0),(0)], AU_2 = [(2),(3),(0)], AU_3 = [(2),(3),(1)]` then the sum of the elements of the matrix `U^(-1)` isA. 6B. 0(zero)C. 1D. `2//3` |
|
Answer» Correct Answer - b |
|
| 878. |
If `I` is a unit matrix of order `10`, then the determinant of `I` is equal toA. 10B. 1C. `(1)/(10)`D. 9 |
| Answer» Correct Answer - B | |
| 879. |
If A is a 3x3 matrix and B is its adjoint matrix the determinant of B is 64 then determinant of A isA. 64B. `pm 64`C. `pm 8`D. 18 |
| Answer» Correct Answer - C | |
| 880. |
If the adjoint of a 3 3 matrix P is1 4 4 2 1 7 1 1 3, then the possible value(s) of thedeterminant of P is (are) (A) 2 (B) 1 (C) 1 (D) 2A. ` pm 2`B. `pm 1`C. `pm 3`D. `pm 4` |
|
Answer» Correct Answer - a |
|
| 881. |
If `d`is the determinant of asquare matrix `A`of order `n`, then the determinantof its adjoint is`d^n`(b) `d^(n-1)`(c) `d^(n+1)`(d) `d`A. `d^(n)`B. `d^(n-1)`C. `d^(n+1)`D. d |
|
Answer» Correct Answer - b |
|
| 882. |
Let `{:A=[(1,0,0),(2,1,0),(3,2,1)]:}and U_1,U_2,U_3` be column matrices satisfying `{:AU_1[(1),(0),(0)],AU_2[(2),(3),(6)],AU_3[(2),(3),(1)]:}`.If U is `3xx3` matrix whose columns are `U_1,U_2,U_3," then "absU=`A. 3B. -3C. `3//2`D. 2 |
|
Answer» Correct Answer - A Let `{:U_1[(a),(b),(c)],U_2[(p),(q),(r)],andU_3[(x),(y),(z)]:}` Then `{:AU_1[(1),(0),(0)]rArr[(a),(2a+b),(3a+2b+c)]=[(1),(0),(0)]rArra=1,b=-2, c=1:}` `{:AU_2[(2),(3),(0)]rArr[(p),(2p+q),(3p+2q+r)]=[(2),(3),(0)]rArrp=2,q=-1,r=4:}` and, `{:AU_2[(2),(3),(1)]rArr[(x),(2x+y),(3x+2y+z)]=[(2),(3),(1)]rArrx=2,y=-1,z=-3:}` `:. U{:abs((U_1,U_2,U_3))=[(1,2,2),(-2,-1,-1),(1,-4,-3)]:}` `rArr absU={:abs((1,2,2),(-2,-1,-1),(1,-4,-3)) =abs((1,2,0),(-2,-1,0),(1,-4,1)):}" Applying " C_3 to C_3-C_2` `rArr absU=3` |
|
| 883. |
If `A=[(alpha,2),(2,alpha)]` and determinant `(A^3)=125,` then the value of `alpha` is (a) `+-1` (b) `+-2` (c) `+-3` (d) `+-5`A. `pm1`B. `pm2`C. `pm3`D. `pm5` |
|
Answer» Correct Answer - C We have, `absA={:abs((alpha,2),(2,alpha))=alpha^2-4:}` `:. Abs(A^3)=125` `rArr abs(A)^3=125` `rArr(alpha-4)^3-125rArr alpha^2=9rArr alpha=pm3`. |
|
| 884. |
If `A=[(1,0,0),(0,1,1),(0,-2,4)],6A^-1=A^2+cA+dI,` then `(c,d)=`A. (-6,11)B. (-11,6)C. (11,6)D. (6,11) |
|
Answer» Correct Answer - A Every square matrix A satisfies its characteristic equation i.e. `abs(A-lambdal)=0`. Here, `abs(A_lambdal)=0` `{:rArr abs((1-lambda,0,0),(0,1-lambda,1),(0,-2,4-lambda))=0:}` `rArr (1-lambda){(1-lambda)(4-lambda)+2}=0` `rArr lambda^3-6lambda^2+11lambda-6=0` `rArr A^3 -6A^2+11A-6l=0` `rArr 6I=A^3-6A^2+11A` `rArr 6A^(-1)=A^2-6A+11I` [Multiplying both sides by `A^(-1)`] `:. 6A^(-1) =A^2=cA+dl rArrc=-6and d=11` |
|
| 885. |
If `{:P=[(sqrt3/2,1/2),(1/2,sqrt3/2)],A=[(1,1),(0,1)]:}and Q=PAP^T," then" P^TQ^2015P`, isA. `{:[(1,2005),(0,1)]:}`B. `{:[(1,2005),(2005,1)]:}`C. `{:[(1,0),(2005,1)]:}`D. `{:[(1,0),(0,1)]:}` |
|
Answer» Correct Answer - A We have, `P P^T={:[(sqrt3/2,1/2),(-1/2,sqrt3/2)][(sqrt3/2,-1/sqrt2),(1/2,sqrt3/2)]=[(1,0),(0,1)]=I:}` `:. P^T=P^(-1)` Now, `Q=PAP^T` `rArr P^TQP=P^T(PAP^T)P` `rArr P^TQP=(P^TP)A(P^TP)=IAI=A={:[(1,1),(0,1)]:}` `and, Q_2=(PAP^T)(PAP^T)=PA(P^TP)AP^T=PA^2P^T` `:. P^TQ^2P=P^T(PA^2P^T)P=(P^TP)A^2(P^TP)=IA^2` `rArr P^TQ^2P=A^2={:[(1,2),(0,1)]:}` Continuing in this manner, we have `P^TQ^2005P=A^2005=`{:[(1,2005),(0,1)]:}` |
|
| 886. |
If `{:P=[(sqrt3/2,1/2),(1/2,sqrt3/2)],A=[(1,1),(0,1)]:}and Q=PAP^T," then" P^TQ^2015P`, isA. `{:[(2015,1),(1,2015)]:}`B. `{:[(1,2015),(0,1)]:}`C. `{:[(0,2015),(0,0)]:}`D. `{:[(2015,0),(1,2015)]:}` |
|
Answer» Correct Answer - B We have, `{:P=[(sqrt3/2,1/2),(-1/2,sqrt3/2)]rArr P^T=[(sqrt3/2,-1/2),(1/2,sqrt3/2)]:}` Clearly, `P^T=P^(-1)` `:. Q=PAP^T=PAP^(-1)` `rArr Q^2015=(PAP^(-1))^2015=PA^2015P^(-1)` `:. P^TQ^2015P=P^(-1)(PA^2015P^(-1))P=A^2015` Now, `A={:[(1,1),(0,1)]:}` `:.{:A^2=[(1,1),(0,1)][(1,1),(0,1)]=[(1,2),(0,1)]:}` `{:A^3=A^2A=[(1,2),(0,1)][(1,1),(0,1)]=[(1,3),(0,1)]:}`and so no. In general `{:A^2015=[(1,2015),(0,1)]:}` Hence, `P^TQ^2015P=A^2015={:[(1,2015),(0,1)]:}`. |
|
| 887. |
if A and B are two matrices such that AB=B and BA=A,then `A^(2)+B^(2)` is equal toA. 2ABB. 2ABC. A+BD. AB |
| Answer» Correct Answer - C | |
| 888. |
If `P= [[sqrt(3)/2, 1/2],[-1/2 , sqrt(3)/ 2]], A = [[1,1],[0,1]]and Q= PAP^(T)` , the ltbr. `P^(T)(Q^(2005)) P` equal toA. `[[1,2005],[0,1]]`B. `[[sqrt(3)//1,2005],[1,0]]`C. `[[1, 2005],[sqrt(3)//2,1]]`D. `[[1, sqrt(3)//2 ],[0, 2005]]` |
|
Answer» Correct Answer - A If `Q = PAP^(T) ` then `P^(T)Q=AP^(T) " "[because PP^(T)=I]` ` rArrP^(T) Q^(2005) P = AP^(T)Q^(2004)P` `= A^(2) P^(T) Q ^(2003) P = A^(3) P^(T) Q^(2002) P ` `= A^(2004)P^(T)(QP)` `= A^(2004)P^(T)(PA)" "[Q= PAP^(T) rArr QP= PA]` `=A^(2005)` `therefore A^(2005) = [[1,2005],[0,1]]` |
|
| 889. |
If `P=[[sqrt(3)/2,1/2],[-1/2,sqrt(3)/2]], A=[[1,1],[0,1]]` and `Q=PAP^T` and `X=P^TQ^(2005)P`, then `X` equal to: |
|
Answer» `P = [[sqrt3/2,1/2],[-1/2, sqrt3/2]]` `:. P^T = [[sqrt3/2,-1/2],[1/2, sqrt3/2]]` Now, `P^TP = [[1,0],[0,1]] = I` Now, `P^TQ^2005P = P^T(PAP^T)^2005P = P^T(PAP^T)(PAP^T)(PAP^T)...P = A^2005` Here, `A = [[1,1],[0,1]]` `A^2 = A.A = [[1,1],[0,1]][[1,1],[0,1]] = [[1,2],[0,1]]` `A^3 = A^2.A = [[1,2],[0,1]][[1,1],[0,1]] = [[1,3],[0,1]]` `:. A^2005 = [[1,2005],[0,1]]` `:. P^TQ^2005P = [[1,2005],[0,1]]` `:. X = [[1,2005],[0,1]]` |
|
| 890. |
If A and B are two matrices such that AB=B and BA=A , then `A^2+B^2=`A. 2 ABB. 2 BAC. A + BD. AB |
|
Answer» Correct Answer - C Proceeding as in Example 2, we have `A^2=Aand B^2=B` `:. A^2+B^2=A+B` |
|
| 891. |
If `x=c y+b z ,y=a z+c x ,z=x+a y ,w h e r ex ,y ,z`are not allzeros, then find the value of `a^2b^2c^2+2a b cdot`A. 2B. -1C. 0D. 1 |
|
Answer» Correct Answer - D It is given that there are x,y,z not all zero satisfying `x =cy +bz, y=az+cxand z=bx+zy`, So, the system of equations `x-cy-bz=0` `cx-y+az=0` `bx+ay-z=0` have non-trivial solutions. `:.{:[(1,-c,-b),(c,-1,a),(b,a,-1)]=0:}` `rArr 1-a^2-c^2-b^2-2abc=0rArr a^2+b^2+c^2+2abc=1` |
|
| 892. |
The number of possible value of `theta` lies in `(0,pi)`, such that system of equation `x+3y+7z=0`, `-x+4y+7z=0`, `xsin3theta+ycos2theta+2z=0` has non trivial solution is/are equal to (a) 2 (b) 3 (c) 5 (d) 4A. oneB. twoC. threeD. none of these |
|
Answer» Correct Answer - D The given system of equations has a non-trivial solution `{:abs((1,3,7),(-1,4,7),(sin3theta,cos2theta,2))=0:}` `rArr{:abs((0,7,14),(-1,4,7),(sin3theta,cos2theta,2))=0:}` [Applying `R_1toR_1+R_2`] `rArr -7(-2-7sin3theta)+14(-cos2theta-4sin3theta)=0` `rArr 2+7sin3theta-2cos2theta-8sin3theta=0` `rArr 2cos2theta +sin3theta=2` `rArr 2(1-2sin^2theta)+3sintheta-4sin^3theta=0` `rArr 4sin^3theta+4sin^2theta-3sintheta=0` `rArr sintheta(4sin^2theta+4sintheta-3)=0` `rArr sintheta (2sintheta-1)(2sintheta+3)=0` `rArr sintheta =0or, sintheta=1/2` `rArr theta=0,pi,pi/6,(5pi)/6` |
|
| 893. |
The system of linear equations`x+lambday-z=0``lambdax-y-z=0``x+y-lambdaz=0`has a non-trivial solution for :(1) infinitely many values of `lambda`.(2) exactly one value of `lambda`.(3) exactly two values of `lambda`.(4) exactlythree values of `lambda`.A. infinitely many value of `lambda`B. exactly one value of `lambda`C. exactly two values `lambda`D. exactly three values of `lambda` |
|
Answer» Correct Answer - D The given system of equations will have infinitely many non-trivial solutions, if `{:abs((1,lambda,-1),(lambda,-1,-1),(1,1 ,lambda))=0:}` `rArr (lambda +1)-lambda (-lambda^2+1)-1(lambda+1)=0` `rArr lambda^3-lambda =0 rArr lambda(lambda^2-1)-0rArrlambda=0,1,-1` |
|
| 894. |
If `x^2+y^2+z^2 ne0, x=cy+bz,y=az+cxandz=bx+ay" then "a^2+b^2+c^2+2abc=`A. 2B. a + b + cC. 1D. ab + bc + ca |
| Answer» Correct Answer - c | |
| 895. |
Consider the system of equations `a_1x+b_1y+c_1z=0` `a_2x+b_2y+c_2z=0` `a_3x+b_3y+c_3z=0` if `{:abs((a_1,b_1,c_1),(a_2,b_2,c_2),(a_3,b_3,c_3)):}=0`, then the system hasA. more than two solutionsB. one trivial and one non-trivial solutionsC. no solutionD. only trivial solution (0,0,0) |
| Answer» Correct Answer - a | |
| 896. |
The inverse of a diagonal matrix isa. a diagonal matrix b. a skew symmetric matrixc. a symmetric matrixd. none of theseA. a symmetric matrixB. a skew-symmetric matrixC. a diagonal matrixD. none of these |
| Answer» Correct Answer - C | |
| 897. |
Let `A ,B`be two matrices such that they commute. Show that for any positiveinteger `n ,`(i) `A B^n=B^n A`(ii) `(A B)^n=A^nB^n` |
|
Answer» Let `P(n):(AB)^(n)=A^(n)B^(n)` `therefore P(1):(AB)^(1)=A^(1)B^(1)rArrAB=AB` So P(1) is true Now, `P(k):(AB)^(k)rArrA^(k)B^(k) ,K in N` So, P(K) is true, whenever `P(k+1)` is true `therefore p(k+1:AB)^(k+1)=A^(k+1)B^(k+1)` `rArr AB^(k),AB^(1)` `rArr A^(k)B^(k).BArArrA^(k)B^(k+1)A` `rArr A^(k).A.B^(k+1)rArrA^(k+1)B^(k+1)` `rArr (A.B)^(k+1)=A^(k+1)B^(k+1)` So, P(k+1) is true for all `n in N` , whenever P(k) is true. By mathematical induction (AB)=`A^(n)B^(n)` is true for all `n in N`. |
|
| 898. |
If `A ` and `B` are two non-singular matrices which commute, then `(A(A+B)^(-1)B)^(-1)(AB)=`A. `A+B`B. `A^(-1)+B^(-1)`C. `A^(-1)+B`D. none of these |
|
Answer» Correct Answer - A `(a)` `(A(A+B)^(-1)B)^(-1)(AB)` `=B^(-1)(A+B^(-1))^(-1)A^(-1)(AB)` `=B^(-1)(A+B)A^(-1)(AB)` `=(B^(-1)A+I)A^(-1)(AB)` `=(B^(-1)A A^(-1)+A^(-1))(AB)` `=(B^(-1)+A^(-1))(AB)` `=B^(-1)AB+A^(-1)AB` `=B^(-1)BA+A^(-1)AB` `=A+B` |
|
| 899. |
If `A`is a skew-symmetric matrix and `n`is odd positive integer, then `A^n`isa skew-symmetric matrixa symmetric matrixa diagonal matrixnone of theseA. a symmetric matrixB. a skew-symmetric matrixC. a diagonal matrixD. none of these |
| Answer» Correct Answer - A | |
| 900. |
If `A_1, A_2, , A_(2n-1)a r en`skew-symmetric matrices of same order, then `B=sum_(r=1)^n(2r-1)(A^(2r-1))^(2r-1)`will besymmetricskew-symmetricneither symmetric nor skew-symmetricdata not adequateA. symmetricB. skew-symmetricC. neither symmetric nor skew-symmetricD. data not adequate |
|
Answer» Correct Answer - B `B=A_(1) +3 A_(3)^(3)+...+ (2n-1) (A_(2n-1))^(2n-1)` `B^(T)=-[A_(1)+3A_(3)^(3)+...+ (2n-1) (A_(2r-1))^(2r-1)]` `=-B` Hence, B is skew-symmetric. |
|