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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.
| 1051. |
If the trace of the matrix A is 5, and the trace of the matrix B is 7, then find the matrix (3A + 2B).A. 12B. 29C. 19D. None of these |
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Answer» Correct Answer - B `"Trace" (kA + mB) = k("Trace" A) + m("Trace" B).` |
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| 1052. |
If `A=[{:(2,-3),(4," "5):}]" and "B=[{:(-1,2),(0,3):}]`, find a matrix X such that `A+2B+X=O.` |
| Answer» `X=[{:(0,-1),(-4,-11):}]` | |
| 1053. |
If `A=[{:(4,2),(1,3):}]" and "B=[{:(-2,1),(3,2):}]`, find a matrix X such that `3A-2B+X=O.` |
| Answer» `X=[{:(-16,-4),(3,-5):}]` | |
| 1054. |
If `f(x)=x^2-4x+1` then find `f(A)` when `A=[[2,3] , [1,2]]` |
| Answer» `f(A)=[{:(0,0),(0,0):}]` | |
| 1055. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement-1 For a singular matrix `A , if AB = AC rArr B = C` Statement-2 If `abs(A) = 0,` thhen` A^(-1)` does not exist.A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-3B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-3C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D `A^(-1) ` exists only for non-singular matrix `AB= AC rArr A^(-1) (AB) = A^(-1) (AC) ` `rArr (A^(-1) A ) B= (A^(-1)A) C` `rArr IB = IC` `rArr B= C, ` if `A^(-1)` exist `therefore abs(A) ne 0` Statement- 2 is false and Statement-2 is true. |
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| 1056. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Let A be a skew-symmetric matrix, `B= (I-A) (I+A)^(-1)` and X and Y be column vectors conformable for multiplication with B. Statement-1 (BX)^(T) (BY) = X^(T) Y Statement- 2 If A is skew-symmetric, then (I+A) is non-singular.A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-5B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-5C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - A `because (BX)^(T)(BY) = {(I-A)(+A)^(-1) X}^(T) (I-A) (I+A)^(-1) Y` `=X^(T){(I+A)^(-1) }^(T) (I-A)^(T)(I-A) (I+A)^(-1) Y` `= X^(T)(I+A^(T))^(-1) (I-A^(T))(I-A) (I+A)^(-1) Y` `= X^(T)(I+A)^(-1) (I+A)(I-A) (I+A)^(-1) Y` `= X^(T)(I+A)^(-1) (I-A)(I+A) (I+A)^(-1) Y` `[because A^(T) = - A and (I-A) (I+A) = (I+A) (I-A)]` `=X^(T) cdot Icdot IcdotIY=X^TY` Both Statements are true, Statement-2 is correct explanation for Statement-1. |
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| 1057. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement- 1 If A, B, C are matrices such that `abs(A_(3xx3))=3, abs(B_(3xx3))= -1 and abs(C_(2xx2)) = 2, abs(2 ABC) = - 12.` Statement - 2 For matrices A, B, C of the same order `abs(ABC) = abs(A) abs(B) abs(C).`A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-1C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D ABC is not defined, as order of A, B and C are such that they are not conformable for multiplication. Hence, Statement-1 is false and Statement -2 is true. |
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| 1058. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement-1 if A and B are two square matrices of order `nxxn` which satisfy `AB= A and BA = B,` then `(A+B) ^(7) = 2^(6) (A+B)` Statement- 2 A and B are unit matrices.A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-2B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-2C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - C `because AB = A, BA = B rArr A^(2) = A and B^(2) = B` `therefore (A+B) ^(2) = A^(2) + B^(2) + AB + BA = A + B + A +B` ` = 2(A+B)` `(A+B)^(3) = (A+B) ^(2) . (A+B)` `=2 (A+B)^(2) = 2^(2) (A+B)` `therefore (A+B)^(7) = 2^(6) (A+B)` Statement - 1 is true and Statement -2 is false. |
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| 1059. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement - 1 `A=[a_(ij)]` be a matrix of order `3xx3` where `a_(ij) = (i-j)/(i+2j)` cannot be expressed as a sum of symmetric and skew-symmetric matrix. Statement-2 Matrix `A= [a_(ij)] _(nxxn),a_(ij) = (i-j)/(i+2j) ` is neither symmetric nor skew-symmetric.A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-1C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D `A= [[0, -1/5, -2/7],[1/4, 0 , -1/8],[2/5, 1/7, 0]]` which is neither symmetric nor skew-symmetric. Infact every square matrix can be expressed as a sum of symmetric and skew-symmetric matrix. Hence, Statement-1 is false and Statement -2 is true. |
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| 1060. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement-1 Let A `2xx2` matrix A has determinant 2. If `B= 9 A^(2)`, the determinant of `B^(T)` is equal to 36. Statement- 2 If A, B and C are three square matrices Such that `C= AB` then `abs(C) = abs(A) abs(B).`A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-1C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D `because abs(A) =2` `and B = 9A^(2)" " (given)` `therefore abs(B) = abs(9A^(2)) = 9^(2) abs(A )^(2)` `= 81 xx4 = 324 rArr abs(B^(T)) = abs(B) = 324` Hence, Statement-1 is false but Statement-2 is true. |
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| 1061. |
Let `Ad nB`be `3xx3`matrtices of ral numbers, where `A`is symmetric, `"B"`is skew-symmetric , and `(A+B)(A-B)=(A-B)(A+B)dot`If `(A B)^t=(-1)^k A B ,w h e r e(A B)^t`is the transpose of the mattix `A B ,`then find the possible values of `kdot` |
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Answer» Correct Answer - B::D `because A^(t) = A, B^(t) = -B` Given, `(A+B) (A-B) = (A-B) (A+B) ` `rArrA^(2) - AB + BA-B^(2) = A^(2) + AB - BA-B^(2)` `rArr AB= BA` Also, given `(AB)^(t)=(-1)^(k)AB` `rArr B^(t) A^(t) = (-1)^(k) AB` `rArr -BA = (-1)^(k) AB` `rArr (-1) = (-1)^(k) [because AB= BA]` `therefore k = 1, 3, 5, ...` |
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| 1062. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement -1 If `A= [[1,-1,-1],[1,-1,0],[1,0,-1]]` then `A^(3) + A^(2) + A= I ` Statement - 2 If `det (A-lambdaI) = C_(0) lambda^(3) + C_(1) lambda^(2) + C_(2)lambda + C_(3) = 0.` then ` C_(0) A^(3) + C_(1)A ^(2) + C_(2)A + C_(3)I = O.`A. Statement-1 is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-1C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D `therefore det (A - lambdaI) = [[1-lambda,-1,-1],[1,-1-lambda,0],[1,0,-1-lambda]]=0` `rArr (1-lambda ) (1+lambda)^(2) - 1-lambda -1 -lambda = 0` `rArr lambda^(3) + lambda^(2) + lambda + 1= 0` ` rArr A^(3) + A^(2) + A + I = 0` `rArr A^(3) + A^(2) + A= -I` Statement -1 is false but Statement -2 is true. |
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| 1063. |
If `A = [[0, 1],[3,0]]and (A^(8) + A^(6) + A^(4) + A^(2) + I) V= [[0],[11]],` where `V` is a vertical vector and `I` is the `2xx2` identity matrix and if `lambda` is sum of all elements of vertical vector `V`, the value of `11 lambda` is |
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Answer» Correct Answer - 1 `because A =[[0,1],[3,0]]` `therefore A^(2) = A cdot A = [[0,1],[3,0]] [[0,1],[3,0]]= [[3,0],[0,3]]= 3I` `rArr A^(4) = (A^(2))^(2) = 9 I , A^(6) = 27I, A^(8) = 81 I` Now, `(A^(8) + A^(6) + A^(4)+ A^(2) + I) V = (121) IV = (121) V" "...(i)` `(A^(8) + A^(6) + A^(4)+ A^(2) + I) V = [[0],[11]]" " (ii) ` From Eqs. (i) and (ii), `(121) V = [[0],[11]]rArr V = [[0],[1/11]]` `therefore` Sun of elements of `V= 0 + 1/11 = 1/11 = lambda` [given] `therefore 11 lambda = 1` |
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| 1064. |
Let the matrix A and B defined as `A=|[3,2] , [2,1]|` and `B=|[3,1] , [7,3]|` Then the value of `|det(2A^9 B^(-1)|=` |
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Answer» Correct Answer - 2 `because A = [[3,2],[2,1]] and B= [[3,1],[7,3]]` `therefore det A =-1 and det B= 2` Now, `det(2A^(9) B^(-1)) = 2^(2) cdot det (A^(9)) cdot det (B^(-1))` ` = 2^(2) cdot (det A)^(9)cdot (det B)^(-1)` ` = 2^(2) cdot (-1)^(9)cdot (2)^(-1)=-2` Hence, absolute value of `det ( 2A^(9)B^(-1))=2` |
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| 1065. |
If U is same as in Example 50, then the value of `{:[(3,2,0)]U[(3),(2),(0)]=:}`A. 5B. `5//2`C. 4D. `3//2` |
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Answer» Correct Answer - A We have, `{:[(3,2,0)]U[(3),(2),(0)]=[(3,2,0)][(1,2,2),(-2,-1,-1),(1,-4,-3)][(3),(2),(0)]:}` `={:[(3,2,0)]:}[(7),(-8),(-5)]=21-16+0=5` |
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| 1066. |
Let ` A= [[5,5alpha,alpha],[0,alpha,5alpha],[0,0,5]]` .` If A^2 = 25,` then `alpha` equals to:A. `5^(2)`B. 1C. `1//5`D. 5 |
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Answer» Correct Answer - C `A = [[5, 5alpha ,alpha ],[0,alpha,alpha5],[0,0,5]]rArr abs(AcdotA)= abs(A) abs(A)= (25alpha)^(2) = 25` `rArr alpha^(2) = 1/25` `rArr alpha = pm1/5` |
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| 1067. |
Statement-1 (Assertion and Statement- 2 (Reason) Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice as given below. Statement - 1 If mateix `A= [a_(ij)] _(3xx3) , B= [b_(ij)] _(3xx3), ` where ` a_(ij) + a_(ji) = 0 and b_(ij) - b_(ji) = 0` then `A^(4) B^(5)` is non-singular matrix. Statement-2 If A is non-singular matrix, then `abs(A) ne 0 .`A. Statement- is true, Statement -2 is true, Statement-2 is a correct explanation for Statement-1B. Statement-1 is true, Statement-2 is true, Sttatement - 2 is not a correct explanation for Stamtement-1C. Statement 1 is true, Statement - 2 is falseD. Statement-1 is false, Statement-2 is true |
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Answer» Correct Answer - D Since, matrix A is skew-symmetric `therefore abs(A) = 0` ` therefore abs(A^(4) B^(5)) = 0` `rArr A^(4) B^(5)` is singular matrix. Statement-1 is false and Statement - 2 is true. |
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| 1068. |
Let `A = [[0, alpha],[0,0]] and(A+I)^(70) - 70 A = [[a-1,b-1],[c-1,d-1]],` the value of ` a + b + c + d ` is |
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Answer» Correct Answer - 6 `because A = [[0 ,alpha],[0,0]]` `therefore A^(2) = Acdot A = [[0 ,alpha],[0,0]][[0 ,alpha],[0,0]]=[[0 ,0],[0,0]]=0` `rArr A^(2) = A^(3) = A^(4) = A^(5) = ...= 0` Now, `(A + I) ^(70) = (I+A)^(70)` `= I + ""^(70)C_(1) A + ""^(70)C_(2) A^(2) + ""^(70)C_(3) A^(3) +...+ ""^(70)C _(70)A^(70` `= I + 70 A + 0 + 0 + ...=I+70A` `rArr (A+I) ^(70) - 70 A = I = [[1,0],[0,1]]= [[a-1,b-1],[c-1, d-1]]` [given ] `therefore a- 1 = 1, b-1 = 0, c-1 = 0, d-1=1` `rArr a=2, b=1, c=, d=2` Hence, `a + b + c + d = 6` |
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| 1069. |
If A and B are square matrices of size `nxxn` such that `A^2-B^2 = (A-B)(A+B)`, then which of the following will be always trueA. A = BB. AB =BAC. either A or B is a zero matrixD. either A or B is an identity matrix |
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Answer» Correct Answer - B We have, `A^2-B^2=(A-B)(A+B)` `rArr A^2-B^2=A(A-B)-B(A+B)` `rArr A^2-B^2=A^2+AB-BA-B^2` `rArr AB=BA` |
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| 1070. |
If det, `(A-B) ne 0, A^(4)=B^(4), C^(3) A=C^(3)B` and `B^(3)A=A^(3)B`, then find the value of det. `(A^(3)+B^(3)+C^(3))`. |
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Answer» `(A^(3)+B^(3)+C^(3)) (A-B)` `=A^(4)-A^(3)B+B^(3)A-B^(4)+C^(3)A-C^(3) B=O` `:.` det. `((A^(2)+B^(3)+C^(3))(A-B))=0` `implies` det. `(A^(3)+B^(3)+C^(3))=0`, as det. `(A-B) ne 0` |
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| 1071. |
If A is non-singular matrix of order, `nxxn,` |
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Answer» Correct Answer - `(A) rarr (r,t); (B) rarr (s); (C) rarr (p); (D) rarr (q)` `(A) rarr (r,t), (B) rarr (s), (C) rarr (p), (D) rarr (q)` (A) `adj (A^(-1)) = (A^(-1))^(-1) det (A^(-1)) = A/(det(A))` `("adj"" adj"A)/((adj A)^(n-1) )=(A[det(A)]^(n-2))/((detA)^(n-1) )=A / (det(A))` (B) `det(adj A^(-1))= (det A^(-1))^(n-1)` `= 1/((det A)^(n-1))=(detA)^(1-n)` (C) `"adj" ["adj "A] = A (det A) ^(n-2)` `"adj" (A det A) = (detA)^(n-1) ("adj 0"A)` |
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| 1072. |
If A and B are square matrices of size `nxxn` such that `A^2-B^2 = (A-B)(A+B)`, then which of the following will be always trueA. A = BB. AB = BAC. Either of A or B is a zero matrixD. Either of A or B is dientity matrix |
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Answer» Correct Answer - B `A^(2) - B^(2) =(A-B)(A+B)` `rArr A^(2) -B^(2) = A^(2) + AB - BA - B^(2)` `rArr AB = BA` |
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| 1073. |
A and B are different matrices of order n satisfying `A^(3)=B^(3)` and `A^(2)B=B^(2)A`. If det. `(A-B) ne 0`, then find the value of det. `(A^(2)+B^(2))`. |
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Answer» `(A^(2)+B^(2))(A-B)=A^(3)-A^(2)B+B^(2)A-B^(3)=O` `:.` det. `[(A^(2)+B^(2))(A-B)]=0` `implies` det. `(A^(2)+B^(2))xx`det. `(A-B)=0` `implies` det. `(A^(2)+B^(2))=0` (as det. `(A-B) ne0`) |
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| 1074. |
If `A` and `B` are different matrices satisfying `A^(3) = B^(3)` and `A^(2) B = B^(2) A`, thenA. det `(A^(3) = B^(3))` must be zeroB. det `(A-B)` must be zeroC. det `(A^(3) = B^(3))` as well as det (A - B) must be zeroD. alteast one of det `(A^(3) = B^(3))` or det (A - B) must be zero |
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Answer» Correct Answer - D `because A^(3) - A^(2) B = B^(3) - B^(2) A ` `rArr A^(2) (A-B) = B^(2)(B-A)` or `(A^(2) + B^(2)) (A-B) =0` or det `(A^(2)+B^(2)) cdot det (A-B) = 0` Either det `(A^(2) + B^(2)) = 0` or det `(A-B) = 0` |
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| 1075. |
Let `{:A=[(0,0,-1),(0,-1,0),(-1,0,0)]:}`. The only correct statement aboul the matrix A isA. `A^2=I`B. A = -I, where I is a unit matrixC. `A^(-1)` does not existD. A is a zero matrix |
| Answer» Correct Answer - c | |
| 1076. |
If A is an orthogonal matrix, thenA. `absA=0`B. `absA = pm 1`C. `absA=pm2`D. none of these |
| Answer» Correct Answer - B | |
| 1077. |
If `{:A=[(1,2,2),(2,3,0),(0,1,2)]and adjA=[(6,-2,-6),(-4,2,x),(y,-1,-1)]:}`,then x + y =A. 6B. -1C. 3D. 1 |
| Answer» Correct Answer - a | |
| 1078. |
If A is a square matrix of order 3 such that `abs(A)=2,` then `abs((adjA^(-1))^(-1))` isA. 1B. 2C. 4D. 8 |
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Answer» Correct Answer - C `because abs(adj A^(-1)) = abs(A^(-1))^(2) = 1/(abs(A))^(2)` `therefore abs((adj A^(-1) )^(-1) ) = 1/(abs(adjA^(-1)))=abs(A)^(2) = 2^(2) = 4` |
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| 1079. |
Let A be a non-singular square matrix of order n. Then; `|adjA| = |A|^(n-1)`A. `absA^n`B. `absA^(n-1)`C. `absA^(n-2)`D. none of these |
| Answer» Correct Answer - B | |
| 1080. |
If A is a square matrix such that `{:A(adjA)=[(4,0,0),(0,4,0),(0,0,4)]:}` then `abs(adjA)`=A. 4B. 16C. 64D. 256 |
| Answer» Correct Answer - b | |
| 1081. |
If A is a square of matrix of order 3 such that |Adj A| = 225, find |A| |
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Answer» Adj A| = 225 ⇒ |A|3-1 = 225 ⇒ |A|2 = 225 ⇒ |A| = 15 |
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| 1082. |
If A is a square matrix and |A| = 2, find the value of |AAT|. |
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Answer» |A| = 2 (Given) |AT| = 2 Now |AAT| = |A| |AT| = 2 × 2 = 4 |
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| 1083. |
If `{:X=[(3,-4),(1,-1)]:}`, the value of `X^n` is equal toA. `{:[(3n,-4n),(n,-n)]:}`B. `{:[(2+n,5-n),(n,-n)]:}`C. `{:[(3^n,(-4)^n),(1^n,(-1)^n)]:}`D. none of these |
| Answer» Correct Answer - d | |
| 1084. |
If `{:A=[(5,2),(3,1)]:}," then "A^(-1)=`A. `{:[(1,0),(0,1)]:}`B. `{:[(1,1),(0,0)]:}`C. `{:[(0,0),(1,1)]:}`D. `{:[(0,1),(1,0)]:}` |
| Answer» Correct Answer - a | |
| 1085. |
The inverse of the matrix `{:[(1,3),(3,10)]:}` is equal toA. `{:[(10,3),(3,1)]:}`B. `{:[(10,-3),(-3,1)]:}`C. `{:[(1,3),(3,10)]:}`D. `{:[(-1,-3),(-3,-10)]:}` |
| Answer» Correct Answer - b | |
| 1086. |
If `{:A=[(a,0,0),(0,b,0),(0,0,c)]:}," then "A^(-1)`, isA. `{:[(1//a,0,0),(0,1//b,0),(0,0,1//c)]:}`B. `{:[(-1//a,0,0),(0,-1//b,0),(0,0,-1//c)]:}`C. `{:[(a,0,0),(0,b,0),(0,0,1//c)]:}`D. none of these |
| Answer» Correct Answer - a | |
| 1087. |
If `A=[(1,3),(2,4)]` and `(AB)^(-1)=[((-1)/(2),(1)/(2)),((1)/(4),0)]`, then `B^(-1). A^(-1)=`A. `[((-5)/(8),(1)/(8)),((3)/(8),(1)/(8))]`B. `[(-1,1),(3,5)]`C. `[((-1)/(2),(1)/(4)),((1)/(2),0)]`D. `[((-1)/(2),(1)/(2)),((1)/(4),0)]` |
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Answer» Correct Answer - d |
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| 1088. |
If `A=[(1,lambda,2),(1,2,5),(2,1,1)]` is not invertible then `lambda`=?A. 2B. 1C. 0D. `-1` |
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Answer» Correct Answer - b |
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| 1089. |
What is the order of the product `[x" "y" "z][{:(a,h,g),(h,b,f),(g,f,c):}][{:(x),(y),(z):}]`?A. `3xx1`B. `1xx1`C. `1xx3`D. `3xx3` |
| Answer» Correct Answer - b | |
| 1090. |
The value of ` lambda` for which the matrix `[(1,-2,-1),(2,lambda,3),(-1,0,3)]` will not be invertible ,A. `-9`B. `(9)/(2)`C. 9D. `(-9)/(2)` |
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Answer» Correct Answer - a |
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| 1091. |
The matrix `[(lamda,-1,4),(-3,0,1),(-1,1,2)]` is invertible ifA. `lambda ne -15`B. `lambda ne -17`C. `lambda ne -16`D. `lambda ne -18` |
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Answer» Correct Answer - b |
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| 1092. |
Matrix `A=[[1, 0, -k], [2, 1, 3], [k, 0, 1]]` is invertible forA. `k=1`B. `k=-1`C. `k=0`D. All real k |
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Answer» Correct Answer - d |
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| 1093. |
If A is an invertible matrix and B is a matrix, thenA. rank (AB) = rank (A)B. rank (AB) = rank (B)C. rank (AB) gt rank (A)D. rank (AB) gt rank (B) |
| Answer» Correct Answer - b | |
| 1094. |
A `3xx3` matrix A, with 1st row elements as 2,-1,-1 respectively, is modified as below to get another matrix B. `R_1` elements of A go to `R_3` of matrix C `R_2` elements of A go to `R_1` of matrix C `R_2` elements of A to `R_1` of matrix C `R_3` elements of A go to `R_2` fo matrix C Now, below operations are done on C as follow, `C_1` elements of C go to `C_3` of B `C_2` elements of C go to `C_1` of B `C_3` elements of C go to `C_2` of B It is found that A = B, thenA. A is symmetric matrixB. A is an upper triangular matrixC. A is singular matrixD. none of these |
| Answer» Correct Answer - C | |
| 1095. |
If adj`B=A ,|P|=|Q|=1,t h e na d j(Q^(-1)B P^(-1))`is`P Q`b. `Q A P`c. `P A Q`d. `P A^1Q`A. `PQ`B. `QAP`C. `PAQ`D. `PA^(-1)Q` |
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Answer» Correct Answer - C adj. `(Q^(-1)BP^(-1))=` adj. `(P^(-1))` adj. `(Q^(-1))` `=P/(|P|) A Q/(|Q|)=PAQ` |
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| 1096. |
If `A`is non-singular and `(A-2I)(A-4I)=O ,t h e n1/6A+4/3A^(-1)`is equal to`O I`b. `2I`c. `6I`d. `I`A. `O`B. `I`C. `2I`D. `6I` |
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Answer» Correct Answer - B We have, `(A-2I) (A-4I)=O` or `A^(2)-2A-4A+8I=O` or `A^(2)-6A+8I=O` or `A^(-1) (A^(2)-6A+8I)=A^(-1)O` or `A-6I+8A^(-1) =O` or `A+8A^(-1)=6I` or `1/6 A+4/3 A^(-1)=1` |
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| 1097. |
If `A`is non-singular and `(A-2I)(A-4I)=O ,t h e n1/6A+4/3A^(-1)`is equal to`O I`b. `2I`c. `6I`d. `I`A. IB. OC. 2ID. 6I |
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Answer» Correct Answer - A We have, `(A-2I)(A-4I)=O` `rArr A^2-2A-4A+8I=O` `rArr A^2-6A+8I=O` `rArr A^(-1) (A^2-6A+8I)=A^9-1)O` `rArr A-6I+8A^(-1)=O` `rArr A+8A^(-1)=6IrArr 1/6A +4/3A^(-1)=I` |
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| 1098. |
If A is non-singular and (A – 2I) (A – 4I) = 0, then 1/6 A + 4/3A-1 = (a) I (b) 0 (c) 2I (d) 6I |
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Answer» Correct option (a) I Explanation: (A – 2I) (A – 4I) = 0 A2 – 2A – 4A + 8I = 0 A2 – 6A + 8I = 0 A –1(A2 – 6A + 8I) = A – 10 (Pre multiply by A–1) A–1A2 – 6A – 1 + 8A – 1I = 0 A – 6I + 8A – 1 = 0 A + 8A – 1 = 6I 1/6A + 4/3A-1 = 1 |
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| 1099. |
If `{:[(a,b^3),(2,0)]=[(1,8),(2,0)]," then " [(a,b),(2,0)]_(-1)^1:}=`A. `{:[(0,-2),(-2,1)]:}`B. `{:[(1,0),(0,1)]:}`C. `{:[(0,-8),(-2,1)]:}`D. `{:[(0,1//2),(1//2,-1//4)]:}` |
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Answer» Correct Answer - D We have, `{:[(a,b^3),(2,0)]=[(1,8),(2,0)]:}rArr a=1,b^3=8rArra=1,b=2` `:. {:[(a,b),(2,0)]=[(1,2),(2,0)]:}` Hence,`{:[(a,b),(2,0)]^(-1)=-1/4[(0,-2),(-2,1)]=[(0,1//2),(1//2,-1//4)]:}` |
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| 1100. |
If `{:A=[(alpha,0),(1,1)]andB=[(1,0),(5,1)]:}`, then te value of `alpha` for which `A^2=B`, isA. 1B. -1C. 4D. no real values |
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Answer» Correct Answer - D We have, `A^2=B` `rArr {:[(alpha,0),(1,1)][(alpha,0),(1,1)]=[(1,0),(5,1)]:}` `rArr {:[(a^2,0),(1,1)]=[(1,0),(5,1)]:}` `rArr alpha^2-1 and alpha+1-5` Clearly, these two equations are inconsistent. Hence, there is no value of `alpha` satisfying `A^2=B`. |
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