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.
| 51. |
If `[{:(2alpha+1," "3beta),(0,beta^(2)-5beta):}]=[{:(beta+3,beta^(2)+2),(0,-6):}]` find the equation whose roots are alpha and beta. |
|
Answer» the given matrices wil be equal, iff `2alpha+1=alpha+3impliesalpha=2` `3beta=beta^(2)+2impliesbeta^(2)-3beta+2=0` :. `beta=1,2 and beta^(2) -5beta=-6` implies ` beta^(2) -5beta+6=0` :. `beta=2,3` from Eqs. (i) and (ii), we get `beta=2` rArr `alpha=2, beta=2` `therefore` "Required equation is `x^(2)-(2+2)x+2.2=0` `x^(2) -4x+4=0` |
|
| 52. |
If `P=[(secalpha,tanalpha),(-cotalpha,cosalpha)] and Q=[(-cosalpha,tanalpha),(-cotalpha,-secalpha)]` , than `2P^-1+Q=`A. `[{:("cos"alpha, -"tan"alpha), ("cot"alpha, "sec"alpha):}]`B. `[{:(0, 0), (0, 0):}]`C. `[{:(-"cos"alpha, "tan"alpha), (-"cot"alpha, -"sec"alpha):}]`D. `[{:(1, 0), (0, 1):}]` |
|
Answer» Correct Answer - B Find `P^(-1)` and then proceed. |
|
| 53. |
A square non-singular matrix A satisfies `A^2-A+2I=0," then "A^(-1)`=A. `I-A`B. `(1)/(2) (I-A)`C. `(1)/(2)(I+A)`D. `I+A` |
|
Answer» Correct Answer - b |
|
| 54. |
If `[[cos theta,sin theta],[-sin theta,cos theta]], ` then `lim _(n rarr infty )A^(n)/n ` is (where `theta in R`)A. a zero matrixB. an identity matrixC. `[[0,1],[-1,0]]`D. `[[0,1],[0,-1]]` |
|
Answer» Correct Answer - A `because A = [[cos theta , sin theta],[-sin theta, cos theta ]]` `therefore A^(n) = [[cos ntheta , sin ntheta],[-sin ntheta, cos ntheta ]]` `rArr A^(n)/n = [[lim_(nrarr infty)(cos ntheta)/n , lim_(nrarr infty)(sin ntheta)/n],[-lim_(nrarr infty)(sin ntheta)/n, lim_(nrarr infty)(cos ntheta)/n ]]= [[0,0],[0,0]]` = a zero matirx `[because - 1 lt sin infty 1 and -1 lt cos infty lt 1]` |
|
| 55. |
The inverse of matrix `A=[[2, -3], [-4, 2]]` isA. `(-1)/(8)[(2,3),(4,2)]`B. `(-1)/(8)[(3,2),(2,4)]`C. `(1)/(8)[(2,3),(4,2)]`D. `(1)/(8)[(3,2),(2,4)]` |
|
Answer» Correct Answer - a |
|
| 56. |
If `A`satisfies the equation `x^3-5x^2+4x+lambda=0`, then `A^(-1)`exists if`lambda!=1`(b) `lambda!=2`(c) `lambda!=-1`(d) `lambda!=0`A. `lambda!=1`B. `lambda!=2`C. `lambda!=-1`D. `lambda!=0` |
| Answer» Correct Answer - D | |
| 57. |
If a square matrix A is involutory, then `A^(2n+1)` is equal toA. IB. AC. `A^(2) `D. `(2n +1) A` |
|
Answer» Correct Answer - B `because A^(2n+1) = (A^(2))^(n) cdot A = (I)^(n) cdot A = IA = A` |
|
| 58. |
If A is a `2xx2` matrix such that `A^(2)-4A+3I=O`, then prove that `(A+3I)^(-1)=7/24 I-1/24 A`. |
|
Answer» We have, `A^(2)-4A+3I=O` `implies A(A+3I)-7A+3I=O` `implies A(A+3I)-7(A+3I)=-24 I` `implies (A+3I) (A-7I)=-24 I` `implies (A+3I) (7/24 I-A/24)=I` `implies (A+3 I)^(-1) =7/24 I- A/24` |
|
| 59. |
`A`is an involuntary matrix given by`A=[0 1-1 4-3 4 3-3 4]`, then the inverse of `A//2`will be`2A`b. `(A^(-1))/2`c. `A/2`d. `A^2`A. 2AB. `A^(-1)/(2)`C. `(A)/(2)`D. `A^(2)` |
| Answer» Correct Answer - A | |
| 60. |
The matrix `A=[-5-8 0 3 5 0 1 2-]`isa. idempotent matrixb. involutory matrixc. nilpotent matrixd. none of these |
|
Answer» `A^(2)=AxxA[(-5,-8,0),(3,5,0),(1,2,-1)]xx[(-5,-8,0),(3,5,0),(1,2,-1)]` `=[(25-24+0,40-40+0,0+0+0),(-15+15+0,-24+25+0,0+0+0),(-5+6-1,-8+10-2,0+0+1)]` `=[(1,0,0),(0,1,0),(0,0,1)]` Hence, the given matrix a is involutory. |
|
| 61. |
`A`is an involuntary matrix given by`A=[0 1-1 4-3 4 3-3 4]`, then the inverse of `A//2`will be`2A`b. `(A^(-1))/2`c. `A/2`d. `A^2`A. `2A`B. `A^(-1)/2`C. `A/2`D. `A^(2)` |
|
Answer» Correct Answer - A A is involuntary. Hence, `A^(2)=I implies A=A^(-1)` Also, `(kA)^(-1)=1/k (A)^(-1)` or `(1/2 A)^(-1) =2(A)^(-1)` or `2A` |
|
| 62. |
If `A` is a skew symmetric matrix, then `B=(I-A)(I+A)^(-1)` is (where `I` is an identity matrix of same order as of `A`)A. idempotent matrixB. symmetric matrixC. orthogonal matrixD. none of these |
|
Answer» Correct Answer - C `(c )` `B=(I-A)(I+A)^(-1)` `impliesB^(T)=(I+A^(T))^(-1)(I-A^(T))` `=(I-A)^(-1)(I+A)` `impliesB B^(T)=(I-A)(I+A)^(-1)(I-A)^(-1)(I+A)` `=(I-A)(I-A)^(-1)(I+A)^(-1)(I+A)` `=I` (As `(I-A).(I+A)=(I+A)(I-A)`) |
|
| 63. |
For two unimobular complex numbers `z_(1)` and `z_(2)`, find `[(bar(z)_(1),-z_(2)),(bar(z)_(2),z_(1))]^(-1) [(z_(1),z_(2)),(-bar(z)_(2),bar(z)_(1))]^(-1)` |
|
Answer» Correct Answer - `[(1//2,0),(0,1//2)]` `[(bar(z)_(1),-z_(2)),(bar(z)_(2),z_(1))]^(-1) [(z_(1),z_(2)),(-bar(z)_(2),bar(z)_(1))]` `=([(z_(1),z_(2)),(-bar(z)_(2),bar(z)_(1))][(bar(z)_(1), -z_(2)),(bar(z)_(2),z_(1))])^(-1)` `=[(z_(1)bar(z)_(1)+z_(2)bar(z)_(2),0),(0,z_(2)bar(z)_(2)+z_(1)bar(z)_(1))]^(-1)` `=[(|z_(1)|^(2)+|z_(2)|^(2),0),(0,|z_(1)|^(2)+|z_(2)|^(2))]^(-1)` `=[(2,0),(0,2)]^(-1)=[(1//2,0),(0,1//2)]` |
|
| 64. |
Let `aa n db`be two real numbers such that `a >1,b > 1.`If `A=(a0 0b)`, then `(lim)_(nvecoo)A^(-n)`isa. unit matrix b. nullmatrixc. `2l`d. none of theseA. unit matrixB. null matrixC. `2I`D. none of these |
|
Answer» Correct Answer - B `A=((a,0),(0,b))` `implies A^(2)=((a,0),(0,b))((a,0),(0,b))=((a^(2),0),(0,b^(2)))` `implies A^(3)=((a^(2),0),(0,b^(2))) ((a,0),(0,b))=((a^(3),0),(0,b^(3)))` `implies A^(n)=((a^(n),0),(0,b^(n)))` `implies (A^(n))^(-1) =1/(a^(n)b^(n)) ((b^(n),0),(0,a^(n)))=((a^(-n),0),(0,b^(-n)))` `implies lim_(n rarr oo) (A^(n))^(-1)=((0,0),(0,0))` as `a gt 1` and `b gt 1` |
|
| 65. |
If s is a real skew-symmetric matrix, the show that `I-S` is non-singular and matrix `A= (I+S) (I-S) ^(-1) = (I-S) ^(-1) (I+S) ` is orthogonal.A.B.C.D. |
|
Answer» `because ` S is skew-symmetric matrix `therefore S^(T) =-S " "…(i) ` Frist we will show that `I-S` is non-singular. The equality `abs(I-S) = 0 rArr I` is a characteristic root of the matrix S but this is not possile, for a real skew-symmetric matrix can have zero or purely imaginary numbers as its characterixtic roots. Thus, `abs(I-S) ne 0` i.e. `I-S` is non-singular. We have, `A^(T) = {(I+S)(I-S)^(-1)}^(T) = {(I-S)^(-1) (I+S)}^(T)` ` = ((I+S)^(-1))^(T)(I+S)^(T) = (I+S)^(T) {(I+S)^(-1)}^(T)` ` = ((I-S)^(T))^(-1)(I+S)^(T) = (I+S)^(T) {(I-S)^(T)}^(-1)` ` = (I^(T)-S^(T))^(-1)(I^(T)+S^(T)) = (I^(T)+S^(T)) (I^(T)-S^(T))^(-1)` `= (I+S)^(-1) (I-S) = (I-S) (I+S)^(-1)` [from Eq. (i)] `therefore A^(T)A= (I+S)^(-1) (I-S) (I+S) (I-S)^(-1)` `= (I-S)(I+S)^(-1) (I-S)^(-1) (I+S)` `= (I+S)^(-1)(I+S) (I-S) (I-S)^(-1) ` `= (I-S) (I-S)^(-1)(I+S)^(-1)(I+S) ` `=Icdot I = Icdot I=I=I` Hence, A is orthogonal. |
|
| 66. |
If A is any square matrix such that `A+I/2` and `A-I/2` are orthogonal matrices, thenA. A is orthogonalB. A is skew- symmetric matrix of even orderC. `A^(2)= 3/4 I`D. None of these |
|
Answer» Correct Answer - B `because (A-1/2I)(A-1/2)^(T)=I ` ...(i) and `because (A+1/2I)(A+1/2)^(T)=I ` ...(ii) `rArr (A-1/2I)(A^(T)-1/2)=I ` and `rArr (A+1/2I)(A^(T)+1/2)=I ` `rArr A + A^(T) = 0 ` [subtracting the two results] ` rArr A^(T) = - A` `therefore` A is skew-symmetric matrix. From first result, we get `A A ^(T) = 3/4 I` `rArr A^(2) = - 3/4 I` `therefore abs(A^(2) ) = abs(-3/4I)` `therefore abs(A)^(2) = (-3/4)^(n)` `rArr n` is even. |
|
| 67. |
Prove that inverse of a skew-symmetric matrix (if it exists) is skew-symmetric. |
|
Answer» A is skew-symmetric, then `A^(T)=-A`. `:. (A^(-1))^(T)=(A^(T))^(-1)=(-A)^(-1) =-A^(-1)` Thus, `A^(-1)` is skew-symmetric. |
|
| 68. |
If `A` is a skew symmetric matrix, then `B=(I-A)(I+A)^(-1)` is (where `I` is an identity matrix of same order as of `A`) |
|
Answer» We have `A=(I+S)(I-S)^(-1)` `:. A^(T)=[(I-S)^(-1)]^(T) (I+S)^(T)=[(I-S)^(T)]^(-1) (I+S)^(T)` But `(I-S)^(T)=I^(T)-S^(T)=I +S" "("as "S^(T)=-S)` and `(I+S)^(T)=I^(T)+S^(T)=I-S` `:. A^(T)=(I+S)^(-1) (I-S)` `:. A^(T)A=(I+S)^(-1) (I-S) (I+S) (I-S)^(-1)` `=(I+S)^(-1) (I+S) (I-S) (I-S)^(-1)` `=I` Thus, a is orthogonal. |
|
| 69. |
In which of the following type of matrix inverse does not exist always?a. idempotent b. orthogonalc. involuntary d. none of theseA. idempotentB. orthogonalC. involuntaryD. none of these |
|
Answer» Correct Answer - A For involuntary matrix, `A^(1)=I` `implies |A^(2)|=|I|implies |A|^(2)=1 implies |A|= pm 1` For idempotent matrix, `A^(2)=A` `implies |A^(2)|=|A|implies |A|^(2)=|A|=0` or 1 for orthogonal matrix, `A A^(T)=I` `implies |AA^(T)|=|I|implies|A||A^(T)|=1implies |A|^(2)=1 implies |A|= pm 1` Thus, if matrix A is idempotent it may not be invertible. |
|
| 70. |
If nth-order square matrix A is a orthogonal, then `|"adj (adj A)"|` isA. always -1 if n is evenB. always 1 if n is oddC. always 1D. none of these |
|
Answer» Correct Answer - B Since A is orthogonal, we have `A A^(T)=I` or `|A A^(T)|=1` or `|A^(2)|=1` or `|A|= pm 1` Now `|"adj. (adj. A)"|=|A|^((n-1)^(2))` |
|
| 71. |
If a square matrix A is orthogonal as well as symmetric, thenA. A is involutory matrixB. A is idempotent matrixC. A is a diagonal matrixD. none of these |
| Answer» Correct Answer - A | |
| 72. |
If A is a square matrix such that `A(adj A)=[(4,0,0),(0,4,0),(0,0,4)]` , then `(|adj (adj A)|)/(|adj A|)` is equal toA. 256B. 16C. 32D. 64 |
|
Answer» Correct Answer - b |
|
| 73. |
If square matrix a is orthogonal, then prove that its inverse is also orthogonal. |
|
Answer» We have `A A^(T)=I` Now, `A^(-1) (A^(-1))^(T)=A^(-1) (A^(T))^(-1)=(A^(T)A)^(-1)=I^(-1)=I` Thus, `A^(-1)` is orthonal. |
|
| 74. |
Let `A=[a_("ij")]` be a matrix of order 2 where `a_("ij") in {-1, 0, 1}` and adj. `A=-A`. If det. `(A)=-1`, then the number of such matrices is ______ . |
|
Answer» Correct Answer - 12 adj. `A=-A` `implies A.` adj. `A=-A^(2)=|A|I=-I` `implies A^(2)=I` Let `A=[(a,b),(c,d)]` `implies A^(2)=[(a^(2)+bc,(a+d)b),((a+b)c,d^(2)+bc)]=[(1,0),(0,1)]` on comparing both sides, we get `a^(2)+bc=1, (a+d)b=0, (a+d)c=0, d^(2)+bc=1` Case I : When `(a+d) ne 0` `implies b=0=c` and `a=1, d=1` or `a=-1, d=-1` `:. A=[(1,0),(0,1)]` or `[(-1,0),(0,-1)]` But both are rejected as det. `A=-1` (given) Case II : When `(a+d)=0 implies d=-a` (i) If `a=1, d=-1 implies bc =0` For `b=0, c` can be `-1, 0, 1` For `b=1, c` can be 0 only. For `b=-1, c` can be 0 only. So, 5 matrices are possible. (ii) If `a=-1, d=1` `implies bc=0` For `b=0, c=-1, 0, 1`. For `b=1, c=0` only. For `b=-1, c=0` only. So, 5 matrices are possible. (iii) If `a=0, d=0` `implies bc=1` `:. A=[(0,1),(1,0)]` or `A=[(0,-1),(-1,0)]` So, 2 matrices are possible. Therefore, total number of matrices is 12. |
|
| 75. |
Let `A=[a_("ij")]_(3xx3), B=[b_("ij")]_(3xx3)` and `C=[c_("ij")]_(3xx3)` be any three matrices, where `b_("ij")=3^(i-j) a_("ij")` and `c_("ij")=4^(i-j) b_("ij")`. If det. `A=2`, then det. `(BC)` is equal to _______ . |
|
Answer» Correct Answer - 4 det. `B=|(a_(11),a_(12)/3,a_(13)/3^(2)),(3a_(21),a_(22),a_(23)/3),(9a_(32),3a_(32),a_(33))|=3xx9|(a_(11),a_(12)/3,a_(13)/3^(2)),(a_(21),a_(22)/3,a_(23)/9),(a_(32),a_(32)/3,a_(33)/9)|` (Dividing `C_(2)` by 3 and `C_(3)` by 9) `=|(a_(11),a_(12),a_(13)),(a_(21),a_(22),a_(23)),(a_(32),a_(32),a_(33))|` `="det. "A=2` Similarly, det, `C=` det. `A=2` `:.` det. `(BC)=` (det. B) (det. C) `=2xx2=4` |
|
| 76. |
If `A=[(costheta,-sintheta),(sintheta,costheta)]` find the values of `theta` satisfying the equation `A^(T)+A =" "I _(2).` |
|
Answer» we have, `A^(T) " " A " " I_(2)` lt brgt `rArrA[(cos0,sin0),(-sin0,cos0)]+[(cos0,-sin0),(sin0,cos0)]=[(1,0),(0,1)]` `rArr[(2cos0,0),(0,2cos0)]=[(1,0),(0,1)]` `rArr cos0 = (1)/(2) = cos((pi)/(3)) rArr 0 = 2npipm (pi)/(3), n in` |
|
| 77. |
If A,B and C are square matrices of order n and det (A)=2, det(B)=3 and det ©=5, then find the value of 10det `(A^(3)B^(2)C^(-1)).` |
|
Answer» Given , `|A|=2,|B|=3 and |c|=5.` Now, 10det `(A^(3)B^(2)C^(-1))=10xx|A^(3)B^(2)C^(1)|` `=10xx|A^(3)|xx|B^(2)|xx|C^(-1)|=10xx|A^(3)|xx|B^(2)|xx|C|^(-1)` `=(10xx|A^(3)|xx|B^(2)|)/(|C|)=(10xx2^(3)xx3^(2))/(5)=144` |
|
| 78. |
If nth-order square matrix A is a orthogonal, then `|"adj (adj A)"|` isA. Statement -1 is True, Statement -2 is true, Statement -2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement -2 is True, Statement -2 is not a correct explanation for Statement -1.C. Statement -1 is True, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - C We have known that `abs (adj(adjA))=absA^((n-1)^2)`. So, statement -2 is false. If A is an orthogonal matrix, then`absA=pm1` `:. Abs(adj(adjA))=absA^((n-1)^2)=(pm1)^((n-1)^2)=pm1`. |
|
| 79. |
Let A be 2 x 2 matrix.Statement I `adj (adj A) = A`Statement II `|adj A| = A`A. Statement -1 is True, Statement -2 is true, Statement -2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement -2 is True, Statement -2 is not a correct explanation for Statement -1.C. Statement -1 is True, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - B For any square matrix a of order n, we have `abs(adjA)=absA^(n-1)and adj(adjA)=absA^(n-2)A` `:. " For a " 2xx2` matrix, we have n=2 `abs(adj A)=absAand adj(adjA)=A` Also, `adj(adjA)=absA^(n-2)` A is obtained by replacing A by adj A in the relation `A (adjA)=absAI_n` Hence, both the statements are true. But, statement -2 is not the correct explanation for statement - 1. |
|
| 80. |
Let A be the set of all `3 xx 3` symmetric matrices all of whose entries are either 0 or 1. Five of these entries are 1 and four of them are 0. The number of matrices in A isA. 12B. 6C. 9D. 3 |
|
Answer» Correct Answer - A A symmetric matrix is symmetric about its diagonal. So, there are even number of 1 and even number of 0 as off diagonal entries. Consequently, there can be either three 1 in the diagonal or one 1 and two zeros. Thus, we have the following cases: CASE I When diagonal elements are 1,1,1. In this case, we have Number of symmetric matrices = Number of arrangements of 1,0,0 as elements above the diagonal `=(3!)/(2!)=3` CASE II When diagonal elements are 1,0,0 ltbr. In this case, we have ltbRgt Number of symmetric matrices =(Number of arrangements of 1,0,0 as entries above the diagonal) `=(3!)/(2!)xx(3!)/(2!)=9` `:. " Total number of matrices in " A=3+9=12` |
|
| 81. |
the square matrix `A=[a_(ij)]_mxxm` given by`a_(ij)=(i-j)^(n),` show that A is symmetic and skew-sysmmetic matrices according as n is even or odd, repectively. |
|
Answer» `therefore " "a_(ij)=(i-j)^(n)=(-1)^(n) (j-i)^(n)` `=(-1)^(n) a_(ij)={{:(a_(ji)", n is even interger"),(-a_(ji)", n is odd integer"):}` Hence , A is symmetric if n is even ana skew-symmetric if n is odd integer. |
|
| 82. |
Let `A=[a_(ij)]` be a square matrix of order n such that `{:a_(ij)={(0," if i ne j),(i,if i=j):}` Statement -2 : The inverse of A is the matrix `B=[b_(ij)]` such that `{:b_(ij)={(0," if i ne j),(1/i,if i=j):}` Statement -2 : The inverse of a diagonal matrix is a scalar matrix.A. Statement -1 is True, Statement -2 is true, Statement -2 is a correct explanation for Statement-1.B. Statement-1 is True, Statement -2 is True, Statement -2 is not a correct explanation for Statement -1.C. Statement -1 is True, Statement -2 is False.D. Statement -1 is False, Statement -2 is True. |
|
Answer» Correct Answer - C We know that the inverse of a diagonal matrix `D=diag (d_1,d_2,d_3,…,d_n)` is a diagonal matrix given by `D^(-1)=diag (d_1^(-1),d_2^(-1),d_3^(-1),…,d_n^(-1))` `:. B=[b_(ij)]` is given by `{:b_(ij)={(0," if i ne j),(1/i,if i=j):}` Hence, statement -1 is true and statement -2 is false. |
|
| 83. |
The square matrix `A=[a_(ij)" given by "a_(ij)=(i-j)^3`, is aA. symmetric matrixB. skew-symmetric matrixC. diagonal matrixD. hermitian matrix |
|
Answer» Correct Answer - B We have, `a_(ij)=(i-j)^3rArra_(ij)=(j-i)^3=-(i-j)^3=-a_(ij)" for all "i,j` `:.` A is a skew-symmetric matrix. |
|
| 84. |
Elements of a matrix A of order 10 x 10 are defined as `a_(ij)=omega^(i+j)` (where omega is cube root unity), then tr(A) of matrix is |
|
Answer» Correct Answer - D tr `(A) = sum_(i=j=1)^(10) a_(ij) = sum _(i=j=1) ^(10) omega^(i+j) = sum_(i=1) ^(10) omega ^(2i) ` `= omega^(2)+omega^(4) + omega^(6) + omega^(8) + ... + omega^(20) ` `= (omega^(2)+omega + 1 )+( omega^(2) + omega+1)+(omega^(2)+omega+1) + omega^(20) ` `= 0 + 0 + 0+ omega ^(2) = omega^(2)` |
|
| 85. |
A square matrix `A=[a_(ij)]` in which `a_(ij)=0` for ` i!=j` and `[a]_(ij)=k` (constant) for `i=j` is called aA. unit matrixB. scalar matrixC. null matrixD. diagonal matrix |
|
Answer» Correct Answer - B in square matrix , Number of rows = number of columns `therefore m=n` |
|
| 86. |
If `A=[a_(ij)]_(2xx3)`, difined as `a_(ij)=i^(2)-j+1`, then find matrix A. |
| Answer» Correct Answer - `A={:[(1,0,-1),(4,3,2)]:}` | |
| 87. |
Lat`A = [a_(ij)]_(3xx 3).` If tr is arithmetic mean of elements of rth row and `a_(ij )+ a_( jk) + a_(ki)=0` holde for all `1 le i, j, k le 3.` Matrix A isA. non- singularB. symmetricC. skew-symmetricD. nether symmetric nor skew-symmetric |
|
Answer» Correct Answer - C `therefore A = [[a_(11) , a_(12),a_(13)],[a_(21),a_(22), a_(23) ],[a_(31), a_(32),a_(33)]]` `rArr t_(1) = (a_(11) + a_(12)+a_(23))/3 = 0, [because a_(ij) + a_(jk) + a_(ki)=0]` `t_(2) = (a_(21) + a_(22) + a_(23))/3 = 0` and `t_(3) = (a_(31) + a_(32) + a_(33))/3 = 0` `because a_(11) + a_(11) + a_(11) = 0, a_(11) + a_(12) + a_(21)= 0,` ` a_(11) + a_(13) + a_(31) = 0, a_(22) + a_(22) + a_(22)= 0, ` ` a_(22) + a_(12) + a_(21) = 0, a_(22) + a_(22) + a_(22)= 0, ` ` a_(33) + a_(13) + a_(31) = 0, a_(33) + a_(23) + a_(32)= 0, ` and ` a_(33) + a_(12) + a_(21) = 0,` we get `a_(11) = a_(22) = a_(33) = 0 ` and `a_(12) =-a_(21), a_(23) = - a_(32), a_(13) = -a_(31)` Hence, A is skew - symmetric matrix. |
|
| 88. |
If `A=[a_(ij)]` is a scalar matrix, then trace of A isA. `undersetisumundersetjsuma_(ij)`B. `underset isuma_(ij)`C. `undersetjsuma_(ij)`D. `undersetisuma_(ij)` |
| Answer» Correct Answer - D | |
| 89. |
If `A=[a_(i j)]`is a scalar matrix oforder `nxxn`such that `a_(i i)=k`for all `i`, then trace of `A`is equal to`n k`(b) `n+k`(c) `n/k`(d) none of theseA. nkB. n+kC. n/kD. none of these |
| Answer» Correct Answer - A | |
| 90. |
The trace of the matrix A = \(\begin{bmatrix}1 & -5 & 7 \\[0.3em]0& 7 &9 \\[0.3em]11 & 8 &9\end{bmatrix}\) is : A. 17 B. 25 C. 3 D. 12 |
|
Answer» (A). 17 As the trace of a matrix is the sum of on – diagonal elements, So, 1 + 7 + 9 = 17 Trace = 17 Option (A) is the answer. |
|
| 91. |
If matrix `A=[a_(ij)]_(2X2)`, where ` a_(ij)={[1,i!=j],[0,i=j]}, then A^2` is equal toA. IB. AC. 0D. none of these |
|
Answer» We have, `A=[a_(ij)]_(2xx2)`, where `a_(ij)=1`, if `inej=0` and if i=j `A=[{:(0,1),(1,0):}]` and `A^(2)=[{:(0,1),(1,0):}][{:(0,1),(1,0):}]=[{:(1,0),(0,1):}]=I` |
|
| 92. |
If `A=[{:(0,1),(1,0):}]` then `A^(2)` is equal toA. `[{:(0,1),(1,0):}]`B. `[{:(1,0),(1,0):}]`C. `[{:(0,1),(0,1):}]`D. `[{:(1,0),(0,1):}]` |
| Answer» `because A^(2)=A.A=[{:(0,1),(1,0):}] [{:(0,1),(1,0):}]=[{:(1,0),(0,1):}]` | |
| 93. |
The matrix A = \(\begin{bmatrix}0 & 0& 4 \\[0.3em]0 & 4 & 0 \\[0.3em]4 & 0 &0\end{bmatrix}\) is a :A. square matrix B. diagonal matrix C. unit matrix D. none of these |
|
Answer» None of these. Option (D) is the answer. |
|
| 94. |
The inverse of a symmetric matrix is :(a) Symmetric (b) Non-symmetric (c) Null matrix (d) Diagonal matrix |
|
Answer» Option : (a) Symmetric |
|
| 95. |
If A and B are square matrices of equal order, then which one is correct among the following? (a) A + B = B + A (b) A + B = A – B (c) A – B = B – A(d) AB = BA |
|
Answer» Correct option is : (a) A + B = B + A Matrix addition is commutative. ∴ A + B = B + A |
|
| 96. |
If` A` and` B `are two matrices of the order` 3 xx m` and `3 xx n`, respectively and `m= n,` then order of matrix `(5A-2B)` is (a) `m xx 3` (b) `3 xx 3` (c) `m xx n` (d) `3 xx n`A. `mxx3`B. `3xx3`C. `mxxn`D. `3xxn` |
|
Answer» Correct Answer - D `A_(3xxm)` and `B_(3xxn)` two matrices. If `m=n`, then A and B have same orders as `3xxn` each. So the order of (5A-2B) should be same as `3xxn`. |
|
| 97. |
If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A – 2B) isA. m × 3B. 3 × 3C. m × nD. 3 × n |
|
Answer» A. m × 3 As order of A is 3 × m and order of B is 3 × n As m = n. So, order of A and B is same = 3 × m ∴ subtraction is possible. And (5A – 3B) also has same order. |
|
| 98. |
If A and B are two matrices of the order 3 × m and 3 × n, respectively, and m = n, then the order of matrix (5A – 2B) is A. m × 3B. 3 × 3C. m × n D. 3 × n |
|
Answer» As order of A is 3 × m and order of B is 3 × n As m = n. So, order of A and B is same = 3 × m ∴ subtraction is possible. And (5A – 3B) also has same order. |
|
| 99. |
`[{:(2x+y,4x),(5x-7,4x):}]=[{:(7,7y-13),(y,x+6):}]` then the value of `x,y` isA. x =3,y=1B. x=2,y=3C. x=2,y=4D. x=3,y=3 |
|
Answer» We have, `4x=x+6rArrx=2` and `4x=7y-13rArr8=7y-13` `rArr 7y=21 rArr y=3` `therefore x+yuu=2+3=5` |
|
| 100. |
The number of possible matrices of order 3 × 3 with each entry 2 or 0 is :A. 9 B. 27 C. 81 D. none of these. |
|
Answer» (D). none of these. Let A = \(\begin{bmatrix}a_{11} & a_{12} & a_{13} \\[0.3em]a_{21} & a_{22} & a_{23} \\[0.3em]a_{31} &a_{32} & a_{33}\end{bmatrix}\) Elements = 9 Order = 3 × 3 Every item in this matrix can be filled in two ways either by 0 or by 2. Possible Matrices = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 = 512 Option (D) is the answer. |
|