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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.
| 1. |
If `A=[[2,-2,-4],[-1,3,4],[1,-2,-3]]` then A is `1) an idempotent matrix 2) nilpotent matrix 3) involutary 4) orthogonal matrixA. nonsingularB. idempotentC. nilpoptentD. orthogonal |
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Answer» Correct Answer - B `A^2=A rArr A ` is idempotent |
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| 2. |
Use matrix method to show that the system that the system of equation `2x+5y=7` `6x+15y=13` is inconsistent |
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Answer» The given equation are `2x+5y=7 " " …(i)` `6x+15y=13 " " (i)` Let `A=[{:(2" 5 "),(6 " 15 "):}], X=[{:(x),(y):}] and B=[{:(7),(13):}]` Then the given system in matrix from is AX `ne` O. Now, `|A|=[{:(2" 5 "),(6 " 15 "):}]=0` The system will be inconsistent if (adjA) B `ne` O The minors of the elements of |A| are `M_(11)=15," " M_(12)=6` `M_(21)=5 " " M_(22)=2` So the confactors of the elements of |A| are `A_(11)=15," " A_(12)=-6` `A_(21)=-5 " " A_(22)=2` `therefore adjA=[{:(15 " -6") ,(-5 " "2) :}]=[{:(15 " -5") ,(-6 " "2) :}]` `rArr (adjA)B=[{:(15 " -6") ,(-5 " "2) :}] [{:(7),(13):}]=[{:(105 " -65") ,(-42 ""+26) :}]=[{:(" "40),(-16):}] ne O` Thus |A|=0 and (adjA)B `ne` O. Hence , the given system of equation is inconsistant |
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| 3. |
6x-9y-20z=-4 4x-15y+10z=-1 2x-3y-5z=-1 |
| Answer» Correct Answer - `x=1/2,y=1/3,z=1/5` | |
| 4. |
Use matrix method to show that the following system of equation is inconsistent 3x-y+2z=3 2x+y+3z=5 x-2y-z=1 |
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Answer» Let us take `A=[{:(3 " "-1 " "2),(2 " "1 " "3),(1 " -2 "-1):}] , X=[{:(x),(y),(z):}] and B=[{:(3),(5),(1):}]` The given system in matrix from is AX =B Now `|A|=[{:(3 " "-1 " "2),(2 " "1 " "3),(1 " -2 "-1):}]` =3(-1+6)+1.(-2-3)+2.(-4-1) =(15-5-10)=0 So the system will be inconsistent of |A| are `M_(11)=5,M_(12)=-5,M_(13)=-5` `M_(21)=5,M_(22)=-5,M_(23)=-5` `M_(31)=-5,M_(32)=5,M_(33)=5` So the confactors of the elements of |A| are `A_(11)=5,A_(12)=-5,A_(13)=-5` `A_(21)=5,A_(22)=-5,A_(23)=-5` `A_(31)=-5,A_(32)=5,A_(33)=5` `therefore (adjA)==[{:(5 " "5 " "-5),(-5 " "-5 " "5),(-5 " -5 " 5):}]= =[{:(5 " "-5 " "-5),(5 " "-5 " "-5),(-5 " 5 " 5):}]` `rArr (adjA)B =[{:(5 " "-5 " "-5),(5 " "-5 " "-5),(-5 " 5 " 5):}]` `[{:(3),(5),(1):}]` `=[{:(5.3+(-5).5+(-5).1),(5.3+(-5).5+(-5).1),((-5).3+5.5+5.1):}]=[{:(15-25-5),(15-25-5),(-15+25+5):}]` `=[{:(-15),(-15),(15):}]neO` Thus |A|=0and (adjA)B` ne ` O. ltbrrgt Hence the given system of equation is inconsistent |
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| 5. |
An amount of Rs 5000 is put into three investments at the rate of interest of `6%,7% and 8%` per annum respectively. The total annual income is Rs 358. If the combined income from the first taoinvestments is Rs 70 more than the income from the third, find the amount of each investment by matrirx method |
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Answer» Correct Answer - ₹1000,₹2200,₹1800 Let these investment of ₹ x ₹ y and ₹ y and ₹ z respectively . Then `x+y +z s= 5000` `(6x)/(100)+(7y)/(100)+(8z)/(100)=358 rArr 6x + 7y + 8z = 35800` and `(6x)/(100)+(7y)/(100)=(8z)/(100)+ 70 rArr 6x + 7y- 8z = 7000` |
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| 6. |
Given `A=[{:(1 " "-1 " "1),(1 " "-2 " "-2),(2 " "1 " "3):}] and B=[{:(-4 " "4" "4),(-7 " "1 " "3),(5 " "-3 " "-1):}]`, find AB and use this result in solving the following system of equation x-y+z=4, x-2y-2z=9 2x+2y+3z=1 |
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Answer» The given equation are `x-y+z=4," " ...(i)` `x-2y-2z=9" " ...(ii)` `2x+2y+3z=1" " ...(iii)` `Let A=[{:(1 " "-1 " "1),(1 " "-2 " "-2),(2 " "1 " "3):}],X=[{:(x),(y),(z):}]and C=[{:(4),(9),(1):}]` Then the given system of equations is AX=C. `Now , AB=[{:(1 " "-1 " "1),(1 " "-2 " "-2),(2 " "1 " "3):}][{:(-4 " "4" "4),(-7 " "1 " "3),(5 " "-3 " "-1):}]` `=[{:(-4+7+5 " "4-1-3 " "4-3-1),(-4+14-10 " "4-2+6 " "4-6+2),(-8-7+15 " "8+1-9 " "8+3-3):}]=[{:(8 " "0 " "0),(0 " "8 " "0),(0 " "0 " "8):}]=8I` `rArrA.(1/8B)=I` `rArr A^(-1)=1/8B=1/8[{:(-4 " "4" "4),(-7 " "1 " "3),(5 " "-3 " "-1):}]` Now , AX=C `rArr X=A^(-1)C` `rArr[{:(x),(y),(z):}]=1/8.[{:(-4 " "4" "4),(-7 " "1 " "3),(5 " "-3 " "-1):}][{:(4),(9),(1):}]` `1/8.[{:(-16+36+4),(-28+9+3),(20-27-1):}]=1/8.[{:(24),(-16),(-8):}]=[{:(3),(-2),(-1):}]` `rArr x=3, y=-2 and z=-1` Hence , x=3, y=-2 and z=-1 |
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| 7. |
The sum of three numbers is 6. If we multiply the third number 2 andadd the first number to the result, we get 7. Be adding second and thirdnumbers to three times the first number we get 12. Use determinants to findthe numbers. |
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Answer» Let the first , second and third numbers be x,y,z respectively Then `x+y+z " " …(i)` `x+2z=7 " " …(ii)` `3x+y+z=12 " " …(iii)` `Let A=[{:(1 " "1 " "2),(1 " "0 " "2),(3 " "1 " "1):}], X=[{:(x),(y),(z):}] and B=[{:(6),(7),(12):}]` Then , the given system in matrix from is AX=B `Now , |A|=[{:(1 " "1 " "2),(1 " "0 " "2),(3 " "1 " "1):}]=[{:(1 " "1 " "1),(1 " "-1 " "1),(0 " "-2 " "-2):}] " "[{:(R_2rarrR_2-R_1,),(R_3rarrR_3-3R_1):}]` =1,(2+2)=4 `ne` 0 `therefore` A is invertible So, the given system has a unique solution , `X=A^(-1)B` The minors of the elements of |A| are `M_(11)=-2, M_(12)=-5, M_(13)=1,` `M_(21)=0, M_(22)=-2, M_(23)=-2,` `M_(31)=2, M_(32)=1, M_(33)=-1,` The cofactors of the elements of |A| are `A_(11)=-2, A_(12)=-5, A_(13)=1,` `A_(21)=0, A_(22)=-2, A_(23)=-2,` `A_(31)=2, A_(32)=1, A_(33)=-1,` `therefore (adjA)=[{:(-2 " "5 " "1),(0 " " -2 " "2),(2 " "-1 " "-1):}]=[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}]` `A^(-1)=1/(|A|).(adjA)=1/4.[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}]` `rArrX=A^(-1)B` `rArr[{:(x),(y),(z):}]=1/4..[{:(-2 " "0 " "2),(5 " "-2 " "-1),(1 " "2 " "-1):}][{:(6),(7),(12):}]` `=1/4.[{:(-12+0+24),(30-14-12),(6+14-12):}]=1/4.[{:(12),(4),(8):}]=[{:(3),(1),(2):}]` `rArr` x=3, y=1 z=2 Hence , the required numbers are 3,1,2. |
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| 8. |
x-y+z=1 2x+y-z=2 x-2y-z=4 |
| Answer» Correct Answer - x=-1 ,y=-1, z=-1 | |
| 9. |
2x+8y+5z=5 x+y+z=-2 x+2y-z=2 |
| Answer» Correct Answer - x=-3 ,y=2, z=-1 | |
| 10. |
If `A=[(1,2,-3),(2,3,2),(3,-3,-4)]` then find `A^-1` and hence solve the follwoing equations: `x+2y-3z=4, 2x+3y+2z=2 and 3x-3y-4z=11` |
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Answer» The given equation are `x+2y-3z=-4. " "…(i)` `2x+3y+2z=2," "…(ii)` `3x-3y-4z=11." "…(iii)` `Let A=[{:(1 " "2 " "-3),(2 " "3 " " 2),(3 " -3 " -4):}],X=[{:(x),(y),(z):}] and B=[{:(-4),(2),(11):}]` So , the given system in matrix form is AX=B. `Now, |A|=[{:(1 " "2 " "-3),(2 " "3 " " 2),(3 " -3 " -4):}]=[{:(1 " "2 " "-3),(0 ""-1 " " 8),(0 " -9 " 5):}]` `[R_2rarrR_2-2R_1 and R_3rarrR_3-3R_1]=1.(-5+72)=67 ne 0` Thus , A is invertible So the has a unique solution , `X=A^(-1)B` Now , the confactors of the elements of |A| are `A_(11)=-6, A_(12)=14, A_(13)=-15` `A_(21)=17, A_(22)=5, A_(23)=9` `A_(31)=13, A_(32)=-8, A_(33)=-1` `therefore adjA=[{:(-6 " "14 " "-15),(17 " "5 " "9),(13 " "-8 " "-1):}]=[{:(-6 " "17 " "13),(14 " "5 " "-8),(-15 " "9 " "-1):}]` So,`A^(-1)=1/|A|.adjA=1/(67).[{:(-6 " "17 " "13),(14 " "5 " "-8),(-15 " "9 " "-1):}]` `thereforeX=A^(-1)B` or `[{:(x),(y),(z):}]=1/(67).[{:(-6 " "17 " "13),(14 " "5 " "-8),(-15 " "9 " "-1):}].[{:(-4),(2),(11):}]` `=1/(67).[{:(201),(-134),(67):}]=[{:(3),(-2),(1):}]` `therefore` x=3 , y=-2 and z=1 |
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| 11. |
x+2y+z=7 x+3z=11 2x-3y=1 |
| Answer» Correct Answer - x=2 ,y=1, z=3 | |
| 12. |
2x+3y+3z=5 x-2y+z=-4 3x-2y-z=3 |
| Answer» Correct Answer - x=1 ,y=2, z=-1 | |
| 13. |
Solve the followingsystem of equations by matrix method :`x+y-2=3``2x+3y+z=10``3x-y-7z=1``x+y+z=3``2x-y+z=-1``2x+y-3z=-9``6x-12 y+25 z=4``4x+15 y-20 z=3``2x+18 y+15 z=10``3x+4y+7z=14``2x-y+3z=4``x+2y-3z=0``2/x-3/y+3/z=10``1/x+1/y+1/z=10``3/x-1/y+2/z=13``5x+3y+z=16``2x+y+3z=19``x+2y+4z=25``3x+4y+2z=8``2y-3z=3``x-2y+6z=-2``2x+y+z=2``x+3y-z=5``3x+y-2z=6``2x+6y=2``3x-z=-8``2x-y+z=-3``2y-z=1``x-y+z=2``2x-y=0``8x+4y+3z=18``2x+y+z=5``x+2y+z=5``x+y+z=6``x+2z=7``3x+y+z=12``2/x+3/y+(10)/z=4,``4/x-6/y+5/z=1``6/x+9/y-(20)/z=2; x , y , z!=0``x-y+2z=7``3x+4y-5z=-5``2x-y+3z=12` |
| Answer» Correct Answer - `x=1/2,y=1/3,z=1/5` | |
| 14. |
x-2y+z=0 y-z=2 2x-3z=10 |
| Answer» Correct Answer - x=2 ,y=0, z=-2 | |
| 15. |
`1//x-1/y+1/z=4, 2/x+1/y-3/z=0, 1/x+1/y+1/z=2(x,y,z ne 0)` |
| Answer» Correct Answer - `x=1/2,y=-1,z=1` | |
| 16. |
If A and B are two nonzero square matrices of the same order such that the product `AB=O`, thenA. |A|=0 or |B|=0B. |A|=0 and |B|=0C. |A| `ne` 0 and |B| `ne` 0D. none of these |
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Answer» Correct Answer - B `[AB=0 and A ne 0, B ne 0] rArr |A| =0 and |B|=0` |
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| 17. |
If `A=[(-2,3),(1,1)]` then `|A^(-1)|`=?A. -5B. `(-1)/5`C. `1/(25)`D. 25 |
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Answer» Correct Answer - B `"AA"^(-1)=1rArr |"AA"^(-1)|=|I|rArr|A|.|A^(-1)|=1rArr|A^(-1)|=1/(|A|)` `|A|=|(2,-3),(1,1)|=(-2-3)=(-5)rArr|A^(-1)|=-1/5` |
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| 18. |
If `A=[{:(3,1),(7,5):}]`, find x and y such that `A^(2)+xI=yA.`A. x=6,y=6B. x=8,y=8C. x=5,y=8D. x=6,y=8 |
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Answer» Correct Answer - B `[(3,1),(7,5)][(3,1),(7,5)]+x[(1,0),(0,1)]=y[(3,1),(7,5)]` `rArr [(16,8),(56,32)]+[(x,0),(0,x)]=[(3y,y),(7y,5y)]rArr[(16+x,8),(56,32+x)]=[(3y,y),(7y,5y)]` `therefore y=8 and (16+x=3y=3xx8=24) rArr x=8` |
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| 19. |
If `A=[{:(2x,0),(x,x):}]and A^(-1)=[{:(1,0),(-1,2):}]`, then what is the value of x ?A. 1B. 2C. `1/2`D. -2 |
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Answer» Correct Answer - C Use `AA^(-1)=I` |
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| 20. |
If `A=[(1,k,3),(3,k,-2),(2,3,-4)]` is singular then K=?A. `(16)/3`B. `(34)/3`C. `(33)/3`D. none of these |
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Answer» Correct Answer - C A is singular `iff |A| = 0` |
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| 21. |
In the matrix `A=[(3-2x,x+1),(2,4)]` is singular then X=? |
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Answer» Correct Answer - B A is singular `iff |A| = 0` |
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| 22. |
Solve for x and y, given that `[{:(x,y),(3y,x):}][{:(1),(2):}]=[{:(3),(5):}].`A. x=1,y=2B. x=2,y=1C. x=1,y=1D. none of these |
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Answer» Correct Answer - C Solve x+2y=3 and 3y+2=5 |
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| 23. |
If `A=[(3,4,1),(1,0,-2),(-2,-1,2)] then A^(-1)`=?A. `[(2,9,-8),(-2,8,7),(-1,5,-4)]`B. `[(2,9,-8),(2,8,7),(-1,-5,4)]`C. `[(-2,-9,-8),(2,8,7),(-1,5,-4)]`D. none of these |
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Answer» Correct Answer - C `A^(-1)=1/(|A|).adjA` |
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| 24. |
If `[(x-y,2x-y),(2x+z,3z+w)]=[(-1,0),(5,13)]`A. z=3,w=4B. z=4,w=3C. z=1,w=2D. z=2,w=-1 |
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Answer» Correct Answer - A `(x-y=-1 and 2x-y=0) rArr (x=1,y=2)` `(2x+z=5rArr z=3) and (3z+w=13 rArr w=4)` |
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| 25. |
If `A=[(costheta,-sintheta),(sintheta,costheta)] " then " A^(-1)` =?A. AB. `-A`C. adjAD. `=-adjA` |
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Answer» Correct Answer - C `|A|=1 rArr A^(-1) =1/|A| adjA=(adjA)` |
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| 26. |
If `2[(3,4),(5,x)]+[(1,y),(0,1)]=[(7,0),(10,5)]` thenA. (x=-2,y=8)B. (x=2,y=-8)C. (x=3,y=-6)D. (x=-3,y=6) |
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Answer» Correct Answer - B `[2x+1=5 rArr x=2] and [8+y=0 rArr y=-8]` |
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| 27. |
If `A=[(2,-1),(1,3)], then A^(-1)`=?A. `[(3/7,(-1)/7),(1/7,2/7)]`B. `[(3/7,(1)/7),((-1)/7,2/7)]`C. `[(3/7,(1)/7),(1/7,2/7)]`D. none of these |
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Answer» Correct Answer - B `|A|=|(2,-1),(1,3)|=(6+1)=7 ne 0` `M_(11)=3,M_(12)=1,M_(21)=-1 and M_(22)=2` `therefore C_(11)=3,C_(12)=-1,C_(21)=1 and C_(22)=2` `rArr AdjA=[(3,-1),(1,2)]=[(3,1),(-1,2)]` `rArr A^(-1)=1/(|A|)(adjA)=1/7[(3,1),(-1,2)]=[(3/7,1/7),(-1/7,2/7)]` |
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| 28. |
Find matrices `A` and `B`, if `2A - B = [[6, -6, 0], [-4, 2, 1]]` and `2B + A = [[3, 2, 5], [-2, 1, -7]]`A. `[(-3,2,1),(2,1,-1)]`B. `[(3,2,-1),(2,-1,1)]`C. `[(3,2,-1),(-2,1,-1)]`D. none of these |
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Answer» Correct Answer - C 5A=2(2A-B)+(2B+A) Then , A=1/5(5A). |
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| 29. |
If |A|=3 and `A^(-1)=[(3,-1),((-5)/3,2/3)]` then adjA=?A. `[(9,5),(-5,-2)]`B. `[(9,-3),(-5,2)]`C. `[(-9,3),(5,-2)]`D. `[(9,-3),(5,-2)]` |
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Answer» Correct Answer - B `A^(-1)=1/(|A|).adjA rarr adjA=|A|.A^(-1)=3A^(-1)=[(9,5),(-5,2)]` |
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| 30. |
If `[(3,-2),(5,6)]+2A=[(5,6),(-7,10)]` then A=?A. `[(1,3),(-5,4)]`B. `[(1,5),(-3,4)]`C. `[(1,4),(6,2)]`D. none of these |
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Answer» Correct Answer - C `2A=[(5,6),(-7,10)]-[(3,-2),(5,6)]=[(2,8),(-12,4)]` |
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| 31. |
If `(A-2B)=[(1,-2),(3,0)] and (2A-3B)=[(-2,2),(3,-3)]` then B=?A. `[(6,-4),(-3,3)]`B. `[(-4,6),(-3,-3)]`C. `[(4,-6),(3,-3)]`D. none of these |
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Answer» Correct Answer - B B=(2A-3B)-2(A-2B) |
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| 32. |
If`A=[(2,0),(-3,1)] and B=[(4,-3),(-6,2)]` are such that 4A+3X=5B then x=?A. `[(4,-5),(-6,2)]`B. `[(4,5),(-6,-2)]`C. `[(-4,5),(6,-2)]`D. none of these |
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Answer» Correct Answer - A `4A+3X=5BrArr 3X=(5B-4A) rArr X=1/3(5B-A)` |
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| 33. |
If `A=[(1,4),(4,-3) and f(x) =2x^2-4x+5 then f(A)` =?A. `[(19,-32),(-16,51)]`B. `[(19,-16),(-32,51)]`C. `[(19,-11),(-27,51)]`D. none of these |
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Answer» Correct Answer - B `f(A)=2A^2-4A+5I` |
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| 34. |
If A is an invertible matrix and `A^(-1)=[(3,5),(5,6)]` then A=?A. `[(6,-4),(-5,3)]`B. `[(1/3,1/4),(1/5,1/6)]`C. `[(-3,2),(5/2,(-3)/2)]`D. none of these |
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Answer» Correct Answer - C `A=(A^(-1))^(-1)` . So find the inverse of `A^(-1)` |
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| 35. |
If `A=[(1,lambda,2),(1,2,5),(2,1,1)]` is not invertible then `lambda`=?A. 2B. 1C. -1D. 0 |
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Answer» Correct Answer - B A is not invertible `iff |A|=0` |
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| 36. |
If `A=[(1,2),(4,3)] then A^2-4A`=?A. IB. 5IC. 3ID. 0 |
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Answer» Correct Answer - B `A^2-4A=[(1,4),(2,3)][(1,4),(2,3)]-[(4,16),(8,12)]=[(9,16),(8,17)]-[(4,16),(8,12)]=[(5,0),(0,5)]=5I` |
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| 37. |
Find a matrix X such that `X.[(3,2),(1,-1)]=[(4,1),(2,3)]`.A. `[(1,-1),(1,1)]`B. `[(1,1),(-1,1)]`C. `[(1,1),(1,-1)]`D. none of these |
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Answer» Correct Answer - C Let `[(a,b),(c,d)][(3,2),(1,-1)]=[(4,1),(2,3)]` Find a, b , c and d |
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| 38. |
If A is an invertible square matrix then `|A^(-1)|`=?A. |A|B. `1/(|A|)`C. 1D. 0 |
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Answer» Correct Answer - B `"AA"^(-1)=1rArr|"AA"^-1|=|I|=1` `rArr |A|.|A^(-1)|=1 rArr |A^(-1)|=1/(|A|)` |
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| 39. |
If A is 2-rowed square matrix and |A|=6 then A,adjA=?A. `[(6,0),(0,6)]`B. `[(3,0),(0,3)]`C. `[(1/6,0),(0,1/6)]`D. none of these |
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Answer» Correct Answer - A `A.(adjA)=|A|.I=6.[(1,0),(0,1)]=[(6,0),(0,6)]` |
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| 40. |
If A is a 3-rowed square matrix and |A|=5 then |adjA|=?A. 5B. 25C. 125D. none of these |
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Answer» Correct Answer - B `|adjA|=|A|^(n-2)=|A|^2=5^2=25` |
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| 41. |
If `A=[(3,-4),(-1,2)]` and B is a square matrix of order 2 such that AB=I then B=?A. `[(1,2),(2,3)]`B. `[(1,1/2),(2,3/2)]`C. `[(1,2),(1/2,3/2)]`D. none of these |
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Answer» Correct Answer - C `AB=I rArr B =A^(-1)` |
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| 42. |
If A is a 3-rowed square matrix and |A|=4 then |adjA|=?A. 4AB. 16AC. 64AD. none of these |
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Answer» Correct Answer - A adj(adjA)`=|A|^(n-2)=|A|^(3-2).A=|A|.A=4A` |
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| 43. |
It A is 3-rowed square matrix and |3A|=k|A| then k=?A. 3B. 9C. 27D. 1 |
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Answer» Correct Answer - C `|3A|=(3xx3xx3)|A|=27.|A|` |
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| 44. |
Which one of the following is a scalar matrixA. `[(1,1),(1,1)]`B. `[(6,0),(0,3)]`C. `[(-8,0),(0,-8)]`D. none of these |
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Answer» Correct Answer - C A scalar matrix is a square matrix each of whose non-diagonal elements is 0 and all diiagonal elements are equal |
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| 45. |
If `A=[{:(1,-1),(2,-1):}],B=[{:(a,-1),(b,-1):}]" and "(A+B)^(2)=(A^(2)+B^(2))` then find the values of a and b.A. a=2,b=-3B. a=-2,b=3C. a=1,b=4D. none of these |
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Answer» Correct Answer - C `(A+B)^2=(A^2+B^2) iff A^2+B^2+AB=(A^2+B^2) iff AB=-BA` `AB=-BA iff [(a-b,2),(2a-b,3)]=[(-a-2,a+1),(-b+2,b-1)]` `Now ,(a+1=2 and b-1=3) rArr (a=1 and b=4)` |
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