NCERT Solutions Class 12 Mathematics Chapter 3 – Matrices

NCERT Solutions Class 12 Mathematics Chapter 3 Matrices are available for the students on the Extramarks website. In addition, the students preparing for the Term one exam can refer to the NCERT Solutions for Chapter 3. This chapter provides a simple and clear solution for every complex problem. The solution is built on the CBSE NCERT latest 2022-2023 syllabus, and it offers step-by-step guidance for the students. 

Students can get NCERT Solutions Class 12 Mathematics Chapter 3 on Extramarks website and can prepare for the upcoming exams and tests. They will be able to understand the essentials of matrices, their order, types, and algebra and will be able to score better. Extramarks solutions cover complex theorems, concepts, formulas, and matrices that aid students in preparing for various competitive exams. Students should  check the Extramarks website regularly for the latest updates and notifications regarding CBSE exams, syllabus changes, notes updates, etc. In addition, Extramarks provides NCERT Solutions for all classes from CLass 1 to Class 12.

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 3

Extramarks NCERT Solutions Class 12 Mathematics Chapter 3 covers all topics under Matrices. These are:

  • An introduction to fundamentals of the matrix
  • Order of a matrix
  • Types of matrices
  • Operation on matrices. 

Some of the essential topics for students include 

  • The transpose of a matrix 
  • The elementary operation of a matrix. 

Further to this, the vital subtopics include 

  • Properties of matrix operation 
  • The transpose of a matrix. 

NCERT Solutions Class 12 Mathematics Chapter 3 has approximately 62 questions in four exercises. Along with this, 15 more are provided in various exercises, amongst which 41 questions are short answer type, 11 multiple choice questions, and the others include 25 long answer type questions. 

List of formulas the students must learn in the Class 12 Mathematics NCERT Solutions Class 3 Mathematics include

  • Scalar Matrix: If a matrix M = [aij]n×n, it is a scalar matrix. 
  • Identity Matrix: If a matrix M = M = [aij]n×n, then it is known as identity matrix.
  • Zero Matrix: If all the elements in a matrix are zero, it is known as zero matrices. 
  • Negative of a Matrix: A negative matrix is presented as -A or -A = (-1) A

Students can click here to access the NCERT Solutions Class 12 Mathematics Chapter 3 provided by Extramarks.

Key topics covered in Chapter 3 Matrices include: 

Exercise Key Topics
3.1 Introduction
3.2 Matrix
3.3 Types of Matrices
3.4 Operations of Matrices
3.5 Transpose of a Matrix
3.6 Symmetric and Skew Symmetric Matrices
3.7 Elementary Operation of a Matrix
3.8 Invertible Matrices

A brief on the NCERT Solutions for Class 12 Mathematics Chapter 3

3.1 Introduction

Class 12 Mathematics Chapter 3 Matrices starts with an introductory exercise that helps understand the basic concepts, meaning students will explore concepts of matrices. They will understand the actual application of matrices in various branches of Mathematics. The students will learn about the evolution of the concept of Matrices, which will help them extract compact and simple techniques for solving problems. For a more detailed explanation, students can refer to Extramarks NCERT Solutions, which offer a clear and simple explanation for every complex problem related to Matrices. 

3.2 Matrix 

The second exercise explains the fundamental principles of matrices. The students will get to know the proper definition of a Matrix and learn the different order of a matrix. They will also understand that the horizontal and vertical lines of entries in a matrix are known as rows and columns. Students may refer to Matrix Class 12 on Extramarks, wherein we cover all essential elements in this exercise. 

3.3 Types of Matrices

Different parts and types of Matrices are illustrated with complex examples. The students will get a hold on various types of matrices, including column matrix, row matrix, square matrix, diagonal matrix, and scalar matrix. Further, there are multiple examples of equality of matrices with solutions. The examples provided will help the students boost their calculative minds and help them delve deeper into the concepts. Many questions appear in the entrance exams based on the types of matrices, and it would help if students practise more. Hence, students may refer to NCERT Solutions Class 12 Mathematics Chapter 3 on Extramarks by clicking here.

3.4 Operations on Matrices

NCERT Solutions Class 12 Mathematics Chapter 3 Exercise 3.4 introduces various operations on matrices. It includes the addition, multiplication, and differences of matrices. Extramarks NCERT Solution highlights every key concept on matrices and scalar multiplication properties. Therefore, it is one of the complex exercises in Class 12 Chapter 3, and students may refer to NCERT Solutions to get a profound understanding of how matrices operate. 

3.5 Transpose of a Matrix 

In the transpose of a matrix, students will learn about the unique types of matrices, including symmetric and skew-symmetric matrices. It helps to establish the foundation of the transpose of the matrix, and students can refer to comprehensive examples and solutions to strengthen their knowledge.

3.6 Symmetric and Skew Symmetric Matrices

Towards the end of exercise 3.5, all the necessary concepts would have been covered. Students will learn the critical difference between symmetric and skew-symmetric matrices in this exercise. However, the exercise from the section will only discuss the theorems and complex derivations. 

3.7 Elementary Operation of a Matrix 

This exercise elaborates the six operations on the matrix, amongst which three are due to rows and three due to columns. In addition, students may refer to  NCERT examples and solutions to learn more about the elementary operations of matrices. 

3.8 Invertible Matrices

In this exercise, students will learn all properties of invertible matrices. This section is essential as students get to know about inverse matrix concepts. They will study the nature of matrices and that not all matrices are invertible. It helps shape a better ideology of the complex concepts as diverse examples and solutions are provided. Students may refer to NCERT Solutions Class 12 Mathematics Chapter 3 on Extramarks by clicking here to find more notes on the difficult theorem on invertible matrices. 

NCERT Solutions Class 12 Mathematics Chapter 3 Exercise & Answer Solutions

Matrices are one of the essential and interactive chapters wherein students get to improve their calculations and speed of solving problems. To benefit the students, NCERT Solutions Class 12 Mathematics Chapter 3 Exercise and Answer is made available for students on the Extramarks website. Students can click here to access the study material. It consists of step-by-step solutions with a theoretical explanation of various complex concepts. 

In our NCERT Solutions, Extramarks have covered all essential concepts such as elementary operations and invertible matrices. Students may click on the below links to download exercise specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 3 Matrices:

  • Chapter 3: Exercise 3.1 Solutions: 10 Questions (5 Short Questions, 5 Long Questions)
  • Chapter 3: Exercise 3.2 Solutions: 22 Questions (3 Short Questions, 19 Long Questions)
  • Chapter 3: Exercise 3.2 Solutions: 22 Questions (3 Short Questions, 19 Long Questions)
  • Chapter 3: Exercise 3.4 Solutions: 18 Questions (18 Short Questions)

Students can explore Class 12 Mathematics Solutions through our Extramarks website for all primary and secondary classes by clicking on the respective class link given below. 

  • NCERT Solutions Class 1
  • NCERT Solutions Class 2
  • NCERT Solutions Class 3
  • NCERT Solutions Class 4, 
  • NCERT Solutions Class 5
  • NCERT Solutions Class 6
  • NCERT Solutions Class 7
  • NCERT Solutions Class 8
  • NCERT Solutions Class 9
  • NCERT Solutions Class 10
  • NCERT Solutions Class 11

NCERT Exemplar Class 12 Mathematics

NCERT Exemplar Class 12 Mathematic is an excellent source of information, and it provides insights into NCERT Solutions. It has various examples that are curtailed with detailed solutions. The students gather complete knowledge on the Mathematics concepts and theories. It will help them secure good marks, and it acts as an excellent material for competitive exam preparation such as JEE Main, NEET, etc. 

The exemplar covers all the essential topics and concepts under the NCERT curriculum and CBSE Class 12 Mathematics Syllabus. The language is easy to understand, and students can grasp it quickly. The NCERT Exemplar consists of various types of questions such as multiple-choice questions, short answer type questions, and long answer type questions. The students can refer to the exemplar to understand the concept and improve their performance in the examination. 

Key Features of NCERT Solutions Class 12 Mathematics Chapter 3

NCERT Solutions Class 12 Mathematics Chapter 3 covers all essential topics needed to prepare for the first-term CBSE 12 exam. The students will get to improve their foundational base and the core concepts. The key features are as follows:

  • Class 12 Mathematics NCERT Solutions Chapter 3 Matrices has elaborative concepts that help students understand complex concepts easily.
  • Students can grasp the concepts quickly and easily and, in turn, inculcate strong principles essential to solve the exercises much quicker. 
  • Students can refer to NCERT Solutions Class 12 Mathematics Chapter 3 to obtain more marks as the solutions are presented with step-by-step details. 
  • Chapter 3 Class 12 NCERT solutions offer a fundamental understanding of matrices along with their properties and types. Students who wish to track their answers can practise the exercises mentioned in the chapter. 

The Chapter 3 Mathematics Class 12 NCERT solutions offer many benefits that students can reap if they refer to it while they study. Students can click here to access the NCERT solutions class 12 mathematics chapter 3 by Extramarks. 

Q.1 In a matrix A=[2  5 19 73525 2123  15 17],  write:(i) The order of the matrix(ii) Then umber of elements,(iii) Write the elements a13,a21,a33,a24,a23.

Ans.

(i) The order of matrix A is 3×4.

(ii) The number of elements in matrix A is 12.

(iii) The value of a13 = 19, a21 = 35, a33 = – 5, a24 = 12, a23 = 5/2.

Q.2 If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?

Ans.

Possible order of matrix

1 × 24, 2 × 12, 3 × 8, 4 × 6, 6 × 4, 12 × 2, 24 × 1, 8 × 3.
Possible order for 13 elements: 1 × 13, 13 × 1.

Q.3 If a matrix has 18 elements, what are the possible orders it can have? What, if it has 5 elements?

Ans.

Possible order for 18 elements

1 ×18, 2 × 9, 3 × 6, 6 × 3, 9 × 2, 18 × 1

Possible order for 5 elements = 1 × 5, 5 × 1.

Q.4

Constructa 2×2matrix, A= [aij],whose elementsaregivenby: (i) aij=(i+j)22 (ii) aij=ij (iii) aij=(i+2j)22

Ans.

    A=[aij](i) aij=(i+j)22      A=(a11 a12a21 a22)=(292928)(ii)  aij=ij        A=(a11 a12a21 a22)=(11221)(iii)aij=(i+2j)22        A=(a11 a12a21 a22)=(92252818)

Q.5

Construct a 3 × 4 matrix, whose elements are  given by:(i) aij=12|3i+j| (ii) aij=2ij

Ans.

(i) aij=12|3i+j|      A=[a11a12a13a14a21a22a23a24a31a32a33a34]      A=[112012522321472352](ii) aij=2ij      A=[a11a12a13a14a21a22a23a24a31a32a33a34]          =[101232  1  054  3  2]

Q.6

Find the values of x, y and z from the following equations:(i)    [4 3x 5]=[y z1 5]  (ii) [x+y 35+z xy]=[6 25 8](iii) [x+y+zx+zy+z]=[957]

Ans.

(i)    [4 3x 5]=[y z1 5]x=1,y=4 and z=3(ii) [x+y 35+z xy]=[6 25 8]5+z=5z=55=0    x+y=6, and xy=8y=8xx+y=6    x+8x=6    x2+8x=6x2+8=6xx26x+8=0x24x2x+8=0      (x4)(x2)=0x=2,4If x=2 then y=82=4If  x=4  then  y=84=2(iii) [x+y+zx+zy+z]=[957]x+y+z=9...(i)            x+z=5...(ii)            y+z=7...(iii)From equation (i) and equation (ii), we havey+5=9y=4From equation (i) and equation (iii), we havex+7=9x=2From equation(i), we have2+4+z=9z=96=3Thus,x=2,y=4and  z=3.

Q.7

Find the value ofa, b, c and d from thee quation:[ab 2a+c2ab 3c+d]=[1   5  0 13]

Ans.

[ab 2a+c2ab 3c+d]=[1   5  0 13]ab=1...(i)        2a+c=5...(ii)        2ab=0...(iii)         3c+d=13...(iv)From equation (i) andequation (iii), we havea=1 and b=2Putting value of a in equation(ii), we get2(1)+c=5c=52=3Putting value of c in equation(iv), we get3(3)+d=13d=139=4

Q.8

A= [aij]m×nis a square matrix, if(A) m<n(B) m>n(C) m=n(D) None of these

Ans.

A is a square matrix if m=n.

So, option (C) is correction answer.

Q.9

Which of the given values of x and y make the followingpair of matrice sequal[3x+7 5y+1 23x],[0 y28 4](A) x=13,  y=7(B) Not possiblet of ind(C) y=7,x=23          (D)x=13,y=23

Ans.

If both matrices are equal, then[3x+7 5y+1 2-3x]= [0 y-28 4]    3x+7=0,y2=5and    y+1=8,  23x=4Then,x=73,y=7andx=23Since, there are two different values of x, so it is difficult to find the value of x to make matrices equal.

Q.10 The number of all possible matrices of order 3 × 3 with each entry 0 or 1 is:

(A) 27

(B) 18

(C) 81

(D) 512

Ans.

Number of elements is 9 and each element can be filled by 0 or 1.

The number of all possible matrices of order 3 x 3
with each entry 0 or 1 = 29 = 512

So, the option ‘D’ is correct.

Q.11 

In the matrix A= [2 43 2],B=[  1 32 5],C=[2 5  3 4]Find each of the following:(i) A+B (ii) AB (iii) 3AC (iv) AB(v) BA

Ans.

In the matrix A=[2 43 2], B=[   1 32 5], C=[2 5   3 4](i) A+B=[2 43 2]+[   1 32 5]      =[2+1 4+33+(2) 2+5]      =[3 71 7](ii) AB=[2 43 2][  1 32 5]      =[2 1 433(2) 25]      =[1   153](iii)3AC=3[2 43 2][2 5  3 4]  =[6 129 6][2 5  3 4]  =[6(2) 12 593 6 4]  =[8 76 2](iv)        AB=[2 43 2][   1 32 5]  =[28 6+2034 9+10]  =[6 261 19](v)        BA=[   1 32 5][2 43 2]=[  2+9 4+64+15 8+10]=[11 1011   2]

Q.12

Compute the following:(i)    [  a bb a]+[a bb a](ii) [a2+b2 b2+c2a2+c2 a2+b2](iii) [1 468 5 162 8 5]+[12 7 68 0 53 2 4](iv) [cos2x sin2xsin2x cos2x]+[sin2x cos2xcos2x sin2x]

Ans.

(i)    [  a bb a]+[a bb a]=[  a+a b+bb+b a+a]        =[2a 2b0 2a](ii)[a2+b2 b2+c2a2+c2 a2+b2]+[  2ab 2bc2ac      2ab]    =[a2+b2+2ab b2+c2+2bca2+c22ac a2+b22ab]    =[(a+b)2 (b+c)2(ac)2 (ab)2](iii)[1 4 68 5 162 8  5]+[12 7 68 0 53 2 4]    =[1+12 4+7         6+68+8 5+0 16+52+3 8+2   5+4]    =[11 11 016 5 21  5 10  9](iv) [cos2x sin2xsin2x cos2x]+[sin2x cos2xcos2x sin2x]    =[cos2x+sin2x sin2x+cos2xsin2x+cos2x cos2x+sin2x]    =[1 11 1]

Q.13

Compute the indicated products:(i)    [  a bb a] [abb  a]      (ii) [123] [234](iii) [1 223] [1 2 32 3 1](iv) [2 3 43 4 54 5 6] [1 3 502 43   0 5] ( v )[ 2 1 3 2 1 1 ][ 1 0 1 1 2 1 ] ( vi )[ 3 1 3 1 0 2 ][ 2 3 1 0 3 1 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaamaabmaabaacbeGaa8NDaaGaayjkaiaawMcaaiaaykW7caaMc8+aamWaaqaabeqaaiaaykW7caaMc8UaaGPaVlaa=jdacaWLjaGaa8xmaaqaaiaaykW7caaMc8UaaGPaVlaa=ndacaWLjaGaa8NmaaqaaiabgkHiTiaa=fdacaWLjaGaa8xmaaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaaykW7caaMc8UaaGPaVlaa=fdacaWLjaGaa8hmaiaaxMaacaWFXaaabaGaeyOeI0Iaa8xmaiaaxMaacaWFYaGaaCzcaiaa=fdaaaGaay5waiaaw2faaaqaamaabmaabaGaa8NDaiaa=LgaaiaawIcacaGLPaaadaWadaabaeqabaGaaGPaVlaaykW7caaMc8Uaa83maiaaxMaacqGHsislcaWFXaGaaCzcaiaa=ndaaeaacqGHsislcaWFXaGaaCzcaiaaykW7caaMc8UaaGPaVlaa=bdacaWLjaGaa8NmaaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaa=jdacaWLjaGaeyOeI0Iaa83maaqaaiaa=fdacaWLjaGaaGPaVlaaykW7caWFWaaabaGaa83maiaaxMaacaaMc8UaaGPaVlaa=fdaaaGaay5waiaaw2faaaaaaa@858D@

Ans.

(i)    [  abba][abb  a]=[  a2+b2ab+abba+ab  b2+a2]      =[  a2+b2ab+abba+ab  b2+a2](ii)[123][234]=[2 3 44 6 86 9 12](iii)[122  3][1 2 32 3 1]=[1426322+64+96+3]    =[3 41  8139](iv)[2 3 43 4 54 5 6][13 50   2 43   0 5]=[2+0+126+6+010+12+203+0+159+8+015+16+254+0+1812+10+020+20+30]          =[140421815622270](v)      [  21  3211][  101121]=[210+22+1320+43+2110+21+1]=[123145220](vi) [   3131  02][231 03 1]=[  61+99+0+32+0+6  3+0+2]=[14645]

Q.14

If A= [12 35021     1  1],B=[3 124252   03] and C=[4 120 321 23],then compute (A+B) and (BC). Also, verify thatA+(BC) = (A+B)C.

Ans.

A+B=[12 350  21    1  1]+[312425203]  =[1+32+(1)3+25+40+22+51+2    1+0  1+3]=[41 19   273 14]BC=[3 124252 03][4 120 321 23]  =[3 411224 0   235221   0(2)33]  =[120  413  1   20]A+(BC)=[12350  21    1  1]+([3124  252  03][412032123])    =[12350  21    1  1]+([34  112240    235221  0+333])    =[12350  21    1  1]+[120  413  1  30]    =[1122      3+05+401  2+31+1        1+3  1+0]    =[00      39    1  52        2  1](A+B)C=([12350  21    1  1]+[3124  252  03])[412032123]    =[1+321      3+25+40+22+51+2      1+0  1+3][412032123]    =[41    19273      14][412032123]    =[4411    1290          23      7231      1+2      43]  =[00    39    152        11]Hence,  A+(BC)=(A+B)C.

Q.15

If  A=[23153132343 73223] & B=[25351152545756525], then compute 3A5B.

Ans.

Given, A=[23 1 5313 23 4373 2 23] and B=[25351152545756525]3A5B=3[23 1 5313 23 4373 2 23]5[25 35 115 25 4575 65 25]        =[3×233×13×533×133×233×433×733×23×23][5×255×355×15×155×255×455×755×655×25]      =[2 3 51 2 47 6 2][2 3 51 2 47 6 2]      =[223355112244776622]      =[0 0 00 0 00 0 0]

Q.16

Simplify cosθ [cosθ sinθsinθ cosθ]+sinθ [sinθ    cosθcosθ sinθ].

Ans.

cosθ [cosθ sinθsinθ cosθ]+sinθ[sinθ    cosθcosθ sinθ]=[  cos2θ   cosθ sinθsinθ cosθ cos2θ]+[sin2θ      sinθ cosθsinθ cosθ sin2θ]=[  cos2θ+sin2θ    cosθ sinθsinθ cosθsinθ cosθ+sinθ cosθ cos2θ+ sin2θ]=[1 00 1].

Q.17

Find X and Y, if(i)X+Y=[7 02 5]  and  XY=[3 00 3](ii) 2X+3Y=[2 34 0]  and  3X+2Y=[  2 21 5]

Ans.

(i)      X+Y=[7 02 5] ...(i)  and  XY=[3 00 3]...(ii)  Adding both equations, we getX+Y+XY=[7 02 5]+[3 00 3]        2X=[7+30+02+05+3]    =[10028]X=12[10028]    =[5014]Putting value of X in equatin(i), weget[5014]+Y=[7025]Y=[7025][5014]    =[75002154]    =[2011]Thus,X=[5014] and Y=[2011].(ii) 2X+3Y=[2340]...(i)and  3X+2Y=[   221  5]...(ii)Multiplying equation(i) by 2 and equation(ii) by 3, we get2(2X+3Y)=2[2340]  4X+6Y=[4680]...(iii)3(3X+2Y)=3[   221  5] 9X+6Y=[   663  15]...(iv)Subtracting equation (iii) from equation(iv), we get    5X=[   663  15][4680]      X=15[   21211  15]=[  25125115   3]Putting value of X in equation (i), we have2[  25125115   3]+3Y=[2340]  3Y=[2340][  45245225   6]        =[2453(245)4(225)06]Y=13[653954256]=[251351452]Thus, X=[  25125115   3]and  Y=[251351452].

Q.18

Find X, if Y= [3 24 0] and 2X+ Y = [  2 21   5].

Ans.

Given, Y = [3 21 4] and 2X + Y=[  1 03 2]...(i)Putting value of Y in equation (i),weget2X+[3214]=[  1 03 2]                    2X=[  1 03 2][3 21 4]        =[  13023124]                        X=12[2 24 2]      =[1 12 1]

Q.19

If  x  and y, if 2 [1 30 x]+[y 01 2]=[5 61 8].

Ans.

Given:   2[130x]+[y 01 2]=[5618][2 602 x]+[y 01 2]=[5 61 8]    [2+y  6+00+12x+2]=[5618]          2+y=5  and  2x+2=8      y=3  andx=822=3Thus, x =3 and y=3.

Q.20

Solve thee quation for x, y, z and t, if2[xzyt]+3[110  2]=3[3546].

Ans.

Given:  2[x zy t]+3[1 10 2]=3[3 54 6]          [2x2 z2y 2t]+[330 6]=[9 1 512 1 8]                [2x+32z32y+02t+6]=[9 1 512 1 8]2x+3=9,  2y=12,  2z3=15  and  2t+6=18x=3,y=6,  z=9 and t=6.

Q.21

If  x [23] + y[1  1]=[10  5], find the value of x and y.

Ans.

Goven:x23+y[1   1]=[10  5]  [2x3x]+[y  y]=[10  5]          [2xy3x+y]=[10  5]2xy=10,  3x+y=5x=3  and  y=4

Q.22

Given3[xyzw]=[x612w]+[4  x+yz+w3], f in d thevalue of x, y, z and w.

Ans.

Given:      3[xyzw]=[x612w]+[4  x+yz+w3][3x3y3z3w]=[x+4      6+x+y1+z+w2w+3]  3x=x+4x=2  3y=6+x+yy=43w=2w+3w=3and  3z=1+z+wz=1Therefore,x=2,y=4,z=1and w=3.

Q.23

If F(x) =cosx sinx 0sinx cosx 00   0 1, show that F(x) F(y)=F(x+y).

Ans.

Given: F(x)=cosxsinx 0sinx cosx 00   0 1      F(y)=cosy    siny 0siny cosy 00 0 1L.H.S.:F(x).F(y)=cosxsinx 0sinx cosx 00   0 1cosy      siny   0siny cosy 00   0 1=[cosx cosysinx sinysiny cosxsinx cosy0sinx cosy+cosx sinysinx siny+cosx cosy00  01]=[cos(x+y)sin(x+y)0sin(x+y)cos(x+y)0001]=F(x+y)=R.H.S.

Q.24

Show that(i) [516  7][2134][2134][516  7](ii) [123010110][1100    11234][1100  11234][123010110]

Ans.

(i)  L.H.S.:[516   7][2134]=[1035412+216+28]        =[  7  13334]R.H.S.:[2134][516   7]=[10+62+715+243+28]        =[1653925]Therefore,[516   7][2134][2134][516   7](ii)[123010110][1100   11234][1100   11234][123010110]L.H.S:[123010110][1100   11234]=[1+0+612+90+2+12   0+0+001+0 0+1+01+0+011+0  0+1+0]    =[  5814  01110  1]R.H.S:[1100   11234][123010110]=[1+0+02+1+03+0+00+0+101+10+0+02+0+44+3+46+0+0]    =[113  1  0  0  611  6]  L.H.S.R.H.S

Q.25

Find A 2 5A+6I, if A=[ 2 0 1 2 1 3 11 0 ]. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakeaaieqacaWFgbGaa8xAaiaa=5gacaWFKbGaaGPaVlaaykW7caWFbbWaaWbaaSqabeaacaWFYaaaaOGaeyOeI0Iaa8xnaiaa=feacaWFRaGaa8Nnaiaa=LeacaWFSaGaa8hiaiaa=LgacaWFMbGaa8hiaiaa=bcacaWFbbGaa8xpamaadmaaeaqabeaacaWFYaGaaCzcaiaa=bdacaWLjaGaa8xmaaqaaiaa=jdacaWLjaGaa8xmaiaaxMaacaWFZaaabaGaa8xmaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaeyOeI0Iaa8xmaiaaxMaacaWFWaaaaiaawUfacaGLDbaacaGGUaaaaa@608E@

Ans.

A 2 5A+6I, A=[ 2 0 1 2 1 3 1-1 0 ] A 2 =A×A =[ 2 0 1 2 1 3 11 0 ][ 2 0 1 2 1 3 11 0 ] =[ 4+0+1 0+01 2+0+0 4+2+3 0+13 2+3+0 22+001+0 13+0 ] =[ 5 1 2 9 2 5 0 12 ] Now, A 2 5A+6I=[ 5 1 2 9 2 5 0 12 ]5[ 2 0 1 2 1 3 11 0 ]+6[ 1 0 0 0 1 0 0 0 1 ] =[ 5 1 2 9 2 5 0 12 ][ 10 0 5 10 5 15 55 0 ]+[ 6 0 0 0 6 0 0 0 6 ] =[ 510+6 1 25 910 25+6 515 05 1+5 2+6 ] =[ 1 1 3 1 110 5 4 4 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@AE37@

Q.26

If A = 1 0 2 0 2 12    0 3 ,prove that  A36A2+7A+2I =0.

Ans.

A =1 0 20 2 12    0 3,     A2 =A×A= 1 0 20 2 12  0 3 1 0 20 2 12 0 3= [1+0+4  0+0+02+0+60+0+2  0+4+00+2+32+0+6      0+0+04+0+9]=[5082458013]6A2 = 6[5082458013]  =[30  04812243016  078]A3=A2.A      =[5082458013][102021203]      =[5+0+160+0+010+0+242+0+100+8+04+4+158+0+260+0+016+0+39]      =[210341282334055]7A=7[102021203]          =[7014014714021]2I=2[100010001]      =[200020002]L.H.S:    A36A2+7A+2I      =[210341282334055]6[5082458013]+[7014014714021]+[200020002]      =[210341282334055][3004812243048078]+[7014014714021]+[200020002]      =[2130+7+200+0+03448+14+01212+0+0824+14+22330+7+03448+14+00+0+0+05578+21+2]      =[000000000]=0

Q.27

If A=[3242],  and I=(1001),f indk so thatA2= kA2I.

Ans.

∵ A=[3242]  A2=[3242][3242]        =[986+41288+4]        =[1244]Now,  A2=kA2I[1244]=k[3242]2[1001][1244]=[3k2k4k2k][2002][1244]=[3k22k4k  2k2]      4=4k      k=1Thus, the value of k is 1.

Q.28

If  A=[0tanα2tanα20]and I is the identity matrix of order 2,show that I+A=(IA)[cosαsinαsinα cosα]

Ans.

I+A=[1001]+[0tanα2tanα20]=[1tanα2tanα21]IA=[1001][0tanα2tanα20]=[1tanα2tanα21]R.H.S.=(IA)[cosαsinαsinα  cosα]  =[1tanα2tanα21][cosαsinαsinα  cosα]  =[1sinα2cosα2sinα2cosα21][cosαsinαsinα  cosα]  =[cosα+sinα.sinα2cosα2sinα+cosαsinα2cosα2cosα.sinα2cosα2+sinαsinα.sinα2cosα2+cosα]  =[cosα.cosα2+sinα.sinα2cosα2(sinα.cosα2cosαsinα2)cosα2sinαcosα2cosα.sinα2cosα2cosαcosα2+sinα.sinα2cosα2]  =[cos(αα2)cosα2sin(αα2)cosα2sin(αα2)cosα2cos(αα2)cosα2]  =[cos(α2)cosα2sin(α2)cosα2sin(α2)cosα2cos(α2)cosα2]  =[1tanα2tanα21]=L.H.S.

Q.29 A trust fund has Rs 30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of: (a) Rs 1800 (b) Rs 2000.

Ans.

      Total money to invest= 30,000Money invested in Ist type of bond=xMoney invested in 2nd type of bond=30,000-xRate of interest on first type of bond=5%Rate of interest on secon type of bond=7%i                              Total obtained interest=1800ii                              Total obtained interest=2000          IIIx30,000-x5%7%=18002000i 5%  ofx+7%30,000-x=1800        5100x+710030,000-x=1800                      5x+210000-7x=180000                      210000-180000=2x      30000=2x      x=300002=15000To get annual interest of 1800, he should invest 15,000 in I type of bond and 15,000 in II type of bond.ii 5%  ofx+7%30,000-x=2000        5100x+710030,000-x=2000                      5x+210000-7x=200000                      210000-200000=2x      10000=2x      x =100002= 5000To get annual interest of 2000 he should invest 5000 & 25000 respectively in I and II type of bond.

Q.30

The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are 80, 60 and 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Ans.

Number of chemistry books on bookshop= 10 dozen= 10×12= 120Number of physics books on bookshop= 8 dozen= 8×12= 96Number of economics books on bookshop= 10 dozen= 10×12= 120Selling price of one chemistry book =80Selling price of one physics book =60Selling price of one economics book =40The total amount the bookshop will receive= 12096120806040=9600+5760+4800=20160Thus, the total amount the bookshop will receive from selling all the booksis20,160.

Q.31

Assume X, Y, Z, W and P are matrices of order 2 x n, 3 x k, 2 x p, n x 3 and p x k respectively. Choose the correct answer in Exercises 21 and 22.

The restriction of n, k and p so that PY + WY will be defined are:

(A) k = 3, p = n

(B) k is arbitrary, p = 2

(C) p is arbitrary, k = 3

(D) k = 2, p = 3

Ans.

Matrix Order
X 2 x n
Y 3 x k
Z 2 x p
W n x 3

PY + WY=Order of P× Order of Y+ Order of W× Order of Y

= ( p×k )×( 3×k )+( n×3 )×( 3×k )

= ( p×k )+( n×k )

Multiplication of PY is possible if k=3 and sum of PY and WY is possible if p = n.

Thus, option ( A ) is correct.

Q.32

Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k respectively. Choose the correct answer in Exercises 21 and 22.

If n = p, then the order of the matrix 7X – 5Z is:

(A) P × 2

(B) 2 × n

(C) n × 3

(D) p × n

Ans.

The order of matrix X is 2×nThe order of matrix Z is 2×p7X5Z is defined when X and Z an of the same order.n=p(Given)Thus the order of 7X5Z is 2×n.Thus, option (B) is correct answer.

Q.33

Find the t ranspose of each of the followingmatrices:(i) [  5  121] (ii) [112  3] (iii) [15  635  6  231]

Ans.

(i) Transpose of  [  5  121]= [5121](ii) Transpose of[112  3]=[  1213] (iii) Transpose of[15  635 6  231]=[13  2  5  5  3  6  61]

Q.34

If A=[123  579211]andB=[415  12  0  13  1], then verify that(i) (A+B)=A+B      (ii) (AB)=AB

Ans.

(i) A+B=[123  579211]+[415  12   0  13   1]      =[142+135  5+17+29+02+11+31+1]      =[532  69  914  2](A+B)=[532  69  914  2]      =[561  39  429  2]A+B=[123  579211]+[415  12  0  13  1]    =[152  27  1  39  1]+[411  123501]    =[145+12+1  2+17+2  1+3  359+0  1+1]    =[561  39  429  2]Thus, (A+B)=A+B.(ii) AB=[123  579211][415  12   0  13   1]        =[1(4)213(5)  517290211311]        =[  318  459320](AB)=[  318  459320]        =[34315289  0]AB=[123  579211][415  12  0  13  1]    =[152  27  1  39  1][411  123501]    =[1+45121  2172  13  3+590  11]    =[34315289  0]Thus, (AB)=AB.

Q.35

If A=[  3412  01] and B=[121  123],then verify that(i) (A+B)=A+B      (ii) (AB)=AB

Ans.

Since,   A‘= [  3412  01]A=[3104  21]and  B=[121  123]So,  A+B=[3104  21]+[121  123]=[311+20+14+1    2+21+3]=[2115  43](A+B)=[2115  44]=[251414]      A+B=[3104  21]+[121  123]=[  3412  01]+[11  22  13]=[  314+11+22+2  0+11+3]=[251414]Thus,(A+B)=A+B.(ii) Since,  A‘=[  3412  01]A=[3104  21]and  B=[121  123]So,  AB=[3104  21][121  123]=[3+1120141    2213]=[4313    02](AB)=[4313    02]=[  4  33  012]      AB=[  3412  01][121  123]=[  3412  01][11  22  13]=[  3+1411222  0113]=[  4  33  012]Thus, (AB) =AB.

Q.36

If A’=[23  12] and B =[10  12], then find (A+2B).

Ans.

Since,  A=[23  12]A=[21  32]and B=[10  12]    A+2B=[21  32]+2[10  12]        =[21  32]+[20  24]        =[41  56]        =[45  16] (A+2B)=[45  16]            =[41  56]

Q.37

For the matrices A and B,verify that (AB)=BA, where(i) A=[  14  3],  B=[121]   (ii) A=[012],B=[157]

Ans.

For the matrices A and B, verify that (AB)=BA‘,where(i) A=[   14   3], B=[1 2 1]    AB=[   14   3][1 2 1]            =[1   2   1  4843  6  3](AB)=[1   2   1  4843  6  3]            =[1  43  28  6  14  3]    A=[   14   3]=[143]    B=[121]=[1  2  1]BA=[1  2  1][143]            =[1  43  28  6  14  3]Thus,(AB)=BA.(ii) A=[012], B=[1 5 7]    AB=[012][1 5 7]            =[0 0 01 5 721 0 14](AB)=[0 0 01 5 721 0 14]            =[0 1 20 5 100 7 14]      A=[012]            =[012]      B=[157]            =[157]BA=[157][012]            =[01  205100714]Thus,(AB)=BA

Q.38

If  (i) A=[  cosα  sinαsinαcosα],the nver if y AA=I    (ii) A=[  sinα  cosαcosα  sinα],the nver if y AA=I

Ans.

(i) Given  A=[  cosα  sinαsinαcosα]        A=[  cosα  sinαsinαcosα]    =[cosαsinαsinα  cosα]L.H.S:  AA=[  cosα  sinαsinαcosα][cosαsinαsinα  cosα]            =[cos2α+sin2αcosαsinα+sinαcosαsinαcosα+cosαsinαsin2α+cos2α]            =[1001]            =I=R.H.S.(ii)A=[  sinα  cosαcosα  sinα]        A=[  sinα  cosαcosα  sinα]    =[sinαcosαcosα  sinα]  AA=[  sinα  cosαcosα  sinα][sinαcosαcosα  sinα]    =[sin2α+cos2αsinαcosα+cosαsinαcosαsinα+sinαcosαcos2α+sin2α]    =[1001]    =IThus, AA=I.

Q.39

(i) Show tha t the matrix A=[1151   215   13] is a symmetric matrix.(ii) Show tha tthe matrix A=[   0  111  0   1   11  0] is a skew      symmetric matrix.

Ans.

(i)A =[   1151  21  5  13] then  A=[   1151  21   5  13]      =[1151  21 5  13]=ASince, A=A, so A is symmetric matrix.(ii) A =[  0  111  0  1  11  0]        A=[   0  111 0   1   11 0]    =[  01    1    1  0111    0]=[  0  111  0  1  11  0]    =ASince,A=A, so A is skew symmetric matrix.

Q.40

For the matrix A=[1567], verify that(i) (A+A) is a symmetric matrix.(ii) (AA) is as kew symmetric matrix.

Ans.

Given:  A =[1567]A=[1657](i) A+A=[1567]+[1657]         =[1+15+66+57+7]         =[2111114](A+A)=[2111114]         =[2111114]         =A+ATherefore,(A+A) is symmetric matrix.(ii) AA=[1567][1657]         =[11566577]         =[0110](AA)=[0110]       =[  0110]=[0110]         =(AA)Therefore,(AA) is a skew symmetric matrix.

Q.41

Find   12(A+A) and 12(AA), when A=0 a ba0 cb c 0.

Ans.

Given,  A= 0    a ba   0 cbc 0]so,      A= 0   a ba   0 cbc 0        =0aba   0cb   c    0(i)  A+A= 0     ba   0 cb c 0 +0a ba   0 cb   c    0=0 0 00 0 00 0 012(A+A)=12 0 0 00 0 00 0 0=0 0 00 0 00 0 0(ii)  AA=[  0 aba  0cbc0][0aba  0cb  c   0]=    0 2a 2b2a0 2c2b       2c   012(AA)=12  0 2a 2b2a 0 2c2b  2c 0=  0    a ba   0 cb c 0

Q.42

Express the following matrices as the sum of a symmetricand as kew symmetric matrix:  (i) [3  511]      (ii) [  62  22  31  21  3](iii) [  3  3122  145  2] (iv) [  1512]

Ans.

(i)  Let  A=[3  511]        A=[3  151]12(A+A)=12{[3  511]+[3  151]}  =12[6  662]  =[3  331]=B(Let)                B=[3  331]  =[3  331]So,  12(A+A) is symmetric matrix.12(AA)=12{[3  511][3  151]}  =12[  0  44  0]  =[  0  22  0]=C(Let)                C=[  0  22  0]  =[022    0]  =[  0  22  0]=CSo,  12(AA) is skew  symmetric matrix.Now,  B+C=[3  331]+[  0  22  0]  =[3  511]=A(ii)Let         A=[  62  22  31  21  3]              A=[  62  22  31  21  3]        A+A=[  62  22  31  21  3]+[  62  22  31  21  3]        =[124  44  62  42  6]12(A+A)=[  62  22  31  21  3]=B(Let)        AA=[  62  22  31  21  3][  62  22  31  21  3]        =[0 0 00 0 00 0 0]12(AA)=[0 0 00 0 00 0 0]=C(Let)C is skew symmetric matrix.        B+C=[  62  22  31  21  3]+[0 0 00 0 00 0 0]        =[  62  22  31  21  3]=A(iii)    Let  A=[  3  3122  145  2]    A=[  324  3251  1  2]A+A=[  3  3122  145  2]+[  324  3251  1  2]    =[  6  15  14454  4]12(A+A)=12[  6  15  14454  4]    =[  3  1252  1222522  2]=B  (Let)                  B=[  3  1252  1222522  2]=BSo,12(A+A) is symmetric matrix.          AA=[  3  3122  145  2][  324  3251  1  2]        =[  0  535  06360]12(AA)=12[  0  535  06360]        =[  0  523252  033230]=C  (Let)        C=[  0  523252  033230]=CSo,12(AA) is skew symmetric matrix.B+C=[  3  1252  1222522  2]+[  0  523252  033230]    =[  3  3122  145  2]=A(iv)Let A=[  1512],          A=[115  2]A+A=[  1512]+[115  2]      =[  1+1511+52+2]        =[2444]12(A+A)=12[2444]        =[1222]=B                  B=[1222]        =[1222]=BTherefore,  12(A+A)  is symmetric matrix.AA=[  1512][115  2]    =[  115+11522]    =[  0660]12(AA)    =12[  0660]    =[  0330]=C        C=[  0330]    =[033  0]=CTherefore,  12(AA)  is a skew symmetric matrix.Now, B+C=[1222]+[  0330]  =[1+02+3232+0]  =[  1512]=A

Q.43

If A, B are symmetric matrices of same order, then AB – BA is a

(A) Skew symmetric matrix

(B) Symmetric matrix

(C) Zero matrix

(D) Identity matrix

Ans.

A and B are symmetric matrices then, A’ = A and B’ = B

Now, (AB – BA)’= (AB)’ – (BA)’

= B’A’ – A’B’

= BA – AB

= – (AB–BA)

Thus, AB – BA is a skew symmetric matrix.

Therefore, option (A) is correct solution.

Q.44

IfA=[cosαsinαsinα  cosα], thenA+A=I, if the value of αis(A) π6(B) π3(C) π(D) 3π2

Ans.

Given:          A=cosαsinαsinα  cosα, A=  cosαsinαsinαcosαA+A=cosα+cosαsinα+sinαsinα+sinα          cosα+cosα    =2cosα002cosα=10012cosα=1    cosα=12=cosπ3α=π3The correct option is B.

Q.45

Using elementary transformations, find the inverse of each of the matrices,1-123

Ans.

Let      A=[112   3]Since,   A=IA    [112   3]=[1001]AApply R2R22R1    [110   5]=[  1021]AR215R2[110   1]=[  1 02515]AApply R1R1+R2    [1001]=[  1 02515]A        A1=[  1 02 515]

Q.46

Using elementary transformations, find the inverse of each of the matrices2111

Ans.

Let            A=[2 11 1]Since,   A=IA      [2 11 1]=[1 00 1]AApply R1R1R2      [1 01 1]=[110   1]AR2R2R1  [1 00 1 ]=[  111  2]A              A1=[  111  2]

Q.47

Using elementary transformations, find the inverse of each of the matrices1327

Ans.

LetA=[1 32 7]Since,   A=IA  [1 32 7]=[1 00 1]AApply R2R22R1  [1 30 1]=[  1 02 1]AR1R13R2 [1 00 1]=[  732  1]A              A1=[  732  1]

Q.48

Using elementary transformations, find the inverse of each of the matrices2357

Ans.

Let          A=[2 35 7]Since,   A=IA  [2 35 7]=[1 00 1]AApply R1R2  [5 72 3]=[0 11 0]AApply R1R12R2[1 12 3]=[2 1  1 0]AApply R2R22R1[1 10 1]=[2  1  52]AApply R1R1R2[1 00 1]=[7  3  52]A          A1=[7  3  52]

Q.49

Using elementary transformations, find the inverse ofeach of the matrices2174

Ans.

Let          A=[2 17 4]Since,   A=IA  [2 17 4]=[1 00 1]AApply R2R23R1[2 11 1]=[  1 03 1]AApply R1R1R2[1 01 1]=[  4 13  1]AApply R2R2R1[1 00 1]=[  417 2]A              A1=[  417  2]

Q.50

Using elementary transformations, find the inverse of each of the matrices2513

Ans.

Let          A=[2 51 3]Since,   A=IA  [2 51 3]=[1 00 1]AApply R1R1R2[1 21 3]=[1 10   1]AApply R2R2R1  [1 20 1]=[  111  2]AApply R1R12R2[1 00 1]=[  351  2]A              A1=[  351  2]

Q.51

Using elementary transformations, find the inverse ofeach of the matrices3152

`
Ans.

Let          A=[3 15 2]Since,   A=IA  [3 15 2]=[1 00 1]AApply R12R1R2[1 05 2]=[210  1]AApply R2R25R1[1 00 2]=[  2110  6]AApply R212R2[1 00 1]=[  215  3]A              A1=[  215  3]

Q.52

Using elementary transformations, find the inverse of each of the matrices 4534

Ans.

Let          A=[4 53 4]Since,   A=IA  [4 53 4]=[1 00 1]AApply R1R1R2[1 13 4]=[110  1]AApply R2R23R1[1 10 1]=[  113  4]AApply R1R1R2[1 00 1]=[  453  4]A              A1=[  453  4]

Q.53

Using elementary transformations, find the inverse of each of the matrices31027

Ans.

Let          A=[3 102  7]Since,   A=IA[3 102  7]=[1 00 1]AApply R1R1R2S0,  [1327]=[110  1]AApply R2R22R1[1 30 1]=[  112  3]AApply R1R13R2[1 00 1]=[  7102  3]A            A1=[  7102  3]

Q.54

Using elementary transformations, find the inverse of each of the matrices3-142

Ans.

Let          A=[   314 2]Since,   A=IA[   314 2]=[1 00 1]AApply R1R2[4 2  31]=[0 11 0]AApply R1(1)R1[4 23 1]=[011  0]AApply R1R1R2[1131]=[11  1  0]AApply R2R23R1[110   2]=[11  4  3]AApply R212R2[110   1]=[11  2  32]AApply R1R1+R2[1 00 1]=[1 122 32]A  A1=[1 122 32]

Q.55

Using elementary transformations, find the inverse of each of the matrices2-61-2

Ans.

Let          A=[2612]Since,   A=IA[2612]=[1 00 1]AApply R1R1R2[1412]=[110  1]AApply R2R2R1[140   2]=[   111  2]AApply R212R2[140   1]=[   1112  1]AApply R1R1+4R2[1 00 1]=[1   312  1]A        A1=[1   312  1]

Q.56

Using elementary transformations, find the inverse of each of the matrices 6-321

Ans.

LetA=[6321]Since, A=IA[6321]=[1 00 1]AApply R1R2[2163]=[0 11 0]AApply R2R2+3R1[140 0]=[0 11 3]AIn second row of L.H.S., all elements are zero.A1 does not exist.

Q.57

Using elementary transformations, find the inverse of each of the matrices 2-312

Ans.

LetA=[ 2 3 1 2 ] Since, A=IA [ 2 3 1 2 ]=[ 1 0 0 1 ]A Apply R 1 R 1 + R 2 [ 1 1 1 2 ]=[ 1 1 0 1 ]A Apply R 2 R 2 + R 1 [ 1 1 0 1 ]=[ 1 1 1 2 ]A Apply R 1 R 1 + R 2 [ 1 0 0 1 ]=[ 2 3 1 2 ]A A 1 =[ 2 3 1 2 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@00D9@

Q.58

Using elementary transformations, find the inverse of each of the matrices2142

Ans.

LetA=[ 2 1 4 2 ] Since, A=IA [ 2 1 4 2 ]=[ 1 0 0 1 ]A Apply R 2 R 2 2 R 1 [ 2 1 0 0 ]=[ 1 0 2 1 ]A Since, in second row of L.H.S., all elements are zero. So, A -1 does not exist. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D7CB@

Q.59

Using elementary transformations, findthe inverse of each of the matrices2-3 32 2 33-2 2

Ans.

Let          A=[23 32  2 332 2]Since,   A=IA[23 32  2 332 2]=[1 0 00 1 00 0 1]AApply R1R3[32 22  2 323 3]=[0 0 10 1 01 0 0]AApply R1R1R2[1412  2  323  3]=[01 10  1 01  0 0]AApply R2R22R1andR3R32R1[1410  10  50  5  5]=[01 10  321  22]AApply R2R2R3[1410  5  00  5  5]=[   0 1   11   1   0   1   2 2]AApply R215R2andR315R3[1 4 10 1  00   1   1]=[   01   115  15  0  15  2525]AApply R3R3R2[14 10   1   00   0 1]=[   01 115   15   0  25   15 25]AApply R1R1+4R2[1 010 1 00 0 1]=[4515    115  15    0  25  1525]AApply R1R1+R3[1 0 001 000 1]=[25  0    3515  15    0  25  1525]A A1=[25   0    3515   15    0  25   1525]

Q.60

Using elementary transformations, find the inverse of each of the matrices   1 323 05  2 5   0

Ans.

LetA=[ 1 3 2 3 0 5 2 5 0 ] Since, A=IA [ 1 3 2 3 0 5 2 5 0 ]=[ 1 0 0 0 1 0 0 0 1 ]A Apply R 2 R 2 +3 R 1 and R 3 R 3 2 R 1 ,we get [ 1 3 2 0 9 11 0 1 4 ]=[ 1 0 0 3 1 0 2 0 1 ]A Apply R 2 R 2 +8 R 3 [ 1 3 2 0 1 21 0 1 4 ]=[ 1 0 0 13 1 8 2 0 1 ]A Apply R 1 R 1 3 R 2 and R 3 R 3 + R 2 [ 1 0 65 0 1 21 0 0 25 ]=[ 40 3 24 13 1 8 15 1 9 ]A Apply R 3 1 25 R 3 [ 1 0 65 0 1 21 0 0 1 ]=[ 40 3 24 13 1 8 15 25 1 25 9 25 ]A = 1 25 [ 40 3 24 13 1 8 15 1 9 ]A Apply R 1 R 1 +65 R 3 and R 2 R 2 +21 R 3 ,weget [ 1 0 0 0 1 0 0 0 1 ]= 1 25 [ 25 10 15 10 4 11 15 1 9 ]A =[ 1 10 25 15 25 10 25 4 25 11 25 15 25 1 25 9 25 ]A =[ 1 2 5 3 5 2 5 4 25 11 25 3 5 1 25 9 25 ]A Therefore, A 1 =[ 1 2 5 3 5 2 5 4 25 11 25 3 5 1 25 9 25 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@8D21@

Q.61

Using elementary transformations, find the inverse of each of the matrices2 0 -15 1 00 0 3

Ans.

LetA=[ 2 0 1 5 1 0 0 1 3 ] Since, A=IA [ 2 0 1 5 1 0 0 1 3 ]=[ 1 0 0 0 1 0 0 0 1 ]A Applying R 1 R 2 ,we get [ 5 1 0 2 0 1 0 1 3 ]=[ 0 1 0 1 0 0 0 0 1 ]A Applying R 1 R 1 2 R 2 ,weget [ 1 1 2 2 0 1 0 1 3 ]=[ 2 1 0 1 0 0 0 0 1 ]A Applying R 2 R 2 2 R 1 ,weget [ 1 1 2 0 2 5 0 1 3 ]=[ 2 1 0 5 2 0 0 0 1 ]A Applying R 2 ( 1 ) R 2 ,weget [ 1 1 2 0 2 5 0 1 3 ]=[ 2 1 0 5 2 0 0 0 1 ]A Applying R 2 R 2 R 3 ,weget [ 1 1 2 0 1 2 0 1 3 ]=[ 2 1 0 5 2 1 0 0 1 ]A Applying R 1 R 1 R 2 ,weget [ 1 0 0 0 1 2 0 1 3 ]=[ 3 1 1 5 2 1 0 0 1 ]A Applying R 3 R 3 R 2 ,weget [ 1 0 0 0 1 2 0 0 1 ]=[ 3 1 1 5 2 1 5 2 2 ]A Applying R 2 R 2 2 R 3 ,weget [ 1 0 0 0 1 0 0 0 1 ]=[ 3 1 1 15 6 5 5 2 2 ]A A 1 =[ 3 1 1 15 6 5 5 2 2 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8IrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@709F@

Q.62

Matrices A and B will be inverse of each other only if

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans.

If B is inverse of A then AB = BA = I.

Thus, option (D) is correct.

Q.63

LetA=0 10 0,show that (aI+bA)n=anI+nan1,where is the identity matrix of order 2 and nN.

Ans.

L.H.S.=(aI+bA)n=(a[1 00 1]+b[0 10 0])n=[ab0a]nR.H.S.=anI+nan1bA=an[1 00 1]+nan1b[0 10 0]=[an 00 an]+[0 nan1b0 0]=[an nan1b0 an]So,[a b0 a]n=[an nan1b0 an]Apply principle of Mathematical InductionP(n):[a b0 a]n=[an nan1b0 an]For n=1,L.H.S.=[a b0 a]1=[a b0 a]R.H.S=[a1 1.a11b0 a1]=[a ab0 a]So,[a b0 a]=[a ab0 a]So,[a b0 a]=[a ab0 a]P (n) is true for n=1.Let P (n) be true for n=k.P(k):[a b0 a]k=[ak kak1b0 ak](i)Multiple both sides by[a b0 a][a b0 a]k×[a b0 a]=[ak kak1b0 ak]×[a b0 a]L.H.S.=[a b0 a]k×[a b0 a]=[ak kak1b0 ak]×[a b0 a][From equation(i),weget]=[ak+1akb+kakb0 ak+1]=[ak+1(k+1)akb0ak+1]L.H.S.=[akkak1b0ak]×[ab0a]=[ak+1bak+kakb0 ak+1]=[ak+1 (k+1)akb0 ak+1]This shows P(n) is true for n = k + 1 then by principle of mathematical induction, P(n)is true for all positive integral values of n.

Q.64

If   A=1 1 11 1 11 1 1 , prove that An = [3n1 3n1 3n13n1 3n13n13n13n13n1],nN.

Ans.

Let P(n): An=[3n-1 3n-1 3n-13n-13n-13n-13n-1 3n-13n-1] and A=[1 1 11 1 11 1 1]For n=1L.H.S.=A=[1 1 11 1 11 1 1]R.H.S.=[31-1 31-131-131-1 31-1 31-131-1 31-131-1]=[30 30 3030 30 3030 30 30]=[1 1 11 1 11 1 1]P(n)is true for n=1.Let P(n) be true for n=k, thenP(k):Ak=[3k-1 3k-1 3k-13k-1 3k-13k-13k-1 3k-13k-1]    ...(i)Multiplying both sides by A, we getAk.A=[3k-1 3k-13k-13k-13k-1 3k-13k-13k-13k-1].AAk+1=[3k-13k-13k-13k-13k-13k-13k-13k-13k-1].[1 1 11 1 11 1 1]R.H.S.=[3k-1 3k-13k-13k-13k-13k-13k-1 3k-13k-1].[111111111]    =[3k1+3k1+3k13k1+3k1+3k13k1+3k1+3k13k1+3k1+3k1 3k1+3k1+3k1 3k1+3k1+3k13k1+3k1+3k13k1+3k1+3k1 3k1+3k1+3k1]    =[3.3k1 3.3k13.3k13.3k13.3k13.3k13.3k13.3k13.3k1]    =[3(k1)+1 3(k1)+13(k1)+13(k1)+1 3(k1)+13(k1)+13(k1)+1 3(k1)+13(k1)+1] P (n)is true forn=k+1By principle of ma the matical induction that P (n) is true f oral ln N.

Q.65

If A=3411,then prove tha t An=1+2n4nn12n,where n is any positive integer.

Ans.

Let P(n):An=1+2n4nn12n, where A=3411Put n=1,A=1+2(1)4(1)(1)12(1)=3411P(n) is true for n=1.Let p(n) be true for n=kP(k):Ak=[1+2k4k      k12k]Multiply both sides by A, we getAk.A=Ak+1=[1+2(k+1)4(k+1)      (k+1)12(k+1)]L.H.S.=Ak.A  =[1+2k4k      k12k].[3411]  =3(1+2k)4k4(1+2k)+4k3k+12k4k1(12k)  =3+6k4k48k+4k3k+12k4k1+2k  =3+2k44kk+12k1  =[1+2(1+k)4(1+k)(k+1)          12(k+1)]=R.H.S.Therefore, P(n) is true for n=k+1.Hence, P(n) is true for all nN.

Q.66

If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.

Ans.

A and B are symmetric matrices, so A’ = A, B’ = B and (AB)’ = B’A’.

Now, (AB – BA)’= (AB)’ – (BA)’

= B’A’ – A’B’

= BA – AB

= – (AB – BA)

Hence, (AB – BA) is skew symmetric matrix.

Q.67

Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.

Ans.

(i) When A is symmetric matrix i.e., A’ = A.

Then,

(B’AB)’ = B’(B’A)’

= B’A’(B’)’

= B’A’B

= B’AB [Since, A’ = A]

Therefore, B’AB is symmetric matrix.

(ii) When A is skew symmetric matrix i.e., A’ = – A.

Then,

(B’AB)’ = B’(B’A)’

= B’A’(B’)’

= B’A’B

= B’(– A)B [Since, A’ = A]

= –(B’AB)

Therefore, B’AB is skew symmetric matrix.

Q.68

Find the values of x, y, z if the matrix A=0 2y  zx   y zx y  zsatisfiy the equation AA= I.

Ans.

    A=0 2y  zx    yzxyz    A=0 xx2y  y yz z   zAA=0x   x2y  y yz zz0 2y  zx  yzx y  z          =0+x2+x2 0+xyxy   0xz+xz0+xyxy 4y2+y2+y2 2yzyzyz0xz+xz 2yzzyyzz2+z2+z2          =2x20006y20003z2Since,AA=ISo,2x2    0      00  6y2      00    0  3z2=1 0 00 1 00 0 12x2=1,6 y2=1and 3z2=1x=±12,y =±16  and  z=±13

Q.69

For what value of x: [ 1 2 1 ][ 1 2 0 2 0 1 1 0 2 ][ 0 2 x ]=0 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=vqpWqaaeaabaGaaiaacaqabeaadaabauaaaOqaaerbwvMCKfMBHbacgeGaa8Nraiaa=9gacaWFYbGaa8hiaiaa=DhacaWFObGaa8xyaiaa=rhacaWFGaGaa8NDaiaa=fgacaWFSbGaa8xDaiaa=vgacaWFGaGaa83Baiaa=zgacaWFGaGaa8hEaiaa=PdacaWFGaGaaGPaVlaaykW7daWadaqaaiaa=fdacaWLjaGaa8NmaiaaxMaacaWFXaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaa=fdacaWLjaGaa8NmaiaaxMaacaWFWaaabaGaa8NmaiaaxMaacaWFWaGaaCzcaiaa=fdaaeaacaWFXaGaaCzcaiaa=bdacaWLjaGaa8NmaaaacaGLBbGaayzxaaWaamWaaqaabeqaaiaa=bdaaeaacaWFYaaabaGaa8hEaaaacaGLBbGaayzxaaGaa8xpaiaa=bdaaaa@66FF@

Ans.

[ 1 2 1 ][ 1 2 0 2 0 1 1 0 2 ][ 0 2 x ]=0 [ 1+4+1 2+0+0 0+2+2 ][ 0 2 x ]=0 [ 6 2 4 ][ 0 2 x ]=0 [ 0+4+4x ]=0 4+4x=0 x=1 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@C41C@

Q.70

If A =    3112, show  that A25A+7I=0.

Ans.

Since, A=    3 11 2 ,thenA2=    3 11 2     3 11 2        =  91 3 + 232        1 +4        =[  8 55         3]L.H.S.=A25A+7I      =  8 55        35     3 11 2 + 7 1 00 1      =  8 55        3   15 55 10 + 7 00 7      =  815+7 5 5+ 05+5+0          3 10+ 7      =0 00 0     =0=R.H.S.

Q.71

Find x, if [x-5-1] 1 0 20 2 12 0 3 x41 = 0.

Ans.

              x51102021203x41=0x+02010+02x53x41=0          x2102x8x41=0                        xx240+2x8=0        x22x40+2x8=0    x248=0                x=±48    =±43

Q.72

A manufacturer produces three product x, y, z which he sells in two markets. Annual sales are indicated below:

Market Products

I 10,000 2,000 18,000

II 6,000 20,000 8,000

(a) If unit sales prices of x, y and z are 2.50,
1.50 and 1.00, respectively, find the total revenue in each market with the help of matrix algebra.

(b) If the unit costs of the above three commodities are 2.00, 1.00 and 50 paise respectively. Find the gross profit.

Ans.

a    Matrix corresponding to sale price of each product=2.501.501.00The total revenue collected by each market is given by10,0002,00018,0006,00020,0008,0002.501.501.00=25,000+3,000+18,00015,000+30,000+8,000  =46,00053,000Total cost prices of the commodities each market I and II are ‘ 46,000, ‘ 53,000.Total  revnue by both markets = 46,000+53,000=99,000bThe cost price of commodities x, y and z are respectively ‘ 2.00, 1.00 and ‘ 0.50 cost price of each market is given below10,0002,00018,0006,00020,0008,0002.001.000.50=20,000+2,000+9,00012,000+20,000+4,000=3100036000Total cost prices of the commodities each market I and II are 31,000, ZZ36,000.Total  revnue by both markets = 31,000+36,000= 67,000Gross profit=99,000-67,000=32,000

Q.73

Find the matrix X so that  X 1 2 34 5 6=789  2  4  6

Ans.

  X 1 2 34 5 6=789  2  4  6Let order of matrix X be m×n.Since, the order of 1 2 34 5 6 is2×3and order of matrix Xis m×n.So,n=2.Now,theorder of 7 89  2    4    6 is  2×3 and order of matrix X is m×n.So, m = 2.Thus, order of matrix X is 2×2.Let X= a bc d, then                a bc d 1 2 34 5 6789  2   4    6 =a+4b 2a+5b 3a+6bc+4d 2c+5d 3c+6d =789  2   4   6    a+4b=7    ...(i)  2a+5b=8    ...(ii)          c+4d=2    ...(iii)    2c+5d=4      .  ..(iv)3c+6d=6        ...(v)Solving equation (i) and equation(ii), we geta=1,b=2value of a and b satis fies 3a+6b=9.Solving equation (iii) and equation(iv), we getc=2,d=0value of c and d satisfies 3c+6d=6.Hence,      X=122   0

Q.74

If A and B are square matric e so f thesame that AB=BA,then prove by induction thatABn=BAn. Further, provethat (AB)n=AnBnforall nN.

Ans.

(i) Let P(n): ABn=BnAFor n=1                            P(1):  AB=BA[Given]        P(n) is true for n=1.Let P(n) be true for n=k.                          P(k):  ABk=BkA  ...(i)For n=k+1            ABk+1=Bk+1A    ...(ii)        L.H.S.=ABk+1            =(ABk)B            =B(ABk)            =B(BkA)[From equation (i)]            =Bk+1A=R.H.S.So, P(n) is true for n=k+1.Therefore,  by the Principle of mathematical induction,P(n) is true for all nN.(ii)  LetP(n):  (AB)n=AnBnFor  n=1      L.H.S.=AB    R.H.S.=ABP(n) is true for n=1.Let P(n) be true for n=k.P(k):  (AB)k=AkBk...(i)Multiply both sides by AB, we get(AB)k(AB)=AkBk(AB)          L.H.S.=(AB)k+1          R.H.S.=AkBk(AB)              =AkBk(BA)  [AB=BA]              =Ak(Bk.B)A              =Ak(Bk+1A)              =Ak(ABk+1)[∵ABn=BnA]              =Ak+1Bk+1Therefore,P(n) is true for n=k+1.By the principle of mathematical inducation,P(n)is true for nN.

Q.75

If A=[α    βγα] is such that A2=I, then(A)1+α2+βγ=0(B)1α2+βγ=0(C)1α2βγ=0(D)1+α2βγ=0

Ans.

    A2=[α    βγα][α    βγα]        =[α2+βγ    αβαβαγαγ    βγ+α2]∵       A2=[1 00 1][1 00 1]=[α2+βγ   00     βγ+α2]α2+βγ=11α2βγ=0OptionC is the correct answer.

Q.76

If the matrix A is both symmetric ands kew symmetric, then(A) A is a diagonal matrix (B) A is a zero matrix(C) A is a square matrix(D) None of these

Ans.

A is symmetric matrix then aij=aji

A is skew symmetric matrix then

aij =– aji

This implies that aij = aji = – aji

2aji = 0

Or aji = 0

Therefore, A is a zero matrix.

Option (B) is correct.

Q.77

If A is square matrix such that A2=A, then (I+A)3=7Ais equal to(A)A (B)IA(C)I (D)3A

Ans.

Given,  A2=A,(I+A)37A=I3+3I2A+3IA2+A37A[(a+b)3=a3+3a2b+3ab2+b3]      =I+3A+3A+A2A7A[∵I3=I]      =I+6A+A.A7A[∵A2=A]      =I+A2A      =I+AA                        [∵A2=A]      =IHence, option (C) is correct option.

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