NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants is available online for the students to refer to and study on Extramarks. The chapter introduces determinants, their properties, and their applications. In addition, the chapter also includes various problems and a step-by-step solution to each problem that aids students in understanding the concept better. Furthermore, the solutions are built on the CBSE syllabus of the latest academic year and with reference to NCERT.

A determinant is a number that can only be calculated with the help of square matrices. If the students want to score more in linear equations, then the determinant is essential to learn. It helps to check whether the system of equations has a unique solution or not. Besides, students will get a glimpse of the previous chapter, matrices. This chapter also carries essential elements and helps to improve their calculations. 

In the NCERT Solutions Class 12 Mathematics Chapter 4, students can understand the complex theories and formulas in an easy to understand language. These sub-topics include basic principles of determinant, cofactors, adjoint, and the inverse of a matrix. Besides, this solution has proper explanations and clear concepts to help students understand better.

Students looking for step-by-step solutions to the problems mentioned in ch 4 mathematics class 12 may register in Extramarks. Extramarks not only offers other NCERT Solutions for Class 12 but also from Class 1 to class 11. In addition, students may log in to the website for updated study material, revision notes, CBSE sample papers, important questions, and the latest notifications and news regarding CBSE exams.

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 4

Extramarks NCERT Solutions for Class 12 Mathematics Chapter 4 covers all essential topics under determinants. Students will start by learning how to find the determinant of a matrix of order one and the properties of determinants. More precise calculations and solutions make them understand multiplications in positives instead of negatives. There are four exercises and three subtopics under NCERT Solutions for Class 12 Mathematics Chapter 4. In addition, students can refer to the study material offered by Extramarks as it offers better solutions.

Topics covered in Chapter 4 Determinant include

Exercise  Topics
4.1 Introduction
4.2 Determinant
4.3 Properties of Determinants
4.4 Area of Triangle
4.5 Minors and Cofactors
4.6 The Adjoint and Inverse of a Matrix
4.7 Applications of Determinants and Matrices

4.1 Introduction

NCERT Solutions Class 12 Mathematics Chapter 4 introduces the students to Determinants and their basic concepts and theorems, wherein they will be introduced to the properties of determinants such as minors, cofactors, and applications of determinants. Further, the students will get a hold of topics like the inconsistency of the system of linear equations and solutions of linear equations. 

4.2 Determinant

This is one of the crucial sections. Students will understand the grassroots concepts; as learnt in the previous Chapter 3 Matrices, any matrix inside a mode is read as a determinant of a matrix. In addition, students can learn how to find the determinant of a matrix of order one, making it easy to comprehend. Students may refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 as it provides straightforward and easy-to-understand calculations.

 4.3 Properties of Determinants 

Under this topic, students will observe that prominent examples revolve around the properties of determinants. This section elaborates on how the statement properties are true for determinants of any order. The students will get to explore six different properties of determinants. Students may refer to the exercise problems to understand the properties better. Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 offers step-by-step solutions for many exercises under this topic. 

4.4 Area of Triangle

This section guides students on calculating the area of triangles and the application of vertices. It also elaborates on particular topics such as positive quantity and absolute values of determinants. It also helps the students understand the positive and negative values of the determinant.

 4.5 Minors and Cofactors

In the NCERT Solutions Class 12 Mathematics Chapter 4, minors, cofactors and their main concepts are covered. In addition, this section will teach the students about five more expansion ways that help calculate the area of a triangle. 

4.6 The Adjoint and Inverse of a Matrix

Further in Chapter 4 Class 12, students will learn about the inverse and existence. It contains detailed explanations in terms of finding the adjoint of the matrix. There is a total of five theorems that helps to verify the concepts. 

4.7 Applications of Determinants and Matrices

NCERT Solutions Class 12 Mathematics Chapter 4 covers the essential concepts in applications of determinants and matrices. A student will learn how to check the consistency of the system of linear equations. Under this topic, students will learn and study the solution system of linear equations. 

For detailed study material, students may refer to Extramarks, where step-by-step solutions on essential concepts, including consistent and inconsistent systems, are provided. With the help of Extramarks NCERT Solutions, the student can score more in determinants and matrices. 

NCERT Solutions Class 12 Mathematics Chapter 4 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 4 Determinant is available for students to refer to on the Extramarks website. It contains step-by-step solutions with detailed explanations. The students can solve complex problems with the easiest and most time-saving methods. Referring to our study material, students can make their basics strong and easily solve advanced theorems.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 4 Determinant.

  • Chapter 4: Exercise 4.1 – 8 Questions (3 Short Questions and 5 Long Questions)
  • Chapter 4: Exercise 4.2 – 16 Questions (4 Short Questions 10 Long Questions)
  • Chapter 4:Exercise 4.3 – 5 Questions (2 Short Questions and 3 Long Questions)
  • Chapter 4: Exercise 4.4 – 5 Questions (2 Short Questions and 3 Long Questions)
  • Chapter 4: Exercise 4.4 – 5 Questions (2 Short Questions and 3 Long Questions)
  • Chapter 4: Exercise 4.4 – 5 Questions (2 Short Questions and 3 Long Questions)

In addition to NCERT solutions class 12 mathematics chapter 4, students can also access other chapters of class 4. Furthermore, Extramarks offers NCERT solutions for other classes, which can be accessed by clicking on the respective link. 

  • 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

The NCERT Exemplar Class 12 Mathematics covers the essential subject knowledge. Once the students clear all the necessary topics under determinants, they can start preparing with an exemplar. There is a lot of importance to the topic as it appears in the competitive exams. Students can start preparing for their first-term CBSE exam with the help of this exemplar. It provides a complete explanatory solution for students with good knowledge of the subject. 

At Extramarks, our subject matter experts offer their best solution with a proper step-by-step explanation. The exemplar consists of all the six exercises, i.e., introduction, determinant, properties of determinants, area of a triangle, minors and cofactors. In addition, the NCERT Exemplar contains various questions, multiple-choice type questions, and long answer type questions. 

Key Features of NCERT Solutions Class 12 Mathematics Chapter 4

Class 12 Mathematics Chapter 4 has some essential elements that are a part of competitive exams. The students will get a detailed explanation of the terms, definitions, rules, and properties of determinants. A basic understanding of these complex theories is essential for better understanding. Hence, NCERT Solutions Class 12 Mathematics Chapter 4 offers a precise solution for each example mentioned in the exercise.  

Some of the key features of NCERT Solutions Class 12 Mathematics Chapter 4 include: 

  • The solutions cover the entire syllabus of CBSE Class 12 Mathematics Chapter 4 and help to provide a detailed explanation of all topics. 
  • The format of the NCERT Solutions is self-explanatory and clears all doubts a student may have while learning complex concepts. 
  • To score better in the examination and competitive entrance exams, students must grasp all the topics. Hence NCERT Solutions Class 12 Mathematics Chapter 4 consists of all important aspects of Determinants needed for the competitive entrance exam. 
  • The solution helps increase knowledge in main topics. 

Q.1

Let A=[1sinθ1sinθ1sinθ1sinθ1], where0θ2π.Then(A) Det (A)=0(B) Det (A) (2,)(C) Det (A)(2,4) (D) Det (A)[2,4]

Ans

Let A=[1sinθ1sinθ1sinθ1sinθ1], where 0θ2π. Then      |A|=|1sinθ1sinθ1sinθ1sinθ1|=1(1+sin2θ)sinθ(sinθ+sinθ)+1(sin2θ+1)=2(1+sin2θ)Now,  0θ2π0sinθ10sin2θ111+sin2θ222(1+sin2θ)4Det(A)[2,4]Thus,​​ correct option is D.

Q.2

If x, y, z are non zero real numbers, then the inverse of matrixA=[x000y000z]is(A) [x1000y1000z1] (B) xyz[x1000y1000z1](C) 1xyz[x000y000z] (D)1xyz[100010001]

Ans

Given matrixisA=[x000y000z]|A|=|x000y000z|    =x(yz0)0+0    =xyz0Thus, inverse of matrix A exists.Now cofactors areA11=yz,  A12=0,  A13=0A21=0,    A22=xz,A23=0A31=0,    A32=0,  A33=xyAdj  A=[yz000xz000xy]=[yz000xz000xy]A1=1|A|Adj  A    =1xyz[yz000xz000xy]    =[yzxyz000xzxyz000xyxyz]    =[1x0001y0001z]=[x1000y1000z1]Thus, the correct option is A.

Q.3

If a, b, c are in A.P., then the determinant|x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|is(A) 0 (B)1 (C) x (D) 2x

Ans

|x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|If a, b, c are in A.P., so 2b=a+cThen,|x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|=|x+2x+3x+2ax+3x+4x+(a+c)x+4x+5x+2c|Applying R1R1R2andR3R3R1,we get  =|11acx+3x+4x+(a+c)11ca|Applying R1R1+R3,we get  =|000x+3x+4x+(a+c)11ca|  =0    [Since,all elements of R1 are zero.]Hence,|x+2x+3x+2ax+3x+4x+2bx+4x+5x+2c|=0Therefore, correct option is A.

Q.4

Solve the system of the following equations2x+3y+10z=44x6y+5z=16x+9y20z=1

Ans

2x+3y+10z=4  4x6y+5z=16x+9y20z=1Let p=1x q=1y and r=1z, then2p+3q+10r=4  4p6q+5r=16p+9q20r=1The system can be written as AX=B, whereA=[23104656920],X=[pqr] and B=[411]|A|=|23104656920|      =2(12045)3(8030)+10(36+36)      =150+330+720      =1200Thus, A is nonsingular. Therefore, its inverse exists.Now, A11=75,A12=110,A13=72,A21=150,A22=100,  A23=0,A31=75,A32=30,  A33=24AdjA=[75110721501000753024]            =[75150751101003072024]    A1=1|A|AdjA            =11200[75150751101003072024]Now,  X=A1B[pqr]  =11200[75150751101003072024][412][pqr]  =11200[300+150+150440100+60288+048][pqr]  =11200[600400240][pqr]  =[121315]p=12,q=13andr=15x=1p=2,y=1q=3  and  z=1r=5Hence, the required solution isx= 2, y= 3 and z= 5.

Q.5 Using properties of determinants prove that:

sinαcosαcos(α+β)sinβcosβcos(β+δ)sinγcosγcos(γ+δ)=0

Ans

L.H.S.=|sinαcosαcos(α+δ)sinβcosβcos(β+δ)sinγcosγcos(γ+δ)|Applying C1sinδC1 and C2cosδC2,we get    =1sinδcosδ|sinαsinδcosαcosδcosαcosδsinαsinδsinβsinδcosβcosδcosβcosδsinβsinδsinγsinδcosγcosδcosγcosδsinγsinδ|C1C1+C3,weget    =1sinδcosδ|cosαcosδcosαcosδcosαcosδsinαsinδcosβcosδcosβcosδcosβcosδsinβsinδcosγcosδcosγcosδcosγcosδsinγsinδ|    =1sinδcosδ×0[Δ=0, if any two coloumns are identical.]    =0=R.H.S.Thus, the given result is proved.

Q.6 Using properties of determinants prove that:

11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q=1

Ans

L.H.S.=|11+p1+p+q23+2p4+3p+2q36+3p10+6p+3q|Applying R2R22R1 and R3R33R1,we get    =|11+p1+p+q012+p037+3p|Expanding along C1, we get    =1|12+p37+3p|0+0    =7+3p63p    =1=R.H.S.Thus, the given result is proved.

Q.7 Using properties of determinants prove that:

3aa+ba+cb+a3bb+cc+ac+b3c =3(a+b+c)(ab+bc+ca)

Ans

L.H.S.=|3aa+ba+cb+a3bb+cc+ac+b3c|Applying C1C1+C2+C3, we get    =|a+b+ca+ba+ca+b+c3bb+ca+b+cc+b3c|    =(a+b+c)|1a+ba+c13bb+c1c+b3c|ApplyR2R2R1 and R3R3R1, we get    =(a+b+c)|1a+ba+c0a+2bab0aca+2c|Expanding along C1, we get    =(a+b+c){(a+2b)(a+2c)(ab)(ac)}    =(a+b+c){a2+2ac+2ab+4bc(a2acba+bc)}    =(a+b+c){a2+2ac+2ab+4bca2+ac+babc}    =3(a+b+c)(ab+bc+ca)=R.H.S.Thus, the given result is proved.

Q.8 Using properties of determinants prove that:

xx21+px3yy21+py3zz21+pz3 =(1+pxyz)(xy)(yz)(zx)

Ans

L.H.S.=|xx21+px3yy21+py3zz21+pz3|Applying R3R3R1andR2R2R1,weget=|xx21+px3yxy2x2(1+py3)(1+px3)zxz2x2(1+pz3)(1+px3)|=(yx)(zx)|xx21+px31y+x(py3px3)1z+x(pz3px3)|=(yx)(zx)|xx21+px31y+xp(y2+yx+x2)1z+xp(z2+zx+x2)|ApplyingR3R3R2, we get=(yx)(zx)|xx21+px31y+xp(y2+yx+x2)0zyp(z2y2+zxyx)|=(yx)(zx)(zy)|xx21+px31y+xp(y2+yx+x2)01p(z+y+x)|Expanding along R3, we get=(yx)(zx)(zy){01|x1+px31p(y2+yx+x2)|+p(z+y+x)|xx21y+x|}=(yx)(zx)(zy)[1{xp(y2+yx+x2)(1+px3)}+p(z+y+x){x(y+x)x2}]=(yx)(zx)(zy){pxy2px2ypx3+1+px3+px2y+pxy2+pxyz}=(yx)(zx)(zy)(1+pxyz)=(1+pxyz)(xy)(yz)(zx)=R.H.S.Therefore, the given result is proved.

Q.9 Using properties of determinants prove that:

αα2β+γββ2γ+αγγ2α+β =(βγ)(γα)(αβ)(α+β+γ)

Ans

L.H.S.=|αα2β+γββ2γ+αγγ2α+β|Applying R2R2R1 and R3R3R1,we get    =|αα2β+γβαβ2α2αβγαγ2α2αγ|    =(βα)(γα)|αα2β+γ1β+α11γ+α1|Applying R3R3R2,weget    =(βα)(γα)|αα2β+γ1β+α10γβ0|Expanding along R3, we get    =(βα)(γα){(γβ)(αβγ)}    =(βγ)(γα)(αβ)(α+β+γ)=R.H.S.

Q.10

Evaluate 1xy1x+yy1xx+y .

Ans

Δ=1xy1x+yy1xx+yApply R2R2R1and R3R3R1    =1xy0y000xExpanding​​ along C1, we get    =1(xy0)0+0    =xy

Q.11

Evaluate|xyx+yyx+yxx+yxy|.

Ans

Δ=|xyx+yyx+yxx+yxy|Applying R1R1+R2+R3,we get    =|2x+2y2x+2y2x+2yyx+yxx+yxy|    =2(x +y)|111yx+yxx+yxy|Applying C2C2C1 and C3C3C1, we get    =2(x +y)|100yxxyx+yyx|Expanding along R1, we get    =2(x +y)|xxyyx|    =2(x +y)(x2+xyy2)    =2(x +y)(x2xy+y2)    =2(x3+y3)

Q.12

Let A=[121231115]. Verify that  (i) [adjA]1=adj (A1) (ii) (A1)1=A

Ans

A=[121231115]|A|=|121231115|      =1(151)+2(101)+1(23)      =14225=130Now,cofactors areA11=14,A12=11,A13=5A21=11,A22=4,A23=3A31=5,A32=3,A33=1AdjA=[141151143531]=[141151143531]So,  A1=1|A|AdjA      =113[141151143531]      =113[141151143531](i)|AdjA|=|141151143531|        =14(49)11(1115)5(33+20)        =182+286+65        =169Cofactors of Adj A areA11=13,A12=26,A13=13A21=26,A22=39,A23=13A31=13,A32=13,A33=65Adj(AdjA)=[132613263913131365]  =[132613263913131365](AdjA)1=1|AdjA|Adj(AdjA)  =1169[132613263913131365]  =1169×13[121231115]=113[121231115]  A1=113[141151143531]=[141311135131113413313513313113] Adj A 1 = [ 4 169 9 169 ( 11 169 15 169 ) 33 169 20 169 ( 11 169 15 169 ) 14 169 25 169 ( 42 169 + 55 169 ) 33 169 + 20 169 ( 42 169 + 55 169 ) 56 169 121 169 ] = 1 169 [ 13 26 13 26 39 13 13 13 65 ] = 13 169 [ 1 2 1 2 3 1 1 1 5 ] Adj A 1 = ( AdjA ) 1 Hence, it is proved. ( ii )Since, A 1 = 1 13 [ 14 11 5 11 4 3 5 3 1 ]=[ 14 13 11 13 5 13 11 13 4 13 3 13 5 13 3 13 1 13 ] andAdjA= 1 13 [ 1 2 1 2 3 1 1 1 5 ] | A 1 |=| 14 13 11 13 5 13 11 13 4 13 3 13 5 13 3 13 1 13 | = ( 1 13 ) 3 | 14 11 5 11 4 3 5 3 1 | = ( 1 13 ) 3 { 14( 49 )+11( 1115 )+5( 33+20 ) } = ( 1 13 ) 3 ( 18228665 ) = ( 1 13 ) 3 ( 169 )= 1 13 ( A 1 ) 1 = Adj A 1 | A 1 | = 1 ( 1 13 ) 1 13 [ 1 2 1 2 3 1 1 1 5 ] =[ 1 2 1 2 3 1 1 1 5 ] ( A 1 ) 1 =A Hence, it is proved. MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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Q.13

If A1=|3111565522| and B=[122130021], find (AB)1.

Ans

Since,  (AB)1=B1A1    B=[122130021]  |B|=|122130021|=1(30)2(10)2(20)=3+24=10Thus, B1 exists.Now, cofactors areA11=3,A12=1,A13=2A21=2,A22=1,A23=2A31=6,A32=2,A33=5AdjB=[312212625]  =[326112225]B1=1|B|AdjB  =11[326112225]=[326112225]Now,  (AB)1=B1A1      =[326112225][3111565522]  =[930+303+1212310+12315+101+6415+4630+25212+10210+10]  =[935210102]Thus, required solution is(AB)1=[935210102]

Q.14

Prove that|a2bcac+c2a2+abb2acabb2+bcc2|=4a2b2c2

Ans

L.H.S.=|a2bcac+c2a2+abb2acabb2+bcc2|Taking out common a, b and c from C1, C2 and C3 respectively.=abc|aca+ ca+bbabb+cc|Applying R2R2R1,R3R3R1we get=abc|aca+ cbbccbaba|Applying R2R2+R1, we get=abc|aca+ ca+bbababa|Applying R3R3+R2, we get=abc|aca+ ca+bba2b2b0|=2ab2c|aca+ ca+bba110|Applying C2C2C1, we get=2ab2c|acaa+ ca+baa100|Expanding along R3, we get=2ab2c{1(aca2+a2+ac)0+0}=2ab2c(2ac)=4a2b2c2=R.H.S.

Q.15

Solve thee quation|x+axxxx+axxxx+a|=0, a0.

Ans

Given:|x+axxxx+axxxx+a|=0,a0.              |x+axxxx+axxxx+a|=0Apply R1R1+R2+R3,we have    |3x+a3x+a3x+axx+axxxx+a|=0(3x+a)|111xx+axxxx+a|=0Apply C2C2C1 and C3C3C1, we get(3x+a)|100xa0x0a|=0Expanding along R1, we get(3x+a){1(a20)0+0}=0      a2(3x+a)=0          (3x+a)=0  x=a3Therefore, x =a3.

Q.16

If a,b and c are real numbers, andΔ=|b+cc+aa+bc+aa+bb+ca+bb+cc+a|=0,Show that either a+b+c=0 or a=b=c.

Ans

Δ=|b+cc+aa+bc+aa+bb+ca+bb+cc+a|Applying R1R1+R2+R3, we get    =|2(a+b+c)2(a+b+c)2(a+b+c)c+aa+bb+ca+bb+cc+a|    =2(a+b+c)|111c+aa+bb+ca+bb+cc+a|ApplyC2C2C1,C3C3C1    =2(a+b+c)|100c+abcbaa+bcacb|Expanding along R1, we get    =2(a+b+c).[1{(bc)(cb)(ba)(ca)}0+0]    =2(a+b+c)(bcb2c2+cbbc+ba+aca2)    =2(a+b+c)(a2b2c2+cb+ba+ac)ButΔ=0So,2(a+b+c)(a2b2c2+cb+ba+ac)=0Either  a+b+c=0  or    a2b2c2+cb+ba+ac=0Ifa2b2c2+cb+ba+ac=0Then,    2a22b22c2+2cb+2ba+2ac=02a2+2b2+2c22cb2ba2ac=0(ab)2+(bc)2+(ca)2=0(ab)2=(bc)2=(ca)2=0[(ab)2,(bc)2,(ca)2 are nonnegative.]a=b=cThus, either a+b+c=0 or a=b=c.

Q.17

Evaluate: cosαcosβcosαsinβsinαsinβcosβ 0sinαcosβsinαsinβcosα.

Ans

Δ=|cosαcosβcosαsinβsinαsinβcosβ0sinαcosβsinαsinβcosα|Expanding along C3,weget  =sinα|sinβ cosβsinαcosβ sinαsinβ|0|cosαcosβcosαsinβsinαcosβsinαsinβ|+cosα|cosαcosβcosαsinβsinβcosβ|=sinα(sinαsin2βsinαcos2β)0+cosα(cosαcos2β+cosαsin2β)=sin2αsin2β+sin2αcos2β+cos2αcos2β+cos2αsin2β=sin2α(sin2β+cos2β)+cos2α(cos2β+sin2β)[sin2θ+cos2θ=1]=sin2α(1)+cos2α(1)=sin2α+cos2α=1=R.H.S/Hence,it is proved.

Q.18

Without expanding the determinant, prove that |aa2bcbb2cacc2ab|=|1a2a31b2b31c2c3|.

Ans

L.H.S.=|aa2bcbb2cacc2ab|Apply R1aR1,R2bR2and  R3cR3,we get=1abc|a2a3abcb2b3abcc2c3abc|=1abc.abc|a2a31b2b31c2c31|[Taking out abc from C3.]=(1)|a21a3b21b3c21c3|[C2C3]=(1)2|1a2a31b2b31c2c3|[C1C2]=|1a2a31b2b31c2c3|=R.H.S.Hence, it is proved.

Q.19

Prove that the determinant |xsinθcosθsinθx1cosθ1x| isindependent of θ.

Ans

|xsinθcosθsinθx1cosθ1x|Expanding along R1, we get=x(x21)sinθ(xsinθcosθ)+cosθ(sinθ+xcosθ)=x3x+xsin2θ+sinθcosθcosθsinθ+xcos2θ=x3x+x(sin2θ+cos2θ)=x3x+x(1)[sin2θ+cos2θ=1]=x3, which is independent of θ.Therefore, determinant is independent of θ.

Q.20 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Ans

Let cost of 1 kg onions be ₹ x, 1 kg wheat be ₹ y and 1 kg rice be ₹z. Then, the given situation can be representedby a system of equations as:4x + 3y + 2z = 602x + 4y + 6z = 906x + 2y + 3z = 70This system of equations can be written in the form of AX = B as given below:A=432246623,X=xyzand B=609070A=432246623=412123636+2424=0+9040=500Now,A11=0,A12=30,A13=20A21=5,A22=0,A23=10A31=10,A32=20,A33=10AdjA=030205010102010=051030020201010A1=1AAdjA=150051030020201010Now,X=A1B=150051030020201010609070=1500450+7001800+014001200+900+700=150250400400xyz=588x=5,y=8andz=8.Thus, the cost of 1 kg onions is ₹5, 1 kg wheat is ₹ 8 and that of 1 kg rice is ₹8.

Q.21

If  A=235324112,find A1.UsingA1solve the system ofequations:2x3y+5z=113x+2y4z=5x+y2z=3

Ans

A=[235324112]|A|=|235324112|      =2(4+4)+3(6+4)+5(32)      =06+5      =10Now,A11=0,  A12=2,A13=1A21=1,  A22=9,  A23=5A31=2,    A32=23,  A33=13AdjA=[0211  95223  13]  =[01229231513]  A1=1|A|AdjA          =11[01229231513]  A1=[01229231513]...(i)Now, the given system of equations can be written in the formof AX= B, whereA=[235324112],X=[xyz] and B=[1153]The solution of the system of equations is given byX=A1B=[01229231513][1153][Using equation (i)]=[05+62245+691125+39][xyz]=[123]Hence,x=1,y=2 and z=3.

Q.22 Solve system of linear equations, using matrix method,

x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12

Ans

The given system of equations is:        xy+2z=7    3x+4y5z=5      2xy+3z=12The given system of equations can be written in the form ofAX=B, whereA=[112345213],  X=[xyz] and B=[  7512]Now,|A|=|11  234521  3|      =1(125)+1(9+10)+2(38)      =7+1922=40A is a nonsigular  matrix.Therefore, A1 exists.Now,A11=7,  A12=19,A13=11A21=1,  A22=1,  A23=1A31=3,    A32=11,  A33=7AdjA=[719111  11311  7]  =[713191111117]  A1=1|A|AdjA          =14[713191111117]X=A1B=14[713191111117][  75  12]=14[49536133+5+13277+5+84][xyz]=14[8412]=[213]Hence,x=2,y=1 and z=3.

Q.23 Solve system of linear equations, using matrix method,

2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3

Ans

The given system of equations is:    2x+3y+3z=5        x2y+z=4      3xy2z=3The given system of equations can be written in the form ofAX=B, whereA=[233121312],  X=[xyz] and B=[  54  3]Now,|A|=|233121312|=2(4+1)3(23)+3(1+6)      =10+15+15=400A is a nonsigular  matrix.Therefore, A1 exists.Now,A11=5,  A12=5,A13=5A21=3,  A22=13,  A23=11A31=9,    A32=1,  A33=7AdjA=[5553  1311917]  =[53951315117]  A1=1|A|AdjA          =140[53951315117]X=A1B=140[53951315117][  54  3]=140[2512+2725+52+3254421][xyz]=140[  40  8040]              =[121]Hence,x=1,y=2 and z=1.

Q.24 Solve system of linear equations, using matrix method,

x – y + z = 4
2x + y – 3z = 0
x + y + z = 2

Ans

The given system of equations is:         ​  xy+z=4      2x+y3z=0          x+y+z=2The given system of equations can be written in the form ofAX=B, whereA=[11  1213111],  X=[xyz] and B=[402]Now,|A|=|11  1213111|      =1(1+3)+1(2+3)+1(21)      =4+5+1=100A is a nonsigular  matrix.Therefore, A1 exists.Now,A11=4,A12=5,A13=1A21=2,  A22=0,A23=2A31=2,    A32=5,A33=3AdjA=[4512  02253]  =[422505123]  A1=1|A|[422505123]          =110[422505123]X=A1B=110[422505123][402]=110[16+0+420+0+104+0+6][xyz]=110[  2010  10]              =[211]Hence,x=2,y=1 and z=1.

Q.25 Solve system of linear equations, using matrix method,

2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9

Ans

The given system of equations is: 2x+y+z=1 x2yz= 3 2 3y5z=9 The given system of equations can be written in the form of AX=B, where A=[ 2 1 1 1 2 1 0 3 5 ],X=[ x y z ] and B=[ 1 3 2 9 ] Now, | A |=| 2 1 1 1 2 1 0 3 5 | =2( 10+3 )1( 50 )+1( 30 ) =26+5+3=340 A is a non-sigularmatrix. Therefore, A -1 exists. Now, A 11 =13, A 12 =5, A 13 =3 A 21 =8, A 22 =10, A 23 =6 A 31 =1, A 32 =3, A 33 =5 AdjA= [ 13 5 3 8 10 6 1 3 5 ] =[ 13 8 1 5 10 3 3 6 5 ] A 1 = 1 | A | [ 13 8 1 5 10 3 3 6 5 ] = 1 34 [ 13 8 1 5 10 3 3 6 5 ] X= A 1 B = 1 34 [ 13 8 1 5 10 3 3 6 5 ][ 1 3 2 9 ] = 1 34 [ 13+12+9 515+27 3945 ] [ x y z ]= 1 34 [ 34 17 51 ]=[ 1 1 2 3 2 ] Hence,x=1,y= 1 2 and z= 3 2 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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Q.26 Solve system of linear equations, using matrix method,

5x + 2y = 3
3x + 2y = 5

Ans

The given system of equations is:  5x+2y=3  3x+2y=5The given system of equations can be written in the form ofAX=B, whereA=[5232],  X=[xy] and B=[35]Now,|A|=|5232|      =106      =40A is a nonsigular  matrix.Therefore, A1 exists.Now, Adj A =[  232  5]    =[2235]    A1=1|A|(Adj A)  =14[2235]      X=A1B  =14[2235][35]  =14[6109+25]    [xy]=14[416][xy]=[14]x=1  andy=4.

Q.27 Solve system of linear equations, using matrix method,

4x – 3y = 3
3x – 5y = 7

Ans

The given system of equations is:4x3y=33x5y=7The given system of equations can be written in the form ofAX=B, whereA=[4335],X=[xy] and B=[37]Now,|A|=|4335|=20+9=110A is a non-sigularmatrix.Therefore, A-1 exists.Now, Adj A =[5334]=[5334]A-1=1|A|(Adj A)=111[5334]=111[5334]X=A1B=111[5334][37]=111[1521928][xy]=111[619][xy]=[6111911]x=611andy=1911.

Q.29 Solve system of linear equations, using matrix method,

2x – y = –2
3x + 4y = 3

Ans

The given system of equations is: 2xy=2 3x+4y=3 The given system of equations can be written in the form of AX=B, where A=[ 2 1 3 4 ],X=[ x y ] and B=[ 2 3 ] Now, | A |=| 2 1 3 4 | =8+3 =110 A is a non-sigularmatrix. Therefore, A -1 exists. Now, Adj A = [ 4 3 1 2 ] =[ 4 1 3 2 ] A -1 = 1 | A | ( Adj A ) = 1 11 [ 4 1 3 2 ] X= A 1 B = 1 11 [ 4 1 3 2 ][ 2 3 ] = 1 11 [ 8+3 6+6 ] [ x y ]= 1 11 [ 5 12 ] [ x y ]=[ 5 11 12 11 ] x= 5 11 andy= 12 11 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D3A8@

Q.30 Solve system of linear equations, using matrix method,

5x + 2y = 4
7x + 3y = 5

Ans

The given system of equations is:  5x+2y=4  7x+3y=5The given system of equations can be written in the form ofAX=B, whereA=[5273],  X=[xy] and B=[45]Now,|A|=|5273|      =1514      =10A is a nonsigular  matrix.Therefore, A1 exists.Now, Adj A =[3725]=[3275]    A1=1|A|(Adj A)  =11[3275]  =[3275]      X=A1B  =[3275][45]  =[121028+25]    [xy]=[23] x=2  andy=3.

Q.31 Examine the consistency of the system of equations.

5x – y + 4z = 5
2x + 3y + 5z = 2
5x –2y + 6z = –1

Ans

The given system of equations is:      5xy+4z=5    2x+3y+5z=2  5x2y+6z=1The given system of equations can be written in the form ofAX=B, whereA=[514235526],  X=[xyz] and B=[521]Now,|A|=|514235526|      =5(18+10)+1(1225)+4(415)      =1401376=510 A is a nonsigular  matrix.Therefore, A1 exists.Hence, the given system of equations is consistent.

Q.32 Examine the consistency of the system of equations.

3x – y – 2z = 2
2y – z = – 1
3x –5y = 3

Ans

The given system of equations is:  3xy2z=2  2yz=13x5y=3The given system of equations can be written in the form ofAX=B, whereA=[312021350],  X=[xyz] and B=[213]Now,|A|=|312021350|      =3(05)+1(0+3)2(06)      =15+3+12=0A is a sigular matrix.A11=(1)1+1|2150|=5A12=(1)1+2|0130|=3A13=(1)1+3|0235|=6A21=(1)2+1|1250|=10A22=(1)2+2|3230|=6A23=(1)2+3|3135|=12A31=(1)3+1|12  21|=5A32=(1)3+2|3201|=3A33=(1)3+3|310  2|=6(Adj A)=[53610612536]=[51053636126](Adj A)B=[51053636126][213]=[1010+1566+91212+18]=[536]OThus, the solution of the given system of equations doesnot exist. Hence, the system of equations is inconsistent.

Q.33 Examine the consistency of the system of equations.

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Ans

The given system of equations is:          x+y+z=1  2x+3y+2z=2ax+ay+2az=4The given system of equations can be written in the form ofAX=B, whereA=[111232aa2a],  X=[xyz] and B=[124]Now,|A|=|111232aa2a|      =1(6a2a)1(4a2a)+1(2a3a)      =4a2aa=a0 A is a nonsigular  matrix.Therefore, A1 exists.Hence, the given system of equations is consistent.

Q.34 Examine the consistency of the system of equations.

x + 3y = 5
2x + 6y = 8

Ans

The given system of equations is:  x+3y=52x+6y=8The given system of equations can be written in the form ofAX=B, whereA=[1326],  X=[xy] and B=[58]Now,|A|=|1326|=66      =0A is a sigular matrix.(Adj A)=[6231]=[6321](Adj A)B=[6321][58]=[302410+8]=[62]OThus, the solution of the given system of equations doesnot exist. Hence, the system of equations is inconsistent.

Q.35 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

The given system of equations is:  2xy=5    x+y=4The given system of equations can be written in the form ofAX=B, whereA=[2111],  X=[xy] and B=[54]Now,|A|=|2111|      =2+1      =30A is a nonsigular  matrix.Therefore, A1 exists.Hence, the given system of equations is consistent.

Q.36 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3

Ans

The given system of equations is:  x+2y=22x+3y=3The given system of equations can be written in the form ofAX=B, whereA=[1223],  X=[xy] and B=[23]Now,|A|=|1223|      =34      =10A is a nonsigularmatrix.Therefore, A1 exists.

Hence, the given system of equations is consistent.

Q.37

If A is aninvertible matrix of order 2, then detA1 isequalto    A DetAB1DetA C 1 D 0

Ans

Since, A is an invertible matrix,then A1 exists and A1=1|A|adjA.Let a matrix of order 2 be as A=[abcd],then |A|=adbc and Adj A=[dbca]Now,    A1=1|A|Adj A    =1|A|[dbca]    =[d|A|b|A|c|A|a|A|]          |A1|=|d|A|b|A|c|A|a|A||    =1|A|2|dbca|    =1|A|2(adbc)    =1|A|2|A|    =1|A|det(A1)=1det(A)Thus, the correct option is (B).

Q.38 Let A be a nonsingular square matrix of order 3 x 3. Then |adj A| is equal to

(A) |A|
(B) |A|2
(C) |A|3
(D) 3|A|

Ans

Since, (AdjA) A=|A|I=[|A|000|A|000|A|]            |(AdjA)A|=||A|000|A|000|A||      |(AdjA)||A|=|A|3|100010001|        |(AdjA)||A|=|A|3(I)        |(AdjA)|=|A|2Hence, the correct option is (B).

Q.39

If A=    2111  21   11  2, verify thatA36A2+9A4I=O and hence find A1.

Ans

  A=[     2  1  11   21  11  2]A2=[     2  1  11   21  11  2][     2  1  11   21  11  2]      =[  4+1+1  2212+1+2221  1+4+11222+1+2122  1+1+4]      =[  6  555  65  556]A3=A2A      =[  6  555  65  556][     2  1  11   21  11  2]      =[  12+5+5  61056+5+101065  5+12+5561010+5+65106  5+5+12]=[  22  212121   22212121  22]A36A2+9A4 I= [  22  212121   22212121  22]6[  6  555  65  556]+9[     2  1  11   21  11  2]4[10001000  1]= [  22  212121   22212121  22][3630303036303030  36]+[189  9918999  18][40004000  4]=[  40  303030   40303030  40][4030303040303030  40]= [000000000]=OThus,A36A2+5A+11 I=0.Hence, it is verified.Now,    A36A2+9A4I=O(AAA)A16(AA)A1+9AA14IA1=O[Postmultiplying by A1 as |A|0]AA(AA1)6A(AA1)+9I=4A1  A2I6AI+9I=4A1      14(A26A+9I)=A1  A1=14(A26A+9I)...(i)Now,A26A+9I=[  6  555  65  556]6[     2  1  11   21  11  2]+9[100010001] =[ 6 5 5 5 6 5 5 5 6 ][ 12 6 6 6 12 6 6 6 12 ]+[ 9 0 0 0 9 0 0 0 9 ] =[ 15 5 5 5 15 5 5 5 15 ][ 12 6 6 6 12 6 6 6 12 ]=[ 3 1 1 1 3 1 1 1 3 ] From equation( i ), we have A -1 =[ 3 1 1 1 3 1 1 1 3 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@1B15@

Q.40

For the matrix A=[  11112321  3], show thatA36A2+5A+11I=O. Hence find A1.

Ans

A=[  1   111   2321  3]A2=[  1   111   2321  3][  1   111   2321  3]      =[  1+1+2  1+2113+31+26  1+4+316921+622-3  2+3+9]      =[4  213  8147314]  A3=A2A      =[4  213  8147314][  1   111   2321  3]      =[  4+2+2  4+4146+33+828  3+16+143244273+287614  7+9+42]      =[  8   7123   27693213  58]A36A2+5A+11 I= [871232769321358]6[421381473  14]+5[11  112321  3]+11[10001000  1]= [871232769321358][241261848844218  84]+[55  551015105  15]    +[1100011000  11]=[824+5+11712+5+016+5+023+18+5+02748+10+1169+8415+03242+10+013+185+05884+15+11]= [000000000]=OThus,A36A2+5A+11 I=0.Now,    A36A2+5A+11I=O(AAA)A16(AA)A1+5AA1+11IA1=O[Postmultiplying by A1 as |A|0]AA(AA1)6A(AA1)+5I=11A1  A2I6AI+5I=11A1  111(A26A+5I)=A1  A1=111(A26A+5I)...(i)Now,A26A+5I=[42138147314]6[11  112321  3]+5[100010001]      =[42138147314][66  661218126  18]+[500050005] =[ 9 2 1 3 13 14 7 3 19 ][ 6 6 6 6 12 18 12 6 18 ] =[ 3 4 5 9 1 4 5 3 1 ] From equation( i ), we have: A 1 = 1 11 ( A 2 6A+5I ) = 1 11 [ 3 4 5 9 1 4 5 3 1 ]= 1 11 [ 3 4 5 9 1 4 5 3 1 ] MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=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@D577@

Q.41

For matrix  A=[3 21 1], find the number a and b such thatA2+aA+bI=O.

Ans

Given matrix  A=[3 21 1], A2=[3 21 1][3 21 1]      =[9+2 6+23+1 2+1]      =[11843]Now,   A2+aA+bI=O      AA(A1)+aAA1=bIA1[Postmultiplying by A1 as |A|0]        A(AA1)+aI=bIA1                A(I)+aI=bIA1                A1=1b(A+aI)Now,    |A|=|3211|=32=1                  A1=1|A|AdjA=11[1213]We have          A1=1b([3 21 1]+a[1001])            [1213]=1b[3+a211+a]            [1213]=[3+ab2b1b1+ab]Comparing the corresponding elements of two matrices, we have1=1bb=1and    1=3+abb=3+a    1=3+a  a=4Thus, the required value of a and b are 4 and 1.

Q.42

If  A =  3 11 2 ,show that A25A+7I=O. Hence find A1.

Ans

Given,  A=[  3  11 2], show that A25A+7I=O. Hence find A1.A2=[  3 11 2][  3 11 2]      =[  91  3+232 1+4]      =[  8 55 3]A25A+7I=[  8 55 3]5[  3 11 2]+7[1 00 1]        =[  8 55 3][  15  55 10]+[7 00 7]        =[  815+7 55+05+5+0 310+7]        =[0 00 0]Hence,A25A+7I=OA.A5A=7IA.A(A1)5AA1=7I(A1)[Postmultiplying by A1 as |A|0]A(AA1)5I=7A1AI5I=7A1A1=17(A5I)A1=17(5IA)A1=17(5[1 00 1][  3  11 2])A1=17([5 00 5][  3  11 2])A1=17[53010+152]A1=17[2113]

Q.43

Let  A=  3 11 2and  B=6 87 9 .Verify that (AB)1=B1A1.

Ans

Let A=[372 5]|A|=|3725|=1514=1Now,A11=(1)1+15=5A12=(1)1+22=2A21=(1)2+17=7A22=(1)2+23=3AdjA=[5273]=[5723]A1=1|A|.AdjA=1[5723]=[5723]Let B=[6 879]|B|=|6879|=5456=2Now,B11=(1)1+19=9B12=(1)1+27=7B21=(1)2+18=8B22=(1)2+26=6AdjB=[9786]=[9876]B1=1|B|.AdjB=12[9876]=[924723]B1A1=[924723][5723]=[4528 632+12352+6 4929]=[612872472672](1)AB=[3725][6879]=[18+4924+6312+3516+45]=[67874761]|AB|=|67874761|=40874089=2Adj(AB)=[61478767]=[61874767](AB)1=12[61874767]=[612872472672](2)Thus, from equation (1) and equation(2), we have(AB)1=B1A1Hence proved.

Q.44 Find the inverse of each of the matrices (if it exists)

1 0       20 cosα   sinα3 sinα cosα

Ans

LetA=[1000cosαsinα0sinαcosα]|A|=|1000cosαsinα0sinαcosα|=1(cos2αsin2α)+0(00)+0(00)=10So, the inverse of A exists.Now, A11=(1)1+1|cosα sinαsinαcosα|=cos2αsin2α=1A12=(1)1+2|0 sinα0cosα|=0A13=(1)1+3|0 cosα0 sinα|=0A21=(1)2+1|0 0sinαcosα|=0A22=(1)2+2|100 cosα|=cosαA23=(1)2+3|1 00 sinα|=sinαA31=(1)3+1|0 0cosα sinα|=0A32=(1)3+2|100 sinα|=sinαA33=(1)3+3|1 00 cosα|=cosαAdjA=[1000cosαsinα0sinαcosα]=[1000cosαsinα0sinαcosα]A1=1|A|AdjA=1[1000cosαsinα0sinαcosα]=[1000cosαsinα0sinαcosα]

Q.45 Find the inverse of each of the matrices (if it exists)

1 1 20 2 33 2   4

Ans

Let  A=[112023324]      |A|=|  1120  23324|  =1(86)+1(0+9)+2(06)  =2+912  =10So, the inverse of A exists.Now, A11=(1)1+1|    23 2   4|      =2A12=(1)1+2|033  4|      =9A13=(1)1+3|0   232|      =06      =6A21=(1)2+1|1 22 4|      =(4+4)=0A22=(1)2+2|1 23 4|=2A23=(1)2+3|1 13 2|      =(2+3)  =1A31=(1)3+1|1 2  2 3|  =1A32=(1)3+2|1   20 3|      =3A33=(1)3+3|1 10   2|      =2AdjA=[29602113  2]  =[  20192  361  2]A1=1|A|AdjA  =1[  20192  361  2]  =[20  1  92  3  61  2]

Q.46 Find the inverse of each of the matrices (if it exists)

  2   1 3  4 1 07 2 1

Ans

Let  A=[   2   1 3   41 07   2 1]      |A|=|   2  1 3   41 07   2 1|  =2(10)1(40)+3(87)  =24  +  3  =30So, the inverse of A exists.Now, A11=(1)1+1|1  0   2 1|      =1A12=(1)1+2|  4 07 1|      =4A13=(1)1+3|  4 17 2|      =87      =1A21=(1)2+1|1 32 1|      =(16)      =5A22=(1)2+2|  2 37 1|      =23A23=(1)2+3|  2 17 2|      =(4+7)      =11A31=(1)3+1|  1 31 0|      =3A32=(1)3+2|2 34 0|      =12A33=(1)3+3|2   14 1|      =6AdjA=[141523113126]  =[153423121116]A1=1|A|AdjA  =13[153423121116]

Q.47 Find the inverse of each of the matrices (if it exists)

1 0    03 3 05 2 1

Ans

LetA=[1 003 3 05 2 1]|A|=|1 0 03 3 05 21|=1(30)0(30)+0(615)=30So, the inverse of A exists.Now, A11=(1)1+1|3021|=3A12=(1)1+2|3051|=3A13=(1)1+3|3 35 2|=615=9A21=(1)2+1|0 021|=0A22=(1)2+2|1051|=1A23=(1)2+3|1 05 2|=(20)=2A31=(1)3+1|0 03 0|=0A32=(1)3+2|1 03 0|=0A33=(1)3+3|1 03 3|=3AdjA=[339012003]=[300310923]A1=1|A|AdjA=13[300310923]

Q.48 Find the inverse of each of the matrices (if it exists)

1 2 30 2 40 0 5

Ans

Let  A=[1 2 30 2 40 0 5]      |A|=|1 2 30 2 40 0 5|  =1(100)2(00)+3(00)  =100So, the inverse of A exists.Now, A11=(1)1+1|2 40 5|=10A12=(1)1+2|0 40 5|=0A13=(1)1+3|0 20 0|=0A21=(1)2+1|2 30 5|=10A22=(1)2+2|1 30 5|=5A23=(1)2+3|1 20 0|  =0A31=(1)3+1|1 20 2|=2A32=(1)3+2|1 30 4|=4A33=(1)3+3|1 20 2|=2 AdjA= 10 0 0 10 5 0 2 4 2 = 10 10 2 0 5 4 0 0 2 A 1 = 1 A AdjA = 1 10 10 10 2 0 5 4 0 0 2 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadgeacaWGKbGaamOAaiaaykW7caWGbbGaeyypa0ZaamWa aeaafaqabeWadaaabaGaaGymaiaaicdaaeaacaaIWaaabaGaaGimaa qaaiaaykW7caaMc8UaeyOeI0IaaGymaiaaicdaaeaacaaI1aaabaGa aGimaaqaaiaaikdaaeaacaaMc8UaaGPaVlabgkHiTiaaisdaaeaaca aMc8UaaGPaVlaaikdaaaaacaGLBbGaayzxaaWaaWbaaSqabeaacaGG NaaaaaGcbaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaamWaaeaafaqabe WadaaabaGaaGymaiaaicdaaeaacaaMc8UaaGPaVlabgkHiTiaaigda caaIWaaabaGaaGOmaaqaaiaaicdaaeaacaaI1aaabaGaeyOeI0IaaG inaaqaaiaaicdaaeaacaaIWaaabaGaaGPaVlaaikdaaaaacaGLBbGa ayzxaaaabaGaamyqamaaCaaaleqabaGamGjGgkHiTiacyciIXaaaaO Gaeyypa0ZaaSaaaeaacaaIXaaabaWaaqWaaeaacaWGbbaacaGLhWUa ayjcSdaaaiaadgeacaWGKbGaamOAaiaaykW7caWGbbaabaGaaCzcai aaykW7caaMc8Uaeyypa0ZaaSaaaeaacaaIXaaabaGaaGymaiaaicda aaWaamWaaeaafaqabeWadaaabaGaaGymaiaaicdaaeaacaaMc8UaaG PaVlabgkHiTiaaigdacaaIWaaabaGaaGOmaaqaaiaaicdaaeaacaaI 1aaabaGaeyOeI0IaaGinaaqaaiaaicdaaeaacaaIWaaabaGaaGPaVl aaikdaaaaacaGLBbGaayzxaaaaaaa@918A@

Q.49 Find the inverse of each of the matrices (if it exists)

1 53 2

Ans

Let  A=[1 53 2]      |A|=|1 53 2|  =2+15  =130So, the inverse of A exists.Now, A11=(1)1+12  =2A12=(1)1+2(3)  =3A21=(1)2+1(5)=5A22=(1)2+2(1)=1AdjA=[  2  351]  =[2 53 1]A1=1|A|AdjA=113[2 53 1]

Q.50 Find the inverse of each of the matrices (if it exists)

[224  3]

Ans

Let  A=[2 24   3]      |A|=|2 24   3|  =6+8  =140So, the inverse of A exists.Now, A11=(1)1+13=3A12=(1)1+24=4A21=(1)2+1(2)=2A22=(1)2+22=2AdjA=[3 42   2]  =[  3 24 2]A1=1|A|AdjA=114[  3 24 2]

Q.51 Verify A(adj A)=(adj A)A = |A|I

11    23  021  0  3

Ans

Let A=[112302103]|A|=|112302103|=1(00)+1(9+2)+2(00)=11A11=(1)1+1|0203|=0,A12=(1)1+2|3213|=11,A13=(1)1+3|3010|=0,A21=(1)2+1|1203|=3A22=(1)2+2|1213|=1A23=(1)2+3|1110|=1A31=(1)3+1|1202|=2A32=(1)3+2|1232|=8A33=(1)3+3|1130|=3AdjA=[0110311283]=[0321118013]A(AdjA)=[0321118013][112302103]=[0+9+2 0+0+0 06+611+3+8 11+0+02 22+2403+3 0+0+0 0+2+9]=[11 0 00 11 00 0 11]Also,(AdjA)A=[112302103][0321118013]=[0+11+0 312 28+60+0+0 9+0+2 6+060+0+0 3+03 2+0+9]=[11 0 00 11 00 0 11]|A|I=11[1 0 00 1 00 0 1]=[11 0 00 11 00 0 11]A(AdjA)=(AdjA)A=|A|I

Q.52 Verify A(adj A) = (adj A)A = |A|I

  2   34 6

Ans

Let A=[  2   34 6]      |A|=|    2   34 6|            =12+12=0A11=(1)1+1(6)=6,A12=(1)1+2(4)=4A21=(1)2+13=3A22=(1)2+2(2)=2AdjA=[6 43 2]=[6 3  4   2]A.(AdjA)=[  2   34 6][6 3  4     2]        =[12+126+6    2424  1212]        =[0 00 0]=0(AdjA).A=[6 3  4     2][  2   346]        =[12+12 18+181212 1212]        =[0 00 0]=0      |A|I=0[1 00 1]=0Thus,A.(AdjA)=(AdjA).A=|A|I

Q.53 Find adjoint of each of the matrices

  1 1 22   3 52   0 1

Ans

Let A=[    11 2    2   3 52   0 1]We haveA11=(1)1+1|3 50 1 |        =3A12=(1)1+2|  2 52 1|        =(2+10)        =12A13=(1)1+3|  2 32 0|        =(0+6)        =6A21=(1)2+1|1 2  0 1|        =1A22=(1)2+2|  1 22 1|        =5A23=(1)2+3|    112  0|        =1(02)        =2A31=(1)3+1|1 2  3 5|        =56        =11A32=(1)3+2|1 22 5|        =(54)        =1A33=(1)3+3|1 12   3|        =3+2        =5Adj A=[    3 12 6    1    5 211 1 5]=[    3    11112    51    6   2   5]

Q.54 Find adjoint of each of the matrices

1 23 4

Ans

Let A=[1 23 4]We haveA11=(1)1+14        =4A12=(1)1+23        =3A21=(1)2+12        =2A22=(1)2+21        =1AdjA=[  4 32   1]=[  4 23   1]

Q.55

If Δ=|a11 a12 a13a21 a22 a23a31 a32a33| and Aij is Co factor so faij, then valueof Δ is given by(A) a11A31+a12A32+a13A33 (B)  a11A11+a12A21+a13A31(C)    a21A11+a22A12+a23A13(D)  a11A11+a21A21+a31A31

Ans

Since, we know thatΔ=Sum of product of elements of C1 with their corresponding cofactors    =a11A11+a21A21+a31A31Thus, the optionD is correct.

Q.56

Using Co factors of element sof third column, evaluateΔ=1 x yz1 y zx1 z xy.

Ans

Given determinant isΔ=|1 x yz1 y zx1 z xy|Minors of elements of third column areM13=|1 y1 z|=zyM23=|1 x1 z|=zxM33=|1 x1 y|=yxCofactors of elements of third column areA13=(1)1+3M13        =zyA23=(1)2+3M23        =(zx)        =xzA33=(1)3+3M33        =yxExpanding the determinant along C3, we get    Δ=a13A13+a23A23+a33A33        =yz(zy)+zx(xz)+xy(yx)        =yz2y2z+zx2z2x+xy2x2y        =zx2y2zx2y+xy2+yz2z2x        =z(x2y2)+xy(yx)+z2(yx)        =(xy){z(x+y)xyz2}        =(xy)(zx+zyxyz2)        =(xy)(zxxy+zyz2)        =(xy){x(yz)+z(yz)}        =(xy)(yz)(zx)

Q.57

Using Co factors of element so f second row, evaluateΔ=|5 3 82 0 11 2 3|

Ans

Δ=|5 3 82 0 11 2 3|Minors corresponding to second row:M21=|3 82 3|        =916        =7CofactorAij=(1)i+jMijA21=(1)2+1M21=(1)(7)=7M22=|5 81 3|        =158        =7A22=(1)2+2M22=7M23=|5 31 2|        =103        =7A23=(1)2+3M23        =(1)(7)        =7Expanding the determinant along R2, we get    Δ=a21A21+a22A22+a23A23        =2×7+0×7+1×7        =14+07        =7

Q.58 Write Minors and Cofactors of the elements of following determinants:

(i) 1 0 00 1 00 0 1 (ii) 1 0 43 510 1 2

Ans

(i)  |1 0 00 1 00 0 1|Minor of the element aij is Mij Minorof element a11=M11    =|1 00 1|    =1    Minorof element a12=M12    =|0 00 1|    =0    Minorof element a13=M13    =|0 10 0|    =0  Minorof element a21=M21    =|0 00 1|    =0Minorof element a22=M22    =|1 00 1|    =1    Minorof element a23=M23    =|1 00 0|    =0        Minorof element a31=M31    =|0 01 0|    =0    Minorof element a32=M32    =|1 00 0|    =0    Minorof element a33=M33    =|1 00 1|    =1Now, cofactor of aij is Aij, so  Aij=(1)i+jMijA11=(1)1+1M11        =1A12=(1)1+2M12        =0A13=(1)1+3M13        =0A21=(1)2+1M21        =0 A 22 = ( 1 ) 2+2 M 22 =1 A 23 = ( 1 ) 2+3 M 23 =0 A 31 = ( 1 ) 3+1 M 31 =0 A 32 = ( 1 ) 3+2 M 32 =0 A 33 = ( 1 ) 3+3 M 33 =1 ( ii )| 1 0 4 3 5 1 0 1 2 | Minor of the element a ij is M ij Minor of element a 11 = M 11 =| 5 1 1 2 | =10+1 =11 Minor of element a 12 = M 12 =| 3 1 0 2 | =6 Minor of element a 13 = M 13 =| 3 5 0 1 | =3 Minor of element a 21 = M 21 =| 0 4 1 2 | =4 Minor of element a 22 = M 22 =| 1 4 0 2 | =2 Minor of element a 23 = M 23 =| 1 0 0 1 | =1 Minor of element a 31 = M 31 =| 0 4 5 1 | =20 Minor of element a 32 = M 32 =| 1 4 3 1 | =13 Minor of element a 33 = M 33 =| 1 0 3 5 | =5 Now, cofactor of a ij is A ij , so A ij = ( 1 ) i+j M ij A 11 = ( 1 ) 1+1 M 11 =11 A 12 = ( 1 ) 1+2 M 12 =6 A 13 = ( 1 ) 1+3 M 13 =3 A 21 = ( 1 ) 2+1 M 21 =( 1 )( 4 ) =4 A 22 = ( 1 ) 2+2 M 22 =2 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A 23 = ( 1 ) 2+3 M 23 =1 A 31 = ( 1 ) 3+1 M 31 =20 A 32 = ( 1 ) 3+2 M 32 =( 1 )( 1 ) =13 A 33 = ( 1 ) 3+3 M 33 =5 MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbbjxAHXgarmWu51MyVXgaruWqVvNCPvMCG4uz3bqefqvATv2CG4uz3bIuV1wyUbqee0evGueE0jxyaibaieYBg9LrFfeu0dXdh9vqqj=hEeeu0xXdbba9frFj0=OqFfea0dXdd9vqaq=JfrVkFHe9pgea0dXdar=Jb9hs0dXdbPYxe9vr0=vr0=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@BE6F@

Q.59 Write Minors and Cofactors of the elements of following determinants:

(i) 2 40   3 (ii) a cb d

Ans

(i)  |2 40   3|Minor of the element aij is MijMinorof element a11=M11=3,    Minorof element a12=M12=0,    Minorof element a21=M21=4,    Minorof element a22=M22=2.Cofactor of the element aij is Aij=(1)(i+j)Mij    A11=(1)(1+1)M11=3      A12=(1)(1+2)M12=0      A21=(1)(2+1)M21=4    A22=(1)(2+2)M22=2(ii)  |a cb d|Minor of the element aij is MijMinorof element a11=M11=d,    Minorof element a12=M12=b,    Minorof element a21=M21=c,    Minorof element a22=M22=a.Cofactor of the element aij is Aij=(1)(i+j)Mij    A11=(1)(1+1)M11=d      A12=(1)(1+2)M12=b      A21=(1)(2+1)M21=c    A22=(1)(2+2)M22=a

Q.60 If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Ans

Given: area of triangle =± 35  (Since area of triangle is absolute value of Δ)Then the area of triangle formed by given vertices (2,6),  (5, 4),(k, 4)        Δ=12|2 6 15   4 1k   4 1|    ±35=12{2(44)(6)(5k)+1(204k)}    ±35=12(0+306k+204k)    ±70=10k+5010k+50=70  or  10k+50=70k=2010=2 or k=12010=12Thus, value of k is either 12 or 2.Thus, option (D) is correct.

Q.61 (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Ans

(i)Point A(1,2) and B(3,6) are joining each other and a point P(x,y) lieson it, so A, P and B are collinear and area of triangle formed by thesepoints is zero. ThenAreaoftriangle ABP = 0|1 2 13 6 1x y 1|=012{1(6y)2(3x)+1(3y6x)}=06y6+2x+3y6x=02y4x=0y=2xThus, the equation of line joining two points (1, 2) and (3,6)is y=2x.(ii)Point A(3,1) and B(9,3) are joining each other and a point P(x,y) lieson it, so A, P and B are collinear and area of triangle formed by thesepoints is zero. ThenAreaoftriangle ABP = 0|3 1 19 3 1x y 1|=012{3(3y)1(9x)+1(9y3x)}=093y9+x+9y3x=06y2x=03y=xThus, the equation of line joining two points (1, 2) and (3,6)is x3y=0.

b

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Ans

(i)Given: area of triangle =±4 (Since area of triangle is absolute value of Δ)Then the area of triangle formed by given vertices (k,0),          (4,0),(0,2)        Δ=12|k 0 14 0 10 2 1|    ±4=12{k(02)0(40)+1(80)}    ±4=12(2k0+8)    ±8=2k+82k+8=8  or  2k+8=8k=0 or k=162=8Thus, value of k is either 0 or 8.(ii) Given: area of triangle =±4 (Since area of triangle is absolute value of Δ)Then the area of triangle formed by given vertices(2,0),  (0, 4),(0, k)        Δ=12|2 0 1  0 4 1  0 k 1|    ±4=12{2(4k)0(00)+1(00)}    ±4=12(8+2k0+0)    ±8=2k82k8=8  or  2k8=8k=162=8 or k=02=0Thus, value of k is either 8 or 0.

Q.63 Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Ans

Theare a of triangle formed by given vertices A (a, b+c),B(b, c+a),C(c, a+b)        Δ=12|ab+c1bc+a1ca+b1|=12{a(c+aab)(b+c)(bc)+1(ba+b2c2ca)}=12{acab(b2c2)+(ba+b2c2ca)}=12(acabb2+c2+ba+b2c2ca)=0Since, area of triangle formed by A, B and C is zero, so thesepoints are collinear.

Q.64 Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)

Ans

(i) Theare a of triangle formed by given vertices (1, 0),(6, 0),(4, 3)        Δ=12|1 0 16 0 14 3 1|=12{1(03)0(64)+1(180)}=12(30+18)=12(15)=152sq.units(ii) Theare a of triangle formed by given vertices (2, 7),(1,1),(10, 8)        Δ=12|2 7 11 1 110 8 1|=12{2(18)7(110)+1(810)}=12(14+632)=12(47)=472sq.units(iii) The area of triangle formed by given vertices (2,3),          (3,2),(1,8)        Δ=12|2 3 1    3   2 118 1|=12{2(2+8)(3)(3+1)+1(24+2)}=12(20+1222)=12(30)=15sq.unitsSince, area of triangle can never be negative. So, area of triangle = 15 square units.

Q.65 Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these.

Ans

To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where aij = (i,j)th element of A.
Therefore, option (C) is correct

Q.66 Let A b a square matrix of order 3 x 3, then |kA| is equal to

(A) k|A| (B) k2|A| (C) k3|A| (D) 3k|A|

Ans

LetA be a square matrix of order 3×3,then  A=[a1 a2 a3b1 b2 b3c1 c2 c3]and        kA=[ka1 ka2 ka3kb1 kb2 kb3kc1 kc2 kc3]        |kA|=|ka1 ka2 ka3kb1 kb2 kb3kc1 kc2 kc3|=k3|a1 a2 a3b1 b2 b3c1 c2 c3|[Taking k common from R1, R2 and R3.]=k3|A|Thus, option (C) is correct.

Q.67 By using properties of determinants, show that:

1+a2 ab abab b2+1 bcca cb c2+1=1+a2+b2+c2

Ans

L.H.S.=|1+a2 ab acab b2+1 bcca cb c2+1|Applying  R11aR1,R21bR2,R31cR3  =abc|1a+a b ca b+1bca b c+1c|Applying  R1R1R2,R2R2R3  =abc|1 a1 b 00   1b1ca    b      c+1c|Applying  C1aC1,C2bC2,C3cC3  =abcabc|1 1 00     1 1a2   b2      c2+1|Applying  C2C2+C1  =|1 0 00 1 1a2 a2+b2      c2+1|Expanding along R1,weget  =|1 1a2+b2 c2+1|  =(c2+1)(1)(a2+b2)  =1+c2+a2+b2  =1+a2+b2+c2  =R.H.S.Hence, the given result is proved.

Q.68 By using properties of determinants, show that:

1+a2b2 2ab 2b2ab 1a2+b2 2a2b 2a 1a2b2=(1+a2+b2)3

Ans

L.H.S. =|1+a2b2   2ab    2b2ab 1a2+b2        2a2b 2a 1a2b2|Apply R1R1+bR3, R2R2aR3  =|1+a2+b20b(1+a2+b2)01+a2+b2a(1+a2+b2)2b2a1a2b2|  =(1+a2+b2)2|10b01a2b2a1a2b2|Expanding along R1, we get  =(1+a2+b2)2[1|    1a2a1a2b2|+(b)|012b1a|]  =(1+a2+b2)2(1a2b2+2a2b×2b)  =(1+a2+b2)2(1a2b2+2a2+2b2)  =(1+a2+b2)3=R.H.S.Hence,it is proved.

Q.69 By using properties of determinants, show that:

1 x x2x2 1 xx x2 1=(1x3)2

Ans

L.H.S.=|1 x x2x2 1 xx x2 1|  =|1+x+x2 1+x+x2 1+x+x2x2 1 xx x2 1|[R1R1+R2+R3]  =(1+x+x2) |1 1 1x2 1 xx x2 1|  =(1+x+x2) |1   0 0x2 1x2 xx2x x2x 1x|[C2C2C1,C3C3C1]  =(1+x+x2) |1   0 0x2 (1x)(1+x) x(1x)x x(x1) (1x)|  =(1+x+x2) (1x)(1x)|1   0 0x2 (1+x) xx x 1|[Taking (1x)  common fromC2 and C3.]Expanding along R1, we get  =(1+x+x2)(1x)(1x)(1+x+x2)  ={(1x)(1+x+x2)}{(1x)(1+x+x2)}  =(1x3)(1x3)  =(1x3)2=R.H.S.Hence Proved.

Q.70 By using properties of determinants, show that:

(i) abc 2a     2a2b bca   2b2c 2c cab=(a+b+c)3(ii) x+y+2z x     yz   y+z+2x   yz           x z+x+2y=2(x+y+z)3

Ans

(i)|abc 2a     2a2b bca    2b2c 2c cab|=(a+b+c)3L.H.S.=|abc 2a     2a2b bca     2b2c 2c cab|=|a+b+c a+b+c a+b+c2b bca     2b2c 2c cab|[ApplyR1R1+R2+R3]=(a+b+c)|1 1 12b bca 2b2c 2c cab|=(a+b+c)|1   0 02b bca  02c 0 cab|      [C2C2C1,C3C3C1]=(a+b+c)|bca 00     cab|=(a+b+c)(bca)2=(a+b+c)3=R.H.S.(ii)L.H.S.=|x+y+2z x       yz     y+z+2x      yz           x z+x+2y|          =|2x+2y+2z x y2x+2y+2z y+z+2x            y2x+2y+2z x z+x+2y|  [C1C1+C2+C3]          =(2x+2y+2z)|1 x y1 y+z+2x            y1 x z+x+2y|          =(2x+2y+2z)|1 x y0 y+z+x 00 0 z+x+y|[R2R2R1R3R3R1]          =(2x+2y+2z)3|1 x y0 1  00 0 1|Expanding  along  C1, we get          =(2x+2y+2z)3|1 00 1|          =(2x+2y+2z)3(10)          =(2x+2y+2z)3=R.H.S.         

Q.71 By using properties of determinants, show that:

(i) x+4 2x 2x2x x+4 2x2x 2x x+4=(5x+4)(4x)2 (ii)y+k y yy y+4 yy y y+4=k2(3y+k)

Ans

(i) L.H.S.=|x+4 2x 2x2x x+4 2x2x 2x x+4|        =|5x+4 2x 2x5x+5 x+4 2x5x+5 2x x+4|[ApplyC1C1+C2+C3]        =(5x+4)|1 2x 2x1 x+4 2x1 2x x+4|        =(5x+4)|1 2x 2x0 4x 00 0 4x|  [ApplyR2R2R1,R3R3R1]Expanding along C1, we get        =(5x+4)|4x 00 4x|        =(5x+4)(4x)2      =R.H.S.(ii) L.H.S.=|y+k y yy y+k yy y y+k|      =|3y+k y     y3y+k y+k     y3y+k y y+k|[ApplyC1C1+C2+C3]      =(3y+k)|1 y   y1 y+k     y1 y y+k|[Apply R2R2R1,R3R3R1]      =(3y+k)|1 y y0 k 00 0 k|Expanding along C1, we get      =(3y+k)|k 00 k|      =(3y+k)×k2      = k2(3y+k)

Q.72 By using properties of determinants, show that:

x x2 yzy y2 zxz z2 xy=(xy)(yz)(zx)(xy+yz+zx)

Ans

L.H.S.=|x x2 yzy y2 zxz z2 xy|  =|xy x2y2 yzzxyz y2z2 zxxyz z2 xy|[ApplyingR1R1R2,R2R2R3]  =|(xy) (xy)(x+y) z(xy)(yz) (yz)(y+z) x(yz)    z       z2     xy|  =(xy)(yz)|1 (x+y) z1 (y+z) xz     z2 xy|  [Taking common (xy)and(yz) from R1 and R2respectively]  =(xy)(yz)|0 (xz) (xz)1 (y+z) xz     z2   xy|[ApplyingR1R1R2]  =(xy)(yz)(zx)|0 1 11 (y+z) xz   z2   xy|[Taking common (zx)from R1]  =(xy)(yz)(zx)|0 0 11 (x+y+z)