NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 (Ex 4.6)

Class 12 plays an important role in the life of a student as well as their parents. The hard work that a student puts in all his or her school life comes full circle. The hopes and dreams of millions of children and their parents depend upon the Class 12 board exams. Students spend countless hours to make sure they secure good marks in the examination.

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NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 (Ex 4.6)

The Class 12 Mathematics syllabus is vast and one needs to have an immaculate plan from the beginning. Students are suggested to finish the whole syllabus in a few months to have proper time for revision. Early preparation helps, so it doesn’t feel like a mountain to climb in the end. The NCERT books are your one-stop solution to beat the odds and score stellar marks in Mathematics.

Previously, in Unit 1, students studied Relations and Functions and Inverse Trigonometric Functions. NCERT Solutions for the above and NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 helps to set a base for upcoming topics.

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Further, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 help you get a comprehensive idea regarding the application of Determinants and Matrices. It ensures that you are able to solve the questions with the utmost accuracy and confidence.

Important Topics 

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

The introduction part gives you a brief idea of the various areas where determinants could be useful in real-life situations. The objective behind going through these chapters is provided in detail. Determinants are applicable in the study of Engineering, Science, Economics, and Social Science. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 discusses the condition for the existence of the inverse of a matrix.

Here, students study various properties of Determinants, Minors, Cofactors, and Application Of Determinants in Finding the Area Of a Triangle, Adjoint And Inverse of a Square Matrix. Solutions of Linear Equations in two or three variables using the inverse of a matrix. Consistency and inconsistency of system of linear equations. Four Properties of Determinants are essential for students to understand the topic and are useful to solve complex linear equations.

Extramarks provides comprehensive study material, complete syllabus coverage, and curriculum mapping to make students clear the exam with flying colours.

To get detailed answers, visit the Extramarks website, which provides NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6.

Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

Important Points

NCERT Solutions help the student adapt to the style of writing the answers that give them an approach to perform well in exams. NCERT solutions help students to prepare for board exams. Extramarks provides NCERT Solutions Class 12, NCERT Solutions Class 11, NCERT Solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6, NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2, NCERT Solutions Class 1.

Extramarks provides online classes and gamified learning experiences. Learning, practising, and analysing is the go-to model that helps the students get better at their studies.

Students often seem shy or nervous to clear their doubts, one-on-one live classes help students to get personalised attention to ensure their doubts are clarified.

The majority of students fail to score well in the exams due to a lack of practice. Extramarks provides them with NCERT Solutions Class 12 Maths Chapter 4 Exercise 4.6 to make the best of their time and excel in the board exams.

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Class 12 Maths Chapter 4 Exercise 4.6 Solutions are also provided by the Extramarks experts to help in providing authentic and detailed answers.

NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6

If a student wants to do well in the board exams, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 really helps the student. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 prepared by the experts at Extramarks accelerate their learning and growth. The biggest reason students fail to achieve good marks is the lack of authentic and detailed solutions.

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Apart from the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6, there are several exercises which contain innumerable questions as well. Our in-house subject experts at Extramarks solve these questions with detailed and authentic solutions. To score well in exams, practice hard using the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6.

Related Questions

Determinant is described as the numerical value of the square matrix. If A is a square matrix, which is a matrix that has an equal number of rows and columns, i.e A = aij of order n, then the determinant of this matrix is denoted by det A. The element aij signifies the entry in the ith row or jth column. In simple terms, a determinant is a special number that is calculated from the column.

Apart from the Class 12 Maths Chapter 4 Exercise 4.6, there are other questions for which complete and detailed NCERT Solutions are given, that are authentic and comprehensive.

Apart from NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6, Extramarks provides solutions for other exercises of determinants.

In exercise 4.1, the first exercise deals with evaluating the Determinants. Other than the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6. Extramarks provides complete solutions for this exercise as well.

Exercise 4.2 deals with the properties of Determinants. For questions 1 to 16, it is required to prove using the properties of determinants without expanding. Extramarks provides genuine solutions to these exercises apart from the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6.

Exercise 4.3 deals with the Area of a Triangle. Questions 1 to 5 deal with finding the area of a triangle, showing the points are collinear, finding the value of k, and finding the equation of the line joining using determinants, respectively. Extensive NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 and Exercise 4.3 would prepare students in less time. The questions in these exercises are vital when preparing for Determinants to attain terrific marks in these topics.

Chapter 4 – Determinants of Other Exercises

Chapter 4 – Determinants of Other Exercises
Exercise 4.1
8 Questions & Solutions (3 Short Answers, 5 Long Answers)
Exercise 4.2
10 Questions & Solutions (4 Short Answers, 10 Long Answers)
Exercise 4.3
5 Questions & Solutions (2 Short Answers, 3 Long Answers)
Exercise 4.4
5 Questions & Solutions (2 Short Answers, 3 Long Answers)
Exercise 4.5
18 Questions & Solutions (4 Short Answers, 14 Long Answers)

Q.1 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|      =3−4      =−1≠0∴A is a non–sigular matrix.Therefore, A–1 exists.

Hence, the given system of equations is consistent.

Q.2 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

The given system of equations is:  2x−y=5    x+y=4The given system of equations can be written in the form ofAX=B, whereA=[2−111],  X=[xy] and B=[54]Now,|A|=|2−111|      =2+1      =3≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Hence, the given system of equations is consistent.

Q.3 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| =6−6      =0∴A is a sigular matrix.(Adj A)=[6−2−31]‘=[6−3−21](Adj A)B=[6−3−21][58]=[30−24−10+8]=[6−2]≠OThus, the solution of the given system of equations doesnot exist. Hence, the system of equations is inconsistent.

Q.4 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(6a−2a)−1(4a−2a)+1(2a−3a)      =4a−2a−a=a≠0∴ A is a non–sigular  matrix.Therefore, A–1 exists.Hence, the given system of equations is consistent.

Q.5 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:   3x−y−2z=2   2y−z=−13x−5y=3The given system of equations can be written in the form ofAX=B, whereA=[3−1−202−13−50],  X=[xyz] and B=[2−13]Now,|A|=|3−1−202−13−50|      =3(0−5)+1(0+3)−2(0−6)      =−15+3+12=0∴A is a sigular matrix.A11=(−1)1+1|2−1−50|=−5A12=(−1)1+2|0−130|=−3A13=(−1)1+3|023−5|=−6A21=(−1)2+1|−1−2−50|=10A22=(−1)2+2|3−230|=6A23=(−1)2+3|3−13−5|=12A31=(−1)3+1|−1−2  2−1|=5A32=(−1)3+2|3−20−1|=3A33=(−1)3+3|3−10  2|=6(Adj A)=[−5−3−610612536]‘=[−5105−363−6126](Adj A)B=[−5105−363−6126][2−13]=[−10−10+15−6−6+9−12−12+18]=[−5−3−6]≠OThus, the solution of the given system of equations doesnot exist. Hence, the system of equations is inconsistent.

Q.6 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:      5x−y+4z=5    2x+3y+5z=2   5x−2y+6z=−1The given system of equations can be written in the form ofAX=B, whereA=[5−142355−26],  X=[xyz] and B=[52−1]Now,|A|=|5−142355−26|      =5(18+10)+1(12−25)+4(−4−15)      =140−13−76=51≠0∴ A is a non–sigular  matrix.Therefore, A–1 exists.Hence, the given system of equations is consistent.

Q.7 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|      =15−14      =1≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Now, Adj A =[3−7−25]‘=[3−2−75]     A–1=1|A|(Adj A)   =11[3−2−75]   =[3−2−75]∴       X=A−1B  =[3−2−75][45]  =[12−10−28+25]    [xy]=[2−3]∴ x=2  and y=−3.

Q.8 Solve system of linear equations, using matrix method,

2x – y = –2
3x + 4y = 3

Ans

The given system of equations is:     2x−y=−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       =11≠0 ∴A is a non-sigular  matrix. 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   and y= 12 11 . MathType@MTEF@5@5@+=feaaguart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKfMBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1BTfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakqaabeqaaiaadsfacaWGObGaamyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGZbGaaeyEaiaabohacaqG0bGaaeyzaiaab2gacaqGGaGaae4BaiaabAgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGZbGaaeiiaiaabMgacaqGZbGaaeOoaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaIYaGaamiEaiabgkHiTiaadMhacqGH9aqpcqGHsislcaaIYaaabaGaaGPaVlaaykW7caaIZaGaamiEaiabgUcaRiaaisdacaWG5bGaeyypa0JaaG4maaqaaiaabsfacaqGObGaaeyzaiaabccacaqGNbGaaeyAaiaabAhacaqGLbGaaeOBaiaabccacaqGZbGaaeyEaiaabohacaqG0bGaaeyzaiaab2gacaqGGaGaae4BaiaabAgacaqGGaGaaeyzaiaabghacaqG1bGaaeyyaiaabshacaqGPbGaae4Baiaab6gacaqGZbGaaeiiaiaabogacaqGHbGaaeOBaiaabccacaqGIbGaaeyzaiaabccacaqG3bGaaeOCaiaabMgacaqG0bGaaeiDaiaabwgacaqGUbGaaeiiaiaabMgacaqGUbGaaeiiaiaabshacaqGObGaaeyzaiaabccacaqGMbGaae4BaiaabkhacaqGTbGaaeiiaiaab+gacaqGMbaabaGaaeyqaiaabIfacqGH9aqpcaqGcbGaaeilaiaabccacaqG3bGaaeiAaiaabwgacaqGYbGaaeyzaaqaaiaabgeacqGH9aqpdaWadaqaauaabeqaciaaaeaacaaIYaaabaGaeyOeI0IaaGymaaqaaiaaiodaaeaacaaI0aaaaaGaay5waiaaw2faaiaacYcacaaMc8UaaGPaVlaadIfacqGH9aqpdaWadaabaeqabaGaamiEaaqaaiaadMhaaaGaay5waiaaw2faaiaabccacaqGHbGaaeOBaiaabsgacaqGGaGaaeOqaiabg2da9maadmaaeaqabeaacqGHsislcaaIYaaabaGaaGPaVlaaykW7caaMc8UaaG4maaaacaGLBbGaayzxaaaabaGaamOtaiaad+gacaWG3bGaaiilaaqaamaaemaabaGaamyqaaGaay5bSlaawIa7aiabg2da9maaemaabaqbaeqabiGaaaqaaiaaikdaaeaacqGHsislcaaIXaaabaGaaG4maaqaaiaaisdaaaaacaGLhWUaayjcSdaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGioaiabgUcaRiaaiodaaeaacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIXaGaaGymaiabgcMi5kaaicdaaeaacqGH0icxcaqGbbGaaeiiaiaabMgacaqGZbGaaeiiaiaabggacaqGGaGaaeOBaiaab+gacaqGUbGaaeylaiaabohacaqGPbGaae4zaiaabwhacaqGSbGaaeyyaiaabkhacaaMc8UaaGPaVlaab2gacaqGHbGaaeiDaiaabkhacaqGPbGaaeiEaiaab6caaeaacaqGubGaaeiAaiaabwgacaqGYbGaaeyzaiaabAgacaqGVbGaaeOCaiaabwgacaqGSaGaaeiiaiaabgeadaahaaWcbeqaaiaab2cacaqGXaaaaOGaaeiiaiaabwgacaqG4bGaaeyAaiaabohacaqG0bGaae4Caiaab6caaeaacaqGobGaae4BaiaabEhacaqGSaGaaeiiaaqaaiaabgeacaqGKbGaaeOAaiaabccacaqGbbGaaeiiaiabg2da9maadmaabaqbaeqabiGaaaqaaiaaisdaaeaacqGHsislcaaIZaaabaGaaGymaaqaaiaaykW7caaMc8UaaGOmaaaaaiaawUfacaGLDbaadaahaaWcbeqaaGqabiaa=DcaaaGccqGH9aqpdaWadaqaauaabeqaciaaaeaacaaI0aaabaGaaGymaaqaaiabgkHiTiaaiodaaeaacaaIYaaaaaGaay5waiaaw2faaaqaaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaeyqamaaCaaaleqabaGaiGjGb2cacGaMagymaaaakiabg2da9maalaaabaGaaGymaaqaamaaemaabaGaamyqaaGaay5bSlaawIa7aaaadaqadaqaaiaabgeacaqGKbGaaeOAaiaabccacaqGbbaacaGLOaGaayzkaaaabaGaaCzcaiaaykW7caaMc8UaaGPaVlabg2da9maalaaabaGaaGymaaqaaiaaigdacaaIXaaaamaadmaabaqbaeqabiGaaaqaaiaaisdaaeaacaaIXaaabaGaeyOeI0IaaG4maaqaaiaaikdaaaaacaGLBbGaayzxaaaabaGaeyinIWLaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caWGybGaeyypa0JaamyqamaaCaaaleqabaGaeyOeI0IaaGymaaaakiaadkeaaeaacaWLjaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaaSaaaeaacaaIXaaabaGaaGymaiaaigdaaaWaamWaaeaafaqabeGacaaabaGaaGinaaqaaiaaigdaaeaacqGHsislcaaIZaaabaGaaGOmaaaaaiaawUfacaGLDbaadaWadaabaeqabaGaeyOeI0IaaGOmaaqaaiaaykW7caaMc8UaaGPaVlaaiodaaaGaay5waiaaw2faaaqaaiaaxMaacaWLjaGaaGPaVlaaykW7cqGH9aqpdaWcaaqaaiaaigdaaeaacaaIXaGaaGymaaaadaWadaabaeqabaGaeyOeI0IaaGioaiabgUcaRiaaiodaaeaacaaI2aGaey4kaSIaaGOnaaaacaGLBbGaayzxaaaabaGaaCzcaiaaykW7caaMc8UaaGPaVlaaykW7daWadaabaeqabaGaamiEaaqaaiaadMhaaaGaay5waiaaw2faaiabg2da9maalaaabaGaaGymaaqaaiaaigdacaaIXaaaamaadmaaeaqabeaacqGHsislcaaI1aaabaGaaGymaiaaikdaaaGaay5waiaaw2faaaqaaiaaxMaacaaMc8UaaGPaVlaaykW7caaMc8+aamWaaqaabeqaaiaadIhaaeaacaWG5baaaiaawUfacaGLDbaacqGH9aqpdaWadaabaeqabaWaaSaaaeaacqGHsislcaaI1aaabaGaaGymaiaaigdaaaaabaWaaSaaaeaacaaIXaGaaGOmaaqaaiaaigdacaaIXaaaaaaacaGLBbGaayzxaaaabaGaeyinIWLaamiEaiabg2da9maalaaabaGaeyOeI0IaaGynaaqaaiaaigdacaaIXaaaaiaaykW7caaMc8Uaamyyaiaad6gacaWGKbGaaGPaVlaadMhacqGH9aqpdaWcaaqaaiaaigdacaaIYaaabaGaaGymaiaaigdaaaGaaiOlaaaaaa@D3A8@

Q.9 Solve system of linear equations, using matrix method,

4x – 3y = 3
3x – 5y = 7

Ans

The given system of equations is:   4x−3y=3   3x−5y=7The given system of equations can be written in the form ofAX=B, whereA=[4−33−5],  X=[xy] and B=[37]Now,|A|=|4−33−5|      =−20+9      =−11≠0∴A is a non-sigular  matrix.Therefore, A-1 exists.Now, Adj A =[−5−33  4]‘     =[−53−34]     A-1=1|A|(Adj A)   =1−11[−53−34]   =111[5−33−4]∴       X=A−1B  =111[5−33−4][37]  =111[15−219−28]    [xy]=111[−6−19]    [xy]=[−611−1911]∴x=−611  and y=−1911.

Q.10 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|      =10−6      =4≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Now, Adj A =[  2−3−2  5]‘     =[2−2−35]     A–1=1|A|(Adj A)   =14[2−2−35]∴      X=A−1B   =14[2−2−35][35]   =14[6−10−9+25]    [xy]=14[−416]⇒[xy]=[−14]∴x=−1  and y=4.

Q.11 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        x−2y−z= 3 2             3y−5z=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( −5−0 )+1( 3−0 )       =26+5+3=34≠0 ∴A is a non-sigular  matrix. 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 Adj A= [ 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 5−15+27 3−9−45 ] [ x y z ]= 1 34 [    34    17 −51 ] =[ 1 1 2 − 3 2 ] Hence, x=1, y= 1 2 and z=− 3 2 . 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Q.12 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:         ​  x−y+z=4       2x+y−3z=0           x+y+z=2The given system of equations can be written in the form ofAX=B, whereA=[1−1   121−3111],  X=[xyz] and B=[402]Now,|A|=|1−1   121−3111|      =1(1+3)+1(2+3)+1(2−1)      =4+5+1=10≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Now, A11=4,A12=−5,A13=1A21=2,   A22=0,A23=−2A31=2,    A32=5,A33=3Adj A=[4−512  0−2253]‘  =[422−5051−23]   A−1=1|A|[422−5051−23]          =110[422−5051−23]∴X= A−1B=110[422−5051−23][402]=110[16+0+4−20+0+104+0+6][xyz]=110[   20−10   10]              =[2−11]Hence, x=2, y=−1 and z=1.

Q.13 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         x−2y+z=−4       3x−y−2z=3The given system of equations can be written in the form ofAX=B, whereA=[23 31−213−1−2],  X=[xyz] and B=[   5−4   3]Now,|A|=|23 31−213−1−2|=2(4+1)−3(−2−3)+3(−1+6)      =10+15+15=40≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Now, A11=5,  A12=5,A13=5A21=3,   A22=−13,  A23=11A31=9,    A32=1,  A33=−7Adj A=[5553  −131191−7]‘  =[5395−131511−7]   A−1=1|A|Adj A          =140[5395−131511−7]∴X= A−1B=140[5395−131511−7][   5−4   3]=140[25−12+2725+52+325−44−21][xyz]=140[   40   80−40]              =[12−1]Hence, x=1, y=2 and z=−1.

Q.14 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:         x−y+2z=7    3x+4y−5z=−5       2x−y+3z=12The given system of equations can be written in the form ofAX=B, whereA=[1−1 234−52−13],  X=[xyz] and B=[   7−5 12]Now,|A|=|1−1  234−52−1  3|      =1(12−5)+1(9+10)+2(−3−8)      =7+19−22=4≠0∴A is a non–sigular  matrix.Therefore, A–1 exists.Now, A11=7,  A12=−19,A13=−11A21=1,   A22=−1,  A23=−1A31=−3,    A32=11,  A33=7Adj A=[7−19−111  −1−1−311  7]‘  =[71−3−19−111−11−17]   A−1=1|A|Adj A          =14[71−3−19−111−11−17]∴X= A−1B=14[71−3−19−111−11−17][   7−5   12]=14[49−5−36−133+5+132−77+5+84][xyz]=14[8412] =[213]Hence, x=2, y=1 and z=3.

Q.15

If  A=2−3532−411−2, find A–1. UsingA–1solve the system ofequations:2x−3y+5z=113x+2y−4z=−5x+y−2z=−3

Ans

A=[2−3532−411−2]|A|=|2−3532−411−2|      =2(−4+4)+3(−6+4)+5(3−2)      =0−6+5      =−1≠0Now, A11=0,  A12=2,A13=1A21=−1,   A22=−9,  A23=−5A31=2,    A32=23,  A33=13Adj A=[021−1  −9−5223  13]‘  =[0−122−9231−513]   A−1=1|A|Adj A          =1−1[0−122−9231−513] ∴  A−1=[01−2−29−23−15−13]...(i)Now, the given system of equations can be written in the formof AX= B, whereA=[2−3532−411−2], X=[xyz] and B=[11−5−3]The solution of the system of equations is given by∴X= A−1B=[01−2−29−23−15−13][11−5−3][Using equation (i)]=[0−5+6−22−45+69−11−25+39][xyz]=[123]Hence, x=1, y=2 and z=3.

Q.16 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=xyz and B=609070A=432246623      =412−12−36−36+24−24      =0+90−40=50≠0Now, A11=0,    A12=30,    A13=−20A21=−5, A22=0,      A23=10A31=10, A32=−20,  A33=10∴Adj A=030−20−501010−2010‘      =0−510300−20−201010∴A−1=1AAdj A=1500−510300−20−201010Now,      X=A−1B=1500−510300−20−201010609070=1500−450+7001800+0−1400−1200+900+700=150250400400⇒xyz=588⇒x=5,  y=8 and z=8.Thus, the cost of 1 kg onions is ₹5, 1 kg wheat is ₹ 8 and that of 1 kg rice is ₹8.

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FAQs (Frequently Asked Questions)

If one or more solutions exist in the system of equations, it is said to be consistent. If no solution exists in the system of equations; it is said to be inconsistent. For step-by-step authentic answers, NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 are available at the Extramarks website prepared by experts. Extramarks benefits by interlinking concepts for better retention.

There are four properties of Determinants:

1) If the rows and columns of a determinant are interchanged, the value of the determinants remains unchanged.

2) The sign of the determinant changes, if any two rows of determinants are interchanged. 

3) The value of the determinant is 0 if any two rows or columns of the determinant are equal or identical.

4) The value of the determinant originally obtained is multiplied by k, if each element of row and column is multiplied by a constant value k.

Learn more about the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.6 that are accessible at the Extramarks website prepared by our subject matter experts. There is a curriculum mapping and full syllabus coverage to help you score well.