NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5

These NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 about Determinants are specifically curated by the Extramarks Maths faculty expert which is detailed and easily comprehensible by the Class 12 students. This will help students practice and solve problems associated with the NCERT solutions Class 12 Maths chapter 4 exercise 4.5. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 provide students with simple and effective examples along with solutions that make it easy for the students to grasp the concept efficiently. The students can analyse the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 for better preparation for their board examination which can give the best results.

NCERT Solutions for Class 12 Maths Chapter 4- Determinants Exercise 4.5

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, and NCERT solutions for other classes and subjects like NCERT solutions Class 11, NCERT solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6. It also has NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 till NCERT Solutions Class 1.  These NCERT solutions are for different subjects which are meant for all the NCERT students of different standards from 1 to 12. Extramarks also provides a convenient way for students to access these materials without them having to do any extra hard work.

NCERT Solutions for Class 12 Maths Chapter 4 consists of solutions and elucidation of what Determinants are and what is their significance. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 of Chapter 4 deals with all the conditions and problems related to the Adjoint and Inverse of a Matrix. Extramarks also has other NCERT Solutions for Class 12 Maths Chapter 4 Exercises apart from Exercise 4.5 for Class 12 Maths NCERT, which explains how to evaluate Determinants, to use the properties of Determinants for solving problems, to use Determinants for finding out Areas of Triangles with Vertices, Values, Equations of Lines etc. These solutions will stand out to the Class 12 students because they will give students a proper understanding of Chapter 4: Determinants as a whole. Hence, Extramarks gives importance to each and every topic just like NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, providing study materials for other Exercises included in the NCERT Mathematics Textbook Chapter 4. Students should practice the Ex 4.4 Class 12 Maths NCERT Solutions to have a good understanding of Determinants.

Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, which can be accessed by the NCERT Class 12 students anytime and anywhere. Basically, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 contains the content of the Chapter 4 Determinants. Determinants are studied as a subject of linear algebra and are used for linear equations. In this Chapter 4, Exercise 4.5 Class 12 Maths NCERT Solutions, the students will learn about: the Adjoint and Inverse of a Square Matrix, the condition for the Inverse of a Square Matrix to exist and the use of an Adjoint of a Matrix.

What are Adjoint and Inverse of a Matrix?

The Inverse of a Matrix can be understood only if the student understands what the adjoint of a Matrix is, so they are clearly related to each other and students must give equal attention to both terms and their significance. Also, to clearly understand the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, students must first be clear with what both the Adjoint and Inverse of a Matrix mean and where these concepts apply specifically. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 particularly deals with problems related to Adjoint of a Matrix and Inverse of a Matrix. Extramarks has proper solutions and study material for even different Matrix problems for Class 12 students to go through and solve, which it provides in different chapters including NCERT solutions Maths for Matrix.

Adjoint of a Matrix

The adjoint of a square Matrix is the reverse of its Cofactor Matrix and it is also called an Adjunct Matrix even if this term is seldomly used in today’s time. For instance, say there is a Matrix A, the adjoint will be denoted as adj (A).  The NCERT Maths Textbook defines the adjoint of a Matrix as “The adjoint of a square Matrix A = [aij] n × n is defined as the transpose of the Matrix [Aij] n × n, where Aij is the cofactor of the element aij . Adjoint of the Matrix A is denoted by adj A.” First and foremost, it is important to find the adjoint of a given Matrix to calculate the Inverse Matrix. However, it is to be noted that it can be done only for Square Matrices. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 consists of solutions of problems related to the adjoint of a Matrix such as finding the Adjoint of any given Matrix or Matrices. Therefore, understanding the concept of an Adjoint of a Matrix is of prior importance to the students to understand further solutions of the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5.

Inverse of a Matrix

As numbers have reciprocals for them, similarly, Matrices too have reciprocals. This reciprocal of a Matrix is known as the Inverse of a Matrix. The Inverse of a Matrix is set by any other given Matrix, which when multiplying with the given Matrix results in the multiplicative identity. For instance, the Inverse of a Matrix M will be denoted by M-1. However, to get the Inverse of 3 × 3 matrices, students need to learn about the Determinant and Adjoint of the Matrix first. There are two ways by which the Inverse of the Matrix can be obtained. The Inverse of a Matrix can be computed by elementary operations and also by using the Adjoint of a Matrix. Therefore, it is said that knowing the Adjoint of a Matrix is important for solving further Matrix problems. The simple operations on Matrices can be calculated through transformations of rows or columns. In addition, the Inverse of a Matrix can be found by using the Inverse of the Matrix formula, that is by using the Determinant and the Adjoint of the Matrix. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, consists of solutions based most specifically on an Inverse of a Matrix along with solutions for Adjoint of a Matrix. These NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, clarifies the basic understanding of the Matrices so that the students can use them to solve the Determinant problems. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 is therefore a significant set of NCERT solutions for students of Class 12 to work on to get conceptualised knowledge on the NCERT Class 12 Mathematics Chapter 4.

Maths NCERT Solutions Class 12 Chapter 4 Exercise 4.5

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 available at Extramarks are intensively and authentically solved, also curated by the Extramarks expert faculty. They are based on the NCERT textbook and give out easy solutions for quick concept clearance of the Class 12 NCERT students. Therefore, students are relieved from the worry of any faults or errors in these NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 and can use them for practice and reference, ideally.

NCERT Class 12 students need effective solutions and comprehension for every topic and subtopic of Mathematics to get high scores, as Mathematics as a subject is very high scoring if the solutions and results are on point. Therefore, every NCERT chapter of the subject of Mathematics for Class 12 must be given equal importance by students so that they have no hesitation while solving the problems related to the topics and can easily find solutions to them. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, in this case, emphasises the Adjoint and Inverse of a Matrix, like how to find an Adjoint of a given Matrix or how to find an Inverse of a given Matrix. Based on these problems, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 also deals with the various theorems and definitions of both Adjoint and Inverse of a Matrix to understand both the elements of the Matrix distinctly. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 consists of solutions that are given by using those theorems, formulas, and definitions of the Adjoint and Inverse of a Matrix. The several subtopics determine all the functions and total significance of Determinants in Algebra, like the types of Determinants: the First Order Determinant, Second-Order Determinant, and the Third Order Determinant, many properties of Determinants, How do Minors and Cofactors work for Matrices and Determinants, what are the major applications of Matrices and Determinants, how are they correlated? And even the method of using the Inverse of a Matrix for Solving a System of Linear Equations. Students gain a perspective on how to solve any given question on Matrix or Determinants by going through the Class 12 maths chapter 4 Exercise 4.5 Solutions and the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5.

Out of these subtopics of Class 12 Mathematics Chapter 4, the Adjoint and Inverse of a Matrix, which is the main topic of explanation for the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, it mainly talks about 4 theorems related to them plus specific definitions. The main 4 theorems of Adjoint and Inverse of a Matrix are explained very briefly below:

  • Theorem 1: If M is the given Square Matrix of order x, then M(adj M) = (adj M) M= |A| I , where I is the Identity Matrix of order x.
  • Theorem 2: If M and N are Nonsingular Matrices of the same order, then MN and NM are also nonsingular Matrices of the same order.
  • Theorem 3: The Determinant of the product of Matrices is equal to the product of their Respective Determinants, which means, |MN|= |M| |N|, where M and N are square matrices of the same order.
  • Theorem 4: A Square Matrix M is Invertible if specifically, M is a Nonsingular Matrix.

Some important points that the students should note are:

  1. A Square Matrix, say M is said to be singular only if |M| = 0.
  2. Similarly, a Square Matrix M is said to be Nonsingular if |M|≠ 0

They will be able to solve any problem related to the questions included in Exercise 4.5 or any similar question mentioned in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 for that matter. Moreover, Extramarks prepares important questions and sets different difficulty levels for students to test every concept they have learned to improve their learning skills. The important questions of the practice series vary in levels from an easy level to the difficulty level and also have a medium level in order to not intimidate students. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 provide verifications and examples for all the theorems and definitions related to the Adjoint and Inverse of a Matrix so that Class 12 students understand why the theorems and solutions work the way they do in minute detail. Extramarks provides extra live lessons and discussions of the topics comprising the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. It also provides an analysis of all the solutions given in NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 by the Extramarks experts of the subject.

The questions which will be set for the Class 12 NCERT question paper of Mathematics will follow similar questions that are in the Class 12 Maths Chapter 4 Exercise 4.5. As a consequence, Class 12 Maths Chapter 4 Exercise 4.5 Solutions provided by Extramarks help the Class 12 students become familiar with questions plus solutions and their patterns which may come in their NCERT Mathematics question paper. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 follow the exact same pattern of questions and answers that are set in the NCERT Mathematics question papers and textbook. Extramarks also provides media-rich study material and sources available both in 2D and 3D engaging videos sometimes also in the form of animation in accordance with the Class NCERT Mathematics syllabus and chapters. These resources open up the minds of the students, and they find the learning experience fun and interesting, which helps them clear their concepts more easily and doesn’t intimidate them. Instead, it makes them more friendly towards the subject, even making them willing to learn more about the topics included in the subject. Students can first learn the particular topics they have a problem with, in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. Then they can practice the solutions given in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 for clearing all their doubts. After learning and practising the solutions explained in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, students can take the Extramarks test for the same topic. Through this, they can see how much they have progressed with a certain topic and can track their progress at each given point.

About Extramarks

Extramarks is an advanced education platform that prioritises the fact that each and every student should comprehend the studied topics effectively so that they can solve any problem given to them at any point of time. Therefore, Extramarks emphasizes conceptual learning, so that students are able to learn concepts and ideas for life and ace any examination they take. It provides live classes held by Experts and also provides recorded classes if missed by any student. All the study materials for all subjects and classes are available together at Extramarks’ website so that the students do not face any problems in learning the topics and can access chapters and solutions anywhere and anytime. Extramarks provides The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 and many other similar NCERT solutions for students who follow the NCERT textbooks.

Besides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, Extramarks also provides different chapters of Mathematics included in the NCERT syllabus. It prepares for the students different sets of test series for Class 12 Mathematics like Test Series XII Mathematics Set-1, Test Series XII Mathematics Set-2 and Test Series XII Mathematics Set-3. At Extramarks, Class 12 students can find sets of model papers of past years along with solutions and also model papers which are based on board sanctioned blueprint for self-evaluation and also preparation of examination. As for the NCERT Class 12 Chapter 4 and the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, Extramarks focuses on conceptual learning with interactive as well as engaging lessons. It covers all the topics: Minors and Cofactors, Application of Determinant, the value of Determinant, etc and also discusses topics related to NCERT Solutions like the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 like the adjoint and Inverse of a Matrix. The classes and videos on the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 have all definitions and theory proofing verifications so that the students can understand the exercise competently. There is also detailed in-depth learning of the NCERT Chapter 4 for Class 12, Determinants as a whole at Extramarks. Extramarks provides quick learning options like supplying the students with easy mind maps of the chapters. Thus, the students can take a quick recap of what they studied and learned, after that they can summarise all the topics. Extramarks additionally provides important and high-weightage study material that contains essential content for the best exam preparation possible. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 that are highly recommended by experts also fall under these detailed and accurate study material supplied by Extramarks.

One more extra benefit of applying to Extramarks is that it has a huge number of practice materials like HOTS (High-Order Thinking Skills) questions along with hints and descriptions. They challenge the cognitive and critical thinking skills of the students practising them. Extramarks also plans and gives out students Q&A sessions based on chapters along with model answers for an all-inclusive practice of concepts and ideas. For the best performance in these practice sessions available on Extramarks, students must also be clear with all the given study materials. They should focus equally on conceptual learning because that study material helps students solve and understand the chapter’s questions and hence they can find solutions easily. For example, if a student is clear with all the solutions given in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, students gradually get used to the pattern of questions.

Extramarks also sets various important tests which are specially curated by the Extramarks experts themselves in the form of many MCQ (Multiple Choice Questions) and long/short questions. These tests are very helpful for students as they can self-assess and learn about their weaknesses in the subject. Also, Extramarks provides model answers for the curated test questions for the students to evaluate their answers later. In addition to these, past years’ papers with solutions are available to students so that they get accustomed to examination patterns. Extramarks avails the students of many mock tests which are prepared based on the latest syllabus and pattern. The recent tests taken by the students at Extramarks can also be tracked, viewed, and reports of their performance in the tests are also available for students and parents to evaluate.

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 4.6
16 Questions & Solutions (3 Short Answers, 13 Long Answers)

 

Q.1 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         =1Adj A=[   4 −3−2    1]‘=[   4 −2−3     1]

Q.2 Find adjoint of each of the matrices

  1 −1 2  2    3 5−2    0 1

Ans

Let A=[    1−1 2    2    3 5−2   0 1]We​ haveA11=(−1)1+1|3 50 1 |        =3A12=(−1)1+2|   2 5−2 1|         =−(2+10)         =−12A13=(−1)1+3|   2 3−2 0|         =(0+6)         =6A21=(−1)2+1|−1 2   0 1|        =1A22=(−1)2+2|   1 2−2 1|         =5A23=(−1)2+3|    1−1−2   0|         =−1(0−2)         =2A31=(−1)3+1|−1 2   3 5|        =−5−6        =−11A32=(−1)3+2|1 22 5|        =−(5−4)        =−1A33=(−1)3+3|1 −12    3|        =3+2        =5Adj A=[    3 −12 6    1    5 2−11 −1 5]‘=[     3    1−11−12    5−1     6     2    5]

Q.3 Verify A(adj A) = (adj A)A = |A|I

   2      3−4 −6

Ans

Let​ A=[   2    3−4 −6]      |A|=|    2    3−4 −6|            =−12+12=0A11=(−1)1+1(−6)=−6,A12=(−1)1+2(−4)=4A21=(−1)2+13=−3A22=(−1)2+2(2)=2Adj A=[−6 4−3 2]‘=[−6 −3   4    2]A.(AdjA)=[   2    3−4 −6][−6  −3   4      2]         =[−12+12−6+6    24−24  12−12]         =[0 00 0]=0(AdjA).A=[−6  −3   4      2][   2    3−4−6]         =[−12+12 −18+1812−12 12−12]         =[0 00 0]=0       |A|I=0[1 00 1]=0Thus,A.(AdjA)=(AdjA).A=|A|I

Q.4 Verify A(adj A)=(adj A)A = |A|I

1−1    23   0−21   0   3

Ans

Let​ A=[1−1    23   0−21   0   3]      |A|=|1−1    23   0−21   0   3|            =1(0−0)+1(9+2)+2(0−0)  =11A11=(−1)1+1|0−20   3|=0,A12=(−1)1+2|3−21   3|=−11,A13=(−1)1+3|3010|=0,A21=(−1)2+1|−12   03|=3A22=(−1)2+2|1213|=1A23=(−1)2+3|1−11  0|=−1A31=(−1)3+1|−1   2   0−2|=2A32=(−1)3+2|1   23−2|=8A33=(−1)3+3|1−13   0|=3       Adj A=[0−11   03     1−12    8   3]‘  =[   0    3   2−11     1  8   0 −1   3]A(Adj A)=[   0    3   2−11     1  8   0 −1   3][1−1   23   0−21   0   3]  =[0+9+2 0+0+0 0−6+6−11+3+8 11+0+0−2 2−2+240−3+3 0+0+0 0+2+9]  =[11 0 00 11 00 0 11]Also,(Adj A)A=[1−1   23   0−21   0   3][   0    3   2−11     1  8   0 −1   3]  =[0+11+0 3−1−2 2−8+60+0+0 9+0+2 6+0−60+0+0 3+0−3 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(Adj A)=(Adj A)A=|A|I

Q.5 Find the inverse of each of the matrices (if it exists)

[2−24   3]

Ans

Let  A=[2 −24    3]       |A|=|2 −24    3|   =6+8   =14≠0So, the inverse of A exists.Now, A11=(−1)1+13=3A12=(−1)1+24=−4A21=(−1)2+1(−2)=2A22=(−1)2+22=2Adj A=[3 −42    2]‘  =[   3 2−4 2]A−1=1|A|Adj A =114[   3 2−4 2]

Q.6 Find the inverse of each of the matrices (if it exists)

−1 5−3 2

Ans

Let  A=[−1 5−3 2]       |A|=|−1 5−3 2|   =−2+15   =13≠0So, 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) =−1Adj A=[   2   3−5−1]‘  =[2 −53 −1]A−1=1|A|Adj A =113[2 −53 −1]

Q.7 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(10−0)−2(0−0)+3(0−0)   =10≠0So, 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

Adj A= 10 0 0   −10 5 0 2   −4   2 ‘   = 10   −10 2 0 5 −4 0 0  2 A −1 = 1 A Adj A   = 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.8 Find the inverse of each of the matrices (if it exists)

1 0    03 3   05 2 −1

Ans

Let  A=[1 0   03 3    05 2 −1]       |A|=|1 0   03 3    05 2−1|   =1(−3−0)−0(−3−0)+0(6−15)   =−3≠0So, the inverse of A exists.Now, A11=(−1)1+1|3   02−1|=−3A12=(−1)1+2|3   05−1|=3A13=(−1)1+3|3 35 2|       =6−15 =−9A21=(−1)2+1|0    02−1| =0A22=(−1)2+2|1   05−1| =−1A23=(−1)2+3|1 05 2|       =−(2−0) =−2A31=(−1)3+1|0 03 0| =0A32=(−1)3+2|1 03 0| =0A33=(−1)3+3|1 03 3| =3Adj A=[−33−9  0−1−20  0  3]‘  =[−3  003−10−9−2 3]A−1=1|A|Adj A  =−13[−3  00  3−10−9−2 3]

Q.9 Find the inverse of each of the matrices (if it exists)

   2    1 3   4 −1 0−7    2 1

Ans

Let  A=[   2    1 3   4−1 0−7   2 1]       |A|=|   2   1 3   4−1 0−7   2 1|   =2(−1−0)−1(4−0)+3(8−7)   =−2−4  +  3   =−3≠0So, the inverse of A exists.Now, A11=(−1)1+1|−1  0   2 1|       =−1A12=(−1)1+2|   4 0−7 1|       =− 4A13=(−1)1+3|  4 −1−7    2|       =8−7       =1A21=(−1)2+1|1 32 1|       =−(1−6)       =5A22=(−1)2+2|   2 3−7 1|       =23A23=(−1)2+3|   2 −1−7    2|       =−(4+7)       =−11A31=(−1)3+1|   1 3−1 0|       =3A32=(−1)3+2|2 34 0|       =12A33=(−1)3+3|2    14 −1|       =−6Adj A= [−1−41523−11312−6]‘  =[−153−423121−11−6]A−1=1|A|Adj A  =−13[−153−423121−11−6]

Q.10 Find the inverse of each of the matrices (if it exists)

1 −1  20    2 −33 −2   4

Ans

Let  A=[1−1202−33−24]       |A|=|   1−120  2−33−24|   =1(8−6)+1(0+9)+2(0−6)   =2+9 − 12   =−1≠0So, the inverse of A exists.Now, A11=(−1)1+1|    2 −3− 2   4|       =2A12=(−1)1+2|0−33  4|       =− 9A13=(−1)1+3|0     23− 2|       =0−6       =−6A21=(−1)2+1|−1 2−2 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|       =2Adj A= [2−9−60−2−1−1 3  2]‘  =[   20−1−9−2  3−6−1  2]A−1=1|A|Adj A  =−1[   20−1−9−2  3−6−1  2]  =[−20  1  92  −3  61  −2]

Q.11 Find the inverse of each of the matrices (if it exists)

1   0         20 cosα    sinα3 sinα −cosα

Ans

Let  A=[1000cosαsinα0sinα−cosα]       |A|=|   1000cosαsinα0sinα−cosα|   =1(−cos2α−sin2α)+0(0−0)+0(0−0)   =−1≠0So, the inverse of A exists.Now, A11=(−1)1+1| cosα sinαsinα   −cosα|       =−cos2α−sin2α       =−1A12=(−1)1+2|0 sinα0   −cosα|       =0A13=(−1)1+3|0 cosα0 sinα| =0A21=(−1)2+1|0 0sinα−cosα| =0A22=(−1)2+2|1    00 −cosα| =−cosαA23=(−1)2+3|1 00 sinα|       =−sinαA31=(−1)3+1|0     0cosα sinα|       =0A32=(−1)3+2|1    00 sinα|       =−sinαA33=(−1)3+3|1 00 cosα|       =cosαAdj A= [−1000−cosα−sinα0 −sinα  cosα]‘  =[   −1000−cosα  −sinα0−sinα  cosα]A−1=1|A|Adj A  =−1[   −1000−cosα  −sinα0−sinα  cosα]  =[   1000cosα  sinα0sinα  −cosα]

Q.12

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

Ans

Let A=[372 5]      |A|=|3725|  =15−14=1Now, A11=(−1)1+15=5A12=(−1)1+22=−2A21=(−1)2+17=−7A22=(−1)2+23=3Adj A=[5−2−73]‘   =[5−7−23]    A−1=1|A|.Adj A   =1[5−7−23]   =[5−7−23]Let B=[6 879]      |B|=|6879|  =54−56=−2Now, B11=(−1)1+19=9B12=(−1)1+27=−7B21=(−1)2+18=−8B22=(−1)2+26=6Adj B=[9−7−86]‘   =[9−8−76]    B−1=1|B|.Adj B   =−12[9−8−76]   =[−92  4  72−3]B−1A−1=[−92  4  72−3][   5−7−2   3]    =[−452−8 632+12352+6 −492−9]    =[−612872472−672]…(1)      AB=[3725][6879]    =[18+4924+6312+3516+45]    =[67874761]     |AB|=|67874761|    =4087−4089 =−2Adj(AB)=[61−47−8767]‘        =[61−87−4767](AB)−1=−12[61−87−4767]        =[−612872472−672]…(2)Thus, from equation (1) and equation(2), we have(AB)−1=B−1A−1Hence proved.

Q.13

If  A =  3  1−1 2 ,show that A2−5A+7I=O. Hence find A–1.

Ans

Given,  A=[  3  1−1 2], show that A2–5A+7I=O. Hence find A–1.A2=[  3  1−1 2][  3  1−1 2]      =[  9−1  3+2−3−2 −1+4]      =[   8 5−5 3]∴A2−5A+7I=[   8 5−5 3]−5[  3  1−1 2]+7[1 00 1]        =[   8 5−5 3]−[  15  5−5 10]+[7 00 7]        =[   8−15+7 5−5+0−5+5+0 3−10+7]        =[ 0 00 0]Hence, A2−5A+7I=O∴A.A−5A=−7I⇒A.A(A−1)−5AA−1=−7I(A−1)[Post–multiplying by A–1 as |A|≠0]⇒A(AA−1)−5I=−7A−1⇒AI−5I=−7A−1⇒A−1=1−7(A−5I)⇒A−1=17(5I−A)⇒A−1=17(5[1 00 1]−[  3  1−1 2])⇒A−1=17([5 00 5]−[  3  1−1 2])⇒A−1=17[5−30−10+15−2]⇒A−1=17[2−113]

Q.14

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(A−1)+aAA−1=−bIA−1[Post–multiplying by A–1 as |A|≠0]⇒         A(AA−1)+aI=−bIA−1⇒                 A(I)+aI=−bIA−1⇒                 A−1=1−b(A+aI)Now,    |A|=|3211|=3−2=1⇒                   A−1=1|A|AdjA=11[1−2−13]We​ have          A−1=1−b([3 21 1]+a[1001])⇒             [1−2−13]=1−b[3+a211+a]⇒             [1−2−13]=[3+a−b2−b1−b1+a−b]Comparing the corresponding elements of two matrices, we have−1=1−b⇒b=1and     1=3+a−b⇒−b=3+a⇒    −1=3+a⇒  a=−4Thus, the required value of a and b are −4 and 1.

Q.15

For the matrix A=[  11112−32−1  3], show thatA3−6A2+5A+11I=O. Hence find A–1.

Ans

A=[  1   111   2−32−1  3]A2=[  1   111   2−32−1  3][  1   111   2−32−1  3]      =[  1+1+2   1+2−11−3+31+2−6   1+4+31−6−92−1+62−2-3  2+3+9]      =[4  21−3  8−147−314]  A3=A2A      =[4  21−3  8−147−314][  1   111   2−32−1  3]      =[  4+2+2   4+4−14−6+3−3+8−28   −3+16+14−3−24−427−3+287−6−14  7+9+42]      =[  8   71−23   27−6932−13  58]∴A3−6A2+5A+11 I= [871−2327−6932−1358]−6[421−38−147−3   14]+5[11  112−32−1   3]+11[10001000   1]= [871−2327−6932−1358]−[24126−1848−8442−18   84]+[55  5510−1510−5   15]    +[1100011000   11]=[8−24+5+117−12+5+01−6+5+0−23+18+5+027−48+10+11−69+84−15+032−42+10+0−13+18−5+058−84+15+11]= [000000000]=OThus, A3−6A2+5A+11 I=0.Now,     A3−6A2+5A+11I=O⇒(AAA)A−1−6(AA)A−1+5AA−1+11 IA−1=O[Post–multiplying by A–1 as |A|≠0]⇒AA(AA−1)−6A(AA−1)+5I=−11A−1⇒  A2I−6AI+5I=−11A−1⇒  1−11 (A2−6A+5I)=A−1⇒  A−1=1−11 (A2−6A+5I)...(i)Now,A2−6A+5I=[421−38−147−314]−6[11  112−32−1  3]+5[100010001]      =[421−38−147−314]−[66  6612−1812−6  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.16

If A=     2−11−1   2−1    1−1  2, verify thatA3−6A2+9A−4I=O and hence find A–1.

Ans

  A=[     2   −1  1−1   2−1  1−1  2]A2=[     2   −1  1−1   2−1  1−1  2][     2   −1  1−1   2−1  1−1  2]      =[  4+1+1   −2−2−12+1+2−2−2−1   1+4+1−1−2−22+1+2−1−2−2  1+1+4]      =[   6  −55−5  6−5  5−56]A3=A2A      =[   6  −55−5  6−5  5−56][     2   −1  1−1   2−1  1−1  2]      =[  12+5+5   −6−10−56+5+10−10−6−5   5+12+5−5−6−1010+5+6−5−10−6  5+5+12]=[  22   −2121−21   22−2121−21  22]∴A3−6A2+9A−4 I= [  22   −2121−21   22−2121−21  22]−6[   6  −55−5  6−5  5−56]+9[     2   −1  1−1   2−1  1−1  2]−4[10001000   1]= [  22   −2121−21   22−2121−21  22]−[36−3030−3036−3030−30   36]+[18−9  9−918−99−9   18]−[40004000   4]=[  40   −3030−30   40−3030−30  40]−[40−3030−3040−3030−30   40]= [000000000]=OThus, A3−6A2+5A+11 I=0.Hence, it is verified.Now,     A3−6A2+9A−4I=O⇒(AAA)A−1−6(AA)A−1+9AA−1−4 IA−1=O[Post–multiplying by A–1 as |A|≠0]⇒AA(AA−1)−6A(AA−1)+9I=4A−1⇒  A2I−6AI+9I=4A−1⇒       14 (A2−6A+9I)=A−1⇒  A−1=14 (A2−6A+9I)...(i)Now,A2−6A+9I=[   6  −55−5  6−5  5−56]−6[     2   −1  1−1   2−1  1−1  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.17 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, (Adj A) A=|A|I=[|A|000|A|000|A|]⇒             |(Adj A)A|=||A|000|A|000|A||⇒       |(Adj A)||A|=|A|3|100010001|⇒         |(Adj A)||A| =|A|3(I)∴         |(Adj A)|=|A|2Hence, the correct option is (B).

Q.18

If A is aninvertible matrix of order 2, then   detA–1   isequalto     A DetA B1DetA C 1 D 0

Ans

Since, A is an invertible matrix,then A–1 exists and A–1=1|A|adjA.Let​ a matrix of order 2 be as A=[abcd],then |A|=ad−bc and Adj A=[d−b−ca]Now,     A−1=1|A|Adj A    =1|A|[d−b−ca]    =[d|A|−b|A|−c|A|a|A|]∴           |A−1|=|d|A|−b|A|−c|A|a|A||    =1|A|2|d−b−ca|    =1|A|2(ad−bc)    =1|A|2|A|    =1|A|∴det(A−1)=1det(A)Thus, the correct option is (B).

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FAQs (Frequently Asked Questions)

Yes, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5  are available online at Extramarks which can be accessed by Class 12 students anytime anywhere. The solutions are skillfully curated by the Mathematics experts faculty of Extramarks and are prepared in a way that the students can easily comprehend and solve any problems that were referred into the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. 

Yes, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 is an important part of the Class 12 board examination. In fact, every exercise in the NCERT solutions is essential for the Class 12 students in their preparation for boards. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 are provided by Extramarks so that students can prepare for their examination by referring to these solutions and improving their concepts. Therefore, they can solve any problem that comes in the question paper of the Class 12 board examination. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 consist of a very important topic of what is Adjoint and Inverse of a Matrix and what it is used for, so the NCERT Mathematics Board Question Paper will consist of questions related to those in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. If students have no knowledge of the chapter and its exercises, they will struggle in solving the questions corresponding to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. So, they will have more chances of getting fewer marks for not completing the concerned questions.

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 deals with basically the questions and solutions of the Adjoint and Inverse of a Matrix. Similar to finding either the Adjoint of a Matrix or the Inverse of a given Matrix or finding both. There are solutions for verifying certain equations using the Adjoint or the Inverse of a Matrix. Also, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 include all the theorems and formulas associated with the Adjoint and Inverse of Matrices which makes the students understand the topics in a detailed manner. 

Yes, besides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, Extramarks also provides solutions for all the chapters and exercises of NCERT Mathematics for Class 12. Students are offered the entire syllabus of the NCERT Mathematics textbook for Class 12 for practice and solving problems by Extramarks. Just like the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 which has solutions for their specific topic or subtopic, other NCERT solutions also have conceptualised solutions for the topics they are related to and deal with all the problems, evaluations, and verifications of the topic. For example the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 deal with Adjoint and Inverse of a Matrix and their Mathematical calculations of the main topic Determinants. Likewise, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 deal with Minors and Cofactors of Determinants. All other chapters dealing with different topics of the Class 12 NCERT Mathematics syllabus are available at Extramarks.