# NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 (Ex 3.4)

Mathematics is a subject that requires an ample amount of practice. Students cannot simply learn Mathematics, as it is a conceptual subject; it needs practice. It is essential for them to clear their basic concepts of the subject to score well. Mathematics is not just an academic subject, it is also the fundamental base of many scientific concepts.

Class 12 Mathematics has two volumes in the curriculum, hence containing a wide range of chapters. Since it has a huge syllabus, students are required to work very hard and have a good hold on the concepts of every chapter. The first step for the students to clear their basics is to learn the NCERT Textbook by heart. One of the many chapters in the Mathematics NCERT Textbook is Chapter 3, Matrices. Extramarks provides Class 12 Maths NCERT Solutions Chapter 3 Matrices Exercise 3.4, which are the most simplified and detailed answers cross-checked by the subject experts of Extramarks.

Matrices is not only a chapter of the Class 12 Mathematics curriculum, it is practised in other specified fields of Mathematics. It is one of the most used concepts in the field of Mathematics, and its existence made the solutions to many problems easier. Matrix is not a straight method of Applied Mathematics but has made solutions a lot easier than straightforward methods. That is why it is very essential to learn matrices in Class 12 because the students who have opted for Mathematics in Class 12 are looking forward to learning it for further studies. For higher studies in topics like Optics, Electrical Circuits, Quantum Physics, etc., matrices are very useful. Matrices are also used in day-to-day life. Some of its real-life applications are 3D Games, Economics and Business, Encryption, etc.

## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 (Ex 3.4)

Students can download the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 in PDF Format. The Ex3 4 Class 12 solutions provided by Extramarks are very reliable and can be accessed on any device.

### Key Learnings from NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 are analysed by the subject experts and are available on the Extramarks’ website. The solutions provided by Extramarks are accurate and properly detailed in a step-by-step approach so that students can achieve maximum marks. Class 12 Maths Chapter 3 Exercise 3.4 Solutions are based on the Elementary Operation (transformation) of a matrix, the inverse of a matrix by elementary operations and invertible matrices. Extramarks recommends that students download the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 so that they can understand the concept of the solutions and then practice them by heart again and again. This will help them with their pace and concepts, and then they will be able to find answers to all the difficult problems of Chapter-3 matrices very accurately and fast.

There is a basic format in which matrices are solved, they are written in rows and columns in a format like this-

⎡ 1 9 -13 ⎤

⎣ 20 5 -6 ⎦

This is a two rows and three columns matrix. It can also be referred to as a two-by-three Matrix or a 2×3- Matrix. The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 tells students about the operations performed on matrices. They are Addition, Subtraction, Multiplication and Division. Matrices are mainly used in Linear Algebra. Square matrices are the most important type of matrices, and so they are also the most practised matrices. Square matrices are the matrices with the same number of rows and columns. The determinant of a square matrix is a number associated with the matrix, which is fundamental for the study of a square matrix; for example, a square matrix can be inverted if and only if it has a non-zero determinant

Some students can find it difficult to understand the steps of these operations of matrices. The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 provided by Extramarks makes it easy for students to learn the complicated concepts involved in Chapter 3 Matrices. Extramarks provides students with expert solutions and live problem-solving classes so that students can learn efficiently and score the best in their exams.

### Benefits of NCERT Solutions Class 12 Maths Chapter 3

Students of Class 12 are required to have the NCERT curriculum on their tips. These books are the first step for them to build their fundamentals. It is very important for students to have the solutions to the NCERT questions in hand so that they don’t waste their time looking for solutions. Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 so that they can get authentic solutions without having to look anywhere else.

Chapter 3, Matrices of Class 12 Mathematics, is one of the most scoring chapters in the curriculum of Class 12 Maths. Once students understand the concepts of the chapter, they can solve any question related to that chapter easily.

NCERT books are written by the subject experts, so if students thoroughly go through the NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.4, they acknowledge all the concepts. Class 12 Maths NCERT Solutions Chapter 3 provided by Extramarks, makes students ready for their board examinations and maintains their confidence.

Exercise 3.4 Class 12 Mathematics is about Invertible Matrices. It tells about the Operations of Addition, Subtraction, Multiplication, and Division of the Matrices. With Linear Algebra, Matrices are also used in Abstract Algebra. Matrices can also have an infinite number of rows and columns.

Exercise 3.4 of Chapter 3 Class 12 Mathematics contains a part called Transpose of Matrices. It is the part where students get confused. Extramarks provides students with the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 so that they can understand difficult topics like Transpose of Matrices with ease. All the answers are explained step by step in these solutions with all the details so that the students do not get confused. Extramarks recommends students go through those answers again and again and score their best in the topic Matrices.

### NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

 Chapter 3 Matrix Exercises Exercise 3.1 10 Questions & Solutions (5 Short Answers, 5 Long Answers) Exercise 3.2 22 Questions & Solutions (3 Short Answers, 19 Long Answers) Exercise 3.3 12 Questions & Solutions (4 Short Answers, 8 Long Answers)

### NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4

Matrices are defined as a set of numbers arranged in rows and columns to form a rectangular array. The numbers are referred to as the Elements, or Entries, of the Matrix. Matrices have wide applications in Physics, Engineering, Economics, and Statistics as well as in various branches of Mathematics. Class 12 Maths NCERT Solutions Chapter 3 includes Types of Matrices, Operations, Adjoint and Inverse of a Matrix, Rank of a Matrix and Special Matrices and solving Linear Equations using a Matrix. When learning Matrices there are some basic concepts that students should know about Matrices like Null Matrix, Transpose of a Matrix, Inverse of Matrix A, Diagonal Matrix, Hermitian Matrix, Skew-Symmetric Matrices, etc. For structuring these basic concepts, Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4.

Matrices is a topic that requires an ample amount of practice like any other topic in mathematics, but its fundamentals and calculations are a bit different from the rest of the chapters in Mathematics. So for learning and understanding this chapter, students are advised to go through the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 thoroughly and clear their doubts and queries right away with the help of Extramarks’ guidance.

The  NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4  provided by Extramarks provides the students with an overview of all the processes of Matrices, followed by giving solved answers to the chapter. By going through these solutions, students can achieve their goals and score well in the board examination.

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Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4.

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Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4.

Q.1

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices},â€„\left[\begin{array}{l}1â€„â€„â€„â€„-1\\ 2â€„â€„â€„â€„â€„3\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 2 \mathrm{3}\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 2 \mathrm{3}\end{array}\right]=\left[\begin{array}{l}10\\ 01\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-2{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 0 \mathrm{5}\end{array}\right]=\left[\begin{array}{l} 10\\ -21\end{array}\right]\mathrm{A}\\ {\mathrm{R}}_{2}\to \frac{1}{5}{\mathrm{R}}_{2}\\ \therefore \left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 0 \mathrm{1}\end{array}\right]=\left[\begin{array}{l} \mathrm{1}0\\ -\frac{\mathrm{2}}{5}\frac{\mathrm{1}}{5}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}\\ \mathrm{ }\left[\begin{array}{l}10\\ 01\end{array}\right]=\left[\begin{array}{l} 10\\ -\frac{\mathrm{2}}{5}\frac{\mathrm{1}}{5}\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} \mathrm{1}0\\ -\frac{2}{5}\frac{\mathrm{1}}{5}\end{array}\right]\end{array}$

Q.2

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„â€„1\\ 1â€„â€„â€„â€„1\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{l}\mathrm{2}\mathrm{1}\\ 11\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}2\mathrm{1}\\ \mathrm{1}1\end{array}\right]=\left[\begin{array}{l}10\\ 0\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}1\mathrm{0}\\ 1\mathrm{1}\end{array}\right]=\left[\begin{array}{l}\mathrm{1}-1\\ 0 1\end{array}\right]\mathrm{A}\\ {\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\\ \therefore \mathrm{ }\left[\begin{array}{l}\mathrm{1}0\\ 01\end{array}\right]=\left[\begin{array}{l} \mathrm{1}-1\\ -1 \mathrm{ }2\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} \mathrm{1}-1\\ -1 \mathrm{ }2\end{array}\right]\end{array}$

Q.3

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}1â€„â€„â€„â€„3\\ 2â€„â€„â€„â€„7\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{1}\mathrm{3}\\ \mathrm{2}7\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}\mathrm{3}\\ 27\end{array}\right]=\left[\begin{array}{l}\mathrm{1}0\\ 0\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-2{\mathrm{R}}_{\mathrm{1}}\\ \mathrm{ }\left[\begin{array}{l}1\mathrm{3}\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} \mathrm{ }1\mathrm{}0\\ -21\end{array}\right]\mathrm{A}\\ {\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-3{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}10\\ 0\mathrm{1}\end{array}\right]=\left[\begin{array}{l} \mathrm{7}-3\\ -2 \mathrm{ }1\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} \mathrm{7}-3\\ -2 \mathrm{ }1\end{array}\right]\end{array}$

Q.4

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„â€„3\\ 5â€„â€„â€„â€„7\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{2}\mathrm{3}\\ \mathrm{5}7\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}2\mathrm{3}\\ 57\end{array}\right]=\left[\begin{array}{l}\mathrm{1}0\\ 0\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}↔{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}5\mathrm{7}\\ \mathrm{2}3\end{array}\right]=\left[\begin{array}{l}\mathrm{0}1\\ \mathrm{1}\mathrm{0}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-2{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}11\\ \mathrm{2}\mathrm{3}\end{array}\right]=\left[\begin{array}{l}-21\\ \mathrm{ }10\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-2{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}1\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l}-2 \mathrm{ }1\\ 5-2\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}0\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l}-7 \mathrm{ }3\\ 5-2\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l}-7 \mathrm{ }3\\ 5-2\end{array}\right]\end{array}$

Q.5

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„â€„1\\ 7â€„â€„â€„â€„4\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}2\mathrm{1}\\ 74\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{2}\mathrm{1}\\ 74\end{array}\right]=\left[\begin{array}{l}\mathrm{1}0\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-3{\mathrm{R}}_{\mathrm{1}}\\ \therefore \left[\begin{array}{l}\mathrm{2}1\\ 1\mathrm{1}\end{array}\right]=\left[\begin{array}{l} \mathrm{ }10\\ -31\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}10\\ \mathrm{1}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 4-1\\ -3 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}0\\ 0\mathrm{1}\end{array}\right]=\left[\begin{array}{l} \mathrm{ }4-1\\ -7 \mathrm{ }2\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} \mathrm{ }4-1\\ -7 2\end{array}\right]\end{array}$

Q.6

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„â€„5\\ 1â€„â€„â€„â€„3\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}2\mathrm{5}\\ \mathrm{1}3\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}2\mathrm{5}\\ \mathrm{1}3\end{array}\right]=\left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}12\\ \mathrm{1}\mathrm{3}\end{array}\right]=\left[\begin{array}{l}1\mathrm{}-1\\ 0 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}2\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 1-1\\ -1 \mathrm{ }2\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-2{\mathrm{R}}_{\mathrm{2}}\\ \mathrm{ }\left[\begin{array}{l}10\\ 0\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 3-5\\ -1 \mathrm{ }2\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} 3-5\\ -1 \mathrm{ }2\end{array}\right]\end{array}$

Q.7

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}3â€„â€„â€„â€„1\\ 5â€„â€„â€„â€„2\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}3\mathrm{1}\\ \mathrm{5}2\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}3\mathrm{1}\\ \mathrm{5}2\end{array}\right]=\left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to 2{\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}\mathrm{1}0\\ \mathrm{5}\mathrm{2}\end{array}\right]=\left[\begin{array}{l}2-1\\ 0 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-5\mathrm{ }{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}10\\ 0\mathrm{2}\end{array}\right]=\left[\begin{array}{l} 2-1\\ -10 \mathrm{ }6\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to \frac{1}{2}{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 2-1\\ -5 \mathrm{ }3\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} \mathrm{ }2-1\\ -5 \mathrm{ }3\end{array}\right]\end{array}$

Q.8

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}4â€„â€„â€„â€„5\\ 3â€„â€„â€„â€„4\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}4\mathrm{5}\\ 34\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}4\mathrm{5}\\ \mathrm{3}4\end{array}\right]=\left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}\mathrm{1}1\\ \mathrm{3}\mathrm{4}\end{array}\right]=\left[\begin{array}{l}1-1\\ 0 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-3\mathrm{ }{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}11\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 1-1\\ -3 \mathrm{ }4\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}10\\ 0\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 4-5\\ -3 \mathrm{ }4\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} 4-5\\ -3 \mathrm{ }4\end{array}\right]\end{array}$

Q.9

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\\ \mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}3â€„â€„â€„10\\ 2â€„â€„â€„â€„7\end{array}\right]\end{array}$

Ans

$\begin{array}{c}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{3}\mathrm{10}\\ 2 7\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{3}\mathrm{10}\\ 2 \mathrm{ }7\end{array}\right]=\left[\begin{array}{l}\mathrm{1}0\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \mathrm{S}0, \left[\begin{array}{l}13\\ 27\end{array}\right]=\left[\begin{array}{l}1-1\\ 0 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{\mathrm{2}}-2\mathrm{ }{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}3\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 1-1\\ -2 \mathrm{ }3\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-3{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}0\\ \mathrm{0}\mathrm{1}\end{array}\right]=\left[\begin{array}{l} 7-10\\ -2 \mathrm{ }3\end{array}\right]\mathrm{A}\\ {\mathrm{A}}^{-1}=\left[\begin{array}{l} 7-10\\ -2 \mathrm{ }3\end{array}\right]\end{array}$

Q.10

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}3â€„â€„â€„-1\\ –4â€„â€„â€„2\end{array}\right]\end{array}$

Ans

$\begin{array}{c}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l} 3-\mathrm{1}\\ -4 \mathrm{ }2\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l} 3-\mathrm{1}\\ -4 \mathrm{ }2\end{array}\right]=\left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}↔{\mathrm{R}}_{\mathrm{2}}\\ \therefore \left[\begin{array}{l}-4 \mathrm{ }2\\ \mathrm{3}-\mathrm{1}\end{array}\right]=\left[\begin{array}{l}01\\ 10\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to \left(-1\right){\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}4 -2\\ 3 -\mathrm{1}\end{array}\right]=\left[\begin{array}{l}0-1\\ 1 \mathrm{ }0\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ \mathrm{3}-\mathrm{1}\end{array}\right]=\left[\begin{array}{l}-\mathrm{1}-1\\ \mathrm{ }1 \mathrm{ }0\end{array}\right]\mathrm{A}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-3{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 0 2\end{array}\right]=\left[\begin{array}{l}-\mathrm{1}-1\\ \mathrm{ }4 \mathrm{ }3\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to \frac{1}{2}{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{1}\\ 0 1\end{array}\right]=\left[\begin{array}{l}-\mathrm{1}-1\\ \mathrm{ }2 \mathrm{ }\frac{3}{2}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}10\\ 01\end{array}\right]=\left[\begin{array}{l}\mathrm{1}\frac{1}{2}\\ 2\frac{3}{2}\end{array}\right]\mathrm{A}⇒ {\mathrm{A}}^{-1}=\left[\begin{array}{l}\mathrm{1}\frac{1}{2}\\ 2\frac{3}{2}\end{array}\right]\end{array}$

Q.11

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„-6\\ 1â€„â€„â€„â€„-2\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{2}-\mathrm{6}\\ 1-2\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{2}-\mathrm{6}\\ 1-2\end{array}\right]=\left[\begin{array}{l}10\\ \mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}-4\\ \mathrm{1}-\mathrm{2}\end{array}\right]=\left[\begin{array}{l}1-1\\ 0 \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{\mathrm{1}}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{4}\\ 0 \mathrm{2}\end{array}\right]=\left[\begin{array}{l} 1-1\\ -1 \mathrm{ }2\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to \frac{1}{2}{\mathrm{R}}_{2}\\ \left[\begin{array}{l}\mathrm{1}-\mathrm{4}\\ 0 \mathrm{1}\end{array}\right]=\left[\begin{array}{l} 1-1\\ -\frac{1}{2} \mathrm{ }1\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{\mathrm{1}}+4{\mathrm{R}}_{\mathrm{2}}\\ \left[\begin{array}{l}\mathrm{1}0\\ 0\mathrm{1}\end{array}\right]=\left[\begin{array}{l}-1 3\\ -\frac{1}{2} \mathrm{ }1\end{array}\right]\mathrm{A}\\ \therefore {\mathrm{A}}^{-1}=\left[\begin{array}{l}-1 3\\ -\frac{1}{2} \mathrm{ }1\end{array}\right]\end{array}$

Q.12

$\begin{array}{r}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\\ \mathrm{inverse}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}6â€„â€„â€„-3\\ –2â€„â€„â€„â€„1\end{array}\right]\end{array}$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{6}-\text{3}\\ -2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right]\\ Since,\text{â€‹}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}=\text{IA}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{6}-\text{3}\\ -2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right]=\left[\begin{array}{l}\text{1}0\\ \text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{}{R}_{1}↔{\text{R}}_{\text{2}}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}-\text{2}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}6-\text{3}\end{array}\right]=\left[\begin{array}{l}\text{0}\text{}1\\ \text{1}\text{0}\end{array}\right]A\\ Apply\text{}{R}_{2}\to {R}_{2}+3{\text{R}}_{\text{1}}\\ \left[\begin{array}{l}\text{1}-\text{4}\\ \text{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\end{array}\right]=\left[\begin{array}{l}\text{0}1\\ 1\text{}3\end{array}\right]A\\ \text{In second row of L}\text{.H}\text{.S}\text{., all elements are zero}\text{.}\\ \therefore {\text{A}}^{-\text{1}}\text{does not exist}\text{.}\end{array}$

Q.13

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}2-3\\ –12\end{array}\right]\end{array}$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}-\text{3}\\ -1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}\right]\\ Since,\text{â€‹}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}=\text{IA}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}-\text{3}\\ -1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}0\\ \text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{}{R}_{1}\to {R}_{1}+{\text{R}}_{\text{2}}\\ \left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}-\text{1}\\ -1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}1\\ 0\text{}1\end{array}\right]A\\ Apply\text{}{R}_{2}\to {R}_{2}+{\text{R}}_{\text{1}}\\ \left[\begin{array}{l}\text{1}\text{}-\text{1}\\ 0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}1\\ 1\text{}2\end{array}\right]A\\ Apply\text{}{R}_{1}\to {R}_{1}+{\text{R}}_{\text{2}}\\ \left[\begin{array}{l}\text{1}\text{}0\\ 0\text{}1\end{array}\right]=\left[\begin{array}{l}\text{2}\text{}3\\ 1\text{}2\end{array}\right]A\\ \therefore \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\left[\begin{array}{l}\text{2}\text{}3\\ 1\text{}2\end{array}\right]\end{array}$

Q.14

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\\ \mathrm{inverse}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}21\\ 42\end{array}\right]\end{array}$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{2}\text{}\text{1}\\ 4\text{}2\end{array}\right]\\ Since,\text{â€‹}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}=\text{IA}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{2}\text{}\text{1}\\ 4\text{}2\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}0\\ \text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{}{R}_{2}\to {R}_{2}-2{\text{R}}_{\text{1}}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{2}\text{}\text{1}\\ 00\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}0\\ -2\text{}1\end{array}\right]A\\ Since,\text{in second row of L}\text{.H}\text{.S}\text{., all elements are zero}\text{.}\\ {\text{So, A}}^{\text{-1}}\text{does not exist}\text{.}\end{array}$

Q.15

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l}\mathrm{2}-\mathrm{3}\mathrm{3}\\ 2 \mathrm{2}\mathrm{3}\\ \mathrm{3}-\mathrm{2}\mathrm{2}\end{array}\right]\\ \mathrm{Since},â€‹ \mathrm{ }\mathrm{A}=\mathrm{IA}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{2}-\mathrm{3}\mathrm{3}\\ 2 2\mathrm{3}\\ \mathrm{3}-2\mathrm{2}\end{array}\right]=\left[\begin{array}{l}\mathrm{1}00\\ 01\mathrm{0}\\ \mathrm{0}\mathrm{0}\mathrm{1}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}↔{\mathrm{R}}_{3}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{3}-\mathrm{2}\mathrm{2}\\ 2 \mathrm{2}\mathrm{3}\\ \mathrm{2}-\mathrm{3}\mathrm{3}\end{array}\right]=\left[\begin{array}{l}\mathrm{0}01\\ 0\mathrm{1}\mathrm{0}\\ \mathrm{1}0\mathrm{0}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{2}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{4}-\mathrm{1}\\ 2 2 \mathrm{3}\\ \mathrm{2}-3 \mathrm{3}\end{array}\right]=\left[\begin{array}{l}\mathrm{0}-11\\ 0 1\mathrm{0}\\ 1 \mathrm{0}\mathrm{0}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-2{\mathrm{R}}_{1}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-2{\mathrm{R}}_{1}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{4}-\mathrm{1}\\ 0 10 \mathrm{5}\\ 0 5 \mathrm{5}\end{array}\right]=\left[\begin{array}{l}\mathrm{0}-1 1\\ 0 \mathrm{3}-\mathrm{2}\\ 1 \mathrm{2}-\mathrm{2}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{3}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{4}-\mathrm{1}\\ 0 5 \mathrm{0}\\ 0 5 \mathrm{5}\end{array}\right]=\left[\begin{array}{l} 0-1 1\\ -1 1 0\\ 1 2-\mathrm{2}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{2}\to \frac{1}{5}{\mathrm{R}}_{2}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{3}\to \frac{1}{5}{\mathrm{R}}_{3}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{4}-\mathrm{1}\\ 0 1 \mathrm{0}\\ 0 1 \mathrm{1}\end{array}\right]=\left[\begin{array}{l} 0-1 1\\ -\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }0\\ \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{2}}{5}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{2}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{1}-\mathrm{4}-\mathrm{1}\\ 0 1 \mathrm{0}\\ 0 0 \mathrm{1}\end{array}\right]=\left[\begin{array}{l} 0-1 1\\ -\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }0\\ \mathrm{ }\frac{\mathrm{2}}{5} \mathrm{ }\frac{\mathrm{1}}{5}-\frac{\mathrm{2}}{5}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+4{\mathrm{R}}_{2}\\ \mathrm{ }\left[\begin{array}{l}10-\mathrm{1}\\ 0 10\\ 0 01\end{array}\right]=\left[\begin{array}{l}-\frac{\mathrm{4}}{5}-\frac{\mathrm{1}}{5} \mathrm{ }1\\ -\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }0\\ \mathrm{ }\frac{\mathrm{2}}{5} \mathrm{ }\frac{\mathrm{1}}{5}\mathrm{ }-\frac{\mathrm{2}}{5}\end{array}\right]\mathrm{A}\\ \mathrm{Apply}\mathrm{}{\mathrm{R}}_{1}\to {\mathrm{R}}_{1}+{\mathrm{R}}_{3}\\ \mathrm{ }\left[\begin{array}{l}100\\ 0 10\\ 0 01\end{array}\right]=\left[\begin{array}{l}-\frac{\mathrm{2}}{5} \mathrm{ }0 \mathrm{ }\frac{\mathrm{3}}{5}\\ -\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }0\\ \mathrm{ }\frac{\mathrm{2}}{5} \mathrm{ }\frac{\mathrm{1}}{5}\mathrm{ }-\frac{\mathrm{2}}{5}\end{array}\right]\mathrm{A}\\ \therefore {\mathrm{A}}^{-1}=\left[\begin{array}{l}-\frac{\mathrm{2}}{5} \mathrm{ }0 \mathrm{ }\frac{\mathrm{3}}{5}\\ -\frac{\mathrm{1}}{5} \mathrm{ }\frac{\mathrm{1}}{5} \mathrm{ }0\\ \mathrm{ }\frac{\mathrm{2}}{5} \mathrm{ }\frac{\mathrm{1}}{5}\mathrm{ }-\frac{\mathrm{2}}{5}\end{array}\right]\end{array}$

Q.16

$\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\mathrm{of}\phantom{\rule{0ex}{0ex}}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l} 13-2\\ -30-5\\ 25 0\end{array}\right]$

Ans

$\begin{array}{c}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{3}\text{}-\text{2}\\ -\text{3}\text{}\text{0}\text{}-\text{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\end{array}\right]\\ Since,\text{â€‹}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}=\text{IA}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{3}\text{}-\text{2}\\ -\text{3}\text{}\text{0}\text{}-\text{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}0\text{}0\\ \text{0}\text{}\text{1}\text{}\text{0}\\ \text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}+3{R}_{1}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}{R}_{3}\to {R}_{3}-2{R}_{1},\text{\hspace{0.17em}}\text{we get}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\text{}-\text{2}\\ 0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{9}\text{}-\text{11}\\ \text{0}\text{}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{4}\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}0\text{}0\\ \text{3}\text{}\text{1}\text{}\text{0}\\ -\text{2}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}+8{R}_{3}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\text{}-\text{2}\\ 0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}21\\ \text{0}\text{}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{4}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}0\text{}0\\ -\text{13}\text{}\text{1}\text{}\text{8}\\ -\text{2}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ Apply\text{\hspace{0.17em}}{R}_{1}\to {R}_{1}-3{R}_{2}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{3}\to {R}_{3}+{R}_{2}\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{1}\text{}\text{0}\text{}-65\\ 0\text{}\text{1}\text{}21\\ \text{0}\text{}0\text{}\text{25}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{40}\text{}-\text{3}\text{}-\text{24}\\ -\text{13}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}\\ -\text{15}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{9}\end{array}\right]A\\ Apply\text{\hspace{0.17em}}{R}_{3}\to \frac{1}{25}{R}_{3}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{1}\text{}\text{0}\text{}-65\\ 0\text{}\text{1}\text{}21\\ \text{0}\text{}0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{40}\text{}-\text{3}\text{}-\text{24}\\ -\text{13}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}\\ -\frac{\text{15}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{1}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{9}}{25}\end{array}\right]A\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{25}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{40}\text{}-\text{3}\text{}-\text{24}\\ -\text{13}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{8}\\ -15\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9\end{array}\right]A\\ Apply\text{\hspace{0.17em}}{R}_{1}\to {R}_{1}+65{R}_{3}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}+21{R}_{3},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}\text{1}\text{}\text{0}\text{}0\\ 0\text{}\text{1}\text{}0\\ \text{0}\text{}0\text{}\text{\hspace{0.17em}}\text{1}\end{array}\right]=\frac{1}{25}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{25}\text{}-\text{10}\text{}-\text{15}\\ -\text{10}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{4}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{11}\\ -15\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9\end{array}\right]A\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}-\frac{\text{10}}{25}\text{}-\frac{\text{15}}{25}\\ -\frac{\text{10}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{4}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{11}}{25}\\ -\frac{15}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{9}{25}\end{array}\right]A\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}-\frac{\text{2}}{5}\text{}-\frac{\text{3}}{5}\\ -\frac{\text{2}}{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{4}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{11}}{25}\\ -\frac{3}{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{9}{25}\end{array}\right]A\\ Therefore,\text{\hspace{0.17em}}{A}^{-1}=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}-\frac{\text{2}}{5}\text{}-\frac{\text{3}}{5}\\ -\frac{\text{2}}{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{4}}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{\text{11}}{25}\\ -\frac{3}{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{1}{25}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{9}{25}\end{array}\right]\\ \end{array}$

Q.17

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}â€„\left[\begin{array}{l}2â€„â€„â€„â€„â€„0â€„-1\\ 5â€„â€„â€„â€„â€„1â€„â€„â€„â€„â€„â€„0\\ 0â€„â€„â€„â€„0â€„â€„â€„â€„â€„â€„3\end{array}\right]\end{array}$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}-\text{1}\\ 5\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \text{0}\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\right]\\ Since,\text{â€‹}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{A}=\text{IA}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}-\text{1}\\ 5\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \text{0}\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\right]=\left[\begin{array}{l}\text{1}\text{}0\text{}0\\ \text{0}\text{}\text{1}\text{}\text{0}\\ \text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ Apply\mathrm{in}g\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{1}↔{R}_{2},\text{\hspace{0.17em}}\text{we get}\\ \text{\hspace{0.17em}}\left[\begin{array}{l}5\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \text{2}\text{}\text{0}\text{}-\text{1}\\ \text{0}\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\right]=\left[\begin{array}{l}\text{0}\text{}\text{1}\text{}0\\ \text{1}\text{}\text{0}\text{}\text{0}\\ \text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}{R}_{1}\to {R}_{1}-2{R}_{2},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \text{\hspace{0.17em}}\left[\begin{array}{l}1\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ \text{2}\text{}\text{0}\text{}-\text{1}\\ \text{0}\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\right]=\left[\begin{array}{l}-\text{2}\text{}\text{1}\text{}0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{0}\text{}\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}-2{R}_{1},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \text{\hspace{0.17em}}\left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ \text{0}\text{}-\text{2}\text{}-\text{5}\\ \text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\right]=\left[\begin{array}{l}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{}-\text{2}\text{}\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to \left(-1\right){R}_{2},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{1}\text{}2\\ \text{0}\text{}\text{2}\text{}\text{5}\\ \text{0}\text{}\text{1}\text{}\text{3}\end{array}\right]=\left[\begin{array}{l}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}0\\ -\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}\text{0}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}-{R}_{3},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{1}\text{}2\\ \text{0}\text{}\text{1}\text{}\text{2}\\ \text{0}\text{}\text{1}\text{}\text{3}\end{array}\right]=\left[\begin{array}{l}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ -\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}-\text{1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{1}\to {R}_{1}-{R}_{2},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{0}\text{}0\\ \text{0}\text{}\text{1}\text{}\text{2}\\ \text{0}\text{}\text{1}\text{}\text{3}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ -\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}-\text{1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{1}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{3}\to {R}_{3}-{R}_{2},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{0}\text{}0\\ \text{0}\text{}\text{1}\text{}\text{2}\\ \text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ -\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\text{}-\text{1}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{}\text{\hspace{0.17em}}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\end{array}\right]A\\ \text{Applying}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{R}_{2}\to {R}_{2}-2{R}_{3},\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\\ \left[\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{0}\text{}0\\ \text{0}\text{}\text{1}\text{}\text{0}\\ \text{0}\text{}\text{0}\text{}\text{1}\end{array}\right]=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ -1\text{5}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{6}\text{}-\text{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{}\text{\hspace{0.17em}}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\end{array}\right]A⇒\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ -\text{15}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{6}\text{}-\text{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{5}\text{}\text{\hspace{0.17em}}-\text{2}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\end{array}\right]\end{array}$

Q.18 Matrices A and B will be inverse of each other only if

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans

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

Thus, option (D) is correct.

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### 1. Is it important to practice all the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4?

Yes, for building strong concepts and scoring good marks, it is very important to practice all the questions of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4.

### 2. Is the NCERT book enough for the students to prepare for Exercise 3.4 of Class 12?

The NCERT book in itself is enough for students to strengthen their basics, but NCERT does not contain the solutions to the problems. Extramarks provides students with NCERT solutions for Class 12 Maths Chapter 3 Exercise 3.4 for an in-depth understanding of the topic and conceptual clarity. It is significant for students who have to appear for board examinations to practice extra questions and sample papers, which they can find on Extramarks.

### 3. How can students clear their doubts about NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4?

Students can subscribe to the Extramarks’ website, where they have the access to live doubt-solving sessions. Students can also record their lectures, which will help them in revising those topics further.

### 4. Are the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 available on Extramarks?

Extramarks provides students with NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4. Along with this, it provides students with K12 study material for their boards and live classes with the experts to provide detailed solutions for all the questions for the students.

### 5. How many books of Mathematics are there in the course of Class 12?

There are 2 Volumes of Mathematics NCERT Textbook in the course of Class 12. But the students should not stress about it. If they have a proper timetable, they will be easily able to cover the syllabus and score good marks in their board examinations.

### 6. What is the topic weightage of chapter 3 in the board examination of Mathematics?

Students can download the Extramarks’ application and know about the topic weightage and marks distribution of chapter 3 in Mathematics Class 12. Along with this, students would also get sample papers and past years’ question papers, and study material like NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4. This helps students practice according to the chapter weightage and the blueprint of the question paper.

### 7. Give a preview of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4?

The  NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4 is about invertible matrices and contains 18 questions. The concept and calculations of all these 18 questions are the same in some way or the other.

### 8. Is NCERT Exemplar book needed for the preparation of NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4?

Mathematics is a subject which requires a lot of practice, practising the examples given in the NCERT book and the NCERT Exemplar will definitely help them in building the concepts of Class 12 Maths Chapter 3 Exercise 3.4. The more students will practice, the better they will get.

Click on the given link to download the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.4