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

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

Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 3

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

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

Some of the essential topics for students include

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

Further to this, the vital subtopics include

Properties of matrix operation
The transpose of a matrix.

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

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

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

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

Key topics covered in Chapter 3 Matrices include:

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

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

3.1 Introduction

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

3.2 Matrix

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

3.3 Types of Matrices

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

3.4 Operations on Matrices

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

3.5 Transpose of a Matrix

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

3.6 Symmetric and Skew Symmetric Matrices

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

3.7 Elementary Operation of a Matrix

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

3.8 Invertible Matrices

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

NCERT Solutions Class 12 Mathematics Chapter 3 Exercise & Answer Solutions

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

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

Chapter 3: Exercise 3.1 Solutions: 10 Questions (5 Short Questions, 5 Long Questions)
Chapter 3: Exercise 3.2 Solutions: 22 Questions (3 Short Questions, 19 Long Questions)
Chapter 3: Exercise 3.3 Solutions: 22 Questions (3 Short Questions, 19 Long Questions)
Chapter 3: Exercise 3.4 Solutions: 18 Questions (18 Short Questions)

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

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

NCERT Exemplar Class 12 Mathematics

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

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

Key Features of NCERT Solutions Class 12 Mathematics Chapter 3

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

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

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

Q.1 In a matrix A=[2 5 19 −735−2 5 2123 1−5 17], write:(i) The order of the matrix(ii) Then umber of elements,(iii) Write the elements a13,a21,a33,a24,a23.

Ans.

(i) The order of matrix A is 3×4.

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

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

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

Ans.

Possible order of matrix

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

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

Ans.

Possible order for 18 elements

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

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

Q.4

\begin{array}{l} Constructa 2 \times 2 matrix, A = [a_{ij}], whose elementsare \\ givenby : (i) a_{ij} = \frac{{(i + j)}^{2}}{2} (ii) a_{ij} = \frac{i}{j} (iii) a_{ij} = \frac{{(i + 2 j)}^{2}}{2} \end{array}

Ans.

$\begin{array}{l} A = [a_{ij}] \\ (i) a_{ij} = \frac{{(i + j)}^{2}}{2} \\ A = (\begin{array}{l} a_{11} a_{12} \\ a_{21} a_{22} \end{array}) = (\begin{array}{l} 2 \frac{9}{2} \\ \frac{9}{2} 8 \end{array}) \\ (ii) a_{ij} = \frac{i}{j} \\ A = (\begin{array}{l} a_{11} a_{12} \\ a_{21} a_{22} \end{array}) = (\begin{array}{l} 1 \frac{1}{2} \\ 21 \end{array}) \\ (iii) a_{ij} = \frac{{(i + 2 j)}^{2}}{2} \\ A = (\begin{array}{l} a_{11} a_{12} \\ a_{21} a_{22} \end{array}) = (\begin{array}{l} \frac{9}{2} \frac{25}{2} \\ 818 \end{array}) \end{array}$

Q.5

\begin{array}{l} Construct a 3 \times 4 matrix, whose elements are given by : \\ (i) a_{ij} = \frac{1}{2} | - 3 i + j | (ii) a_{ij} = 2 i - j \end{array}

Ans.

$\begin{array}{l} (i) aij = \frac{1}{2} | - 3 i + j | \\ A = [\begin{array}{l} a_{11} a_{12} a_{13} a_{14} \\ a_{21} a_{22} a_{23} a_{24} \\ a_{31} a_{32} a_{33} a_{34} \end{array}] \\ A = [\begin{array}{l} 1 \frac{1}{2} 0 \frac{1}{2} \\ \frac{5}{2} 2 \frac{3}{2} 1 \\ 4 \frac{7}{2} 3 \frac{5}{2} \end{array}] \\ (ii) a_{ij} = 2 i - j \\ A = [\begin{array}{l} a_{11} a_{12} a_{13} a_{14} \\ a_{21} a_{22} a_{23} a_{24} \\ a_{31} a_{32} a_{33} a_{34} \end{array}] \\ = [\begin{array}{l} 10 - 1 - 2 \\ 32 1 0 \\ 54 3 2 \end{array}] \end{array}$

Q.6

\begin{array}{l} Find the values of x, y and z from the following equations : \\ (i) [\begin{array}{l} 4 3 \\ x 5 \end{array}] = [\begin{array}{l} y z \\ 1 5 \end{array}] \\ (ii) [\begin{array}{l} x + y 3 \\ 5 + z xy \end{array}] = [\begin{array}{l} 62 \\ 58 \end{array}] \\ (iii) [\begin{array}{l} x + y + z \\ x + z \\ y + z \end{array}] = [\begin{array}{l} 9 \\ 5 \\ 7 \end{array}] \end{array}

Ans.

$\begin{array}{l} (i) [\begin{array}{l} 4 3 \\ x 5 \end{array}] = [\begin{array}{l} y z \\ 1 5 \end{array}] \\ \Rightarrow x = 1, y = 4 and z = 3 \\ (ii) [\begin{array}{l} x + y 3 \\ 5 + z xy \end{array}] = [\begin{array}{l} 6 2 \\ 5 8 \end{array}] \\ \Rightarrow 5 + z = 5 \Rightarrow z = 5 - 5 = 0 \\ x + y = 6, and xy =8 \Rightarrow y = \frac{8}{x} \\ \Rightarrow x + y = 6 \\ x + \frac{8}{x} = 6 \\ \frac{x^{2} + 8}{x} = 6 \\ \Rightarrow x^{2} + 8 = 6 x \\ x^{2} - 6 x + 8 = 0 \Rightarrow x^{2} - 4 x - 2 x + 8 = 0 \\ \Rightarrow (x - 4) (x - 2) = 0 \\ \Rightarrow x = 2, 4 \\ If x = 2 then y = \frac{8}{2} = 4 \\ If x = 4 then y = \frac{8}{4} = 2 \\ (iii) [\begin{array}{l} x + y + z \\ x + z \\ y + z \end{array}] = [\begin{array}{l} 9 \\ 5 \\ 7 \end{array}] \\ x + y + z = 9 . .. (i) \\ x + z = 5 . .. (ii) \\ y + z = 7 . .. (iii) \\ From equation (i) and equation (ii), we have \\ y + 5 = 9 \Rightarrow y = 4 \\ From equation (i) and equation (iii), we have \\ x + 7 = 9 \Rightarrow x = 2 \\ From equation (i), we have \\ 2 + 4 + z = 9 \Rightarrow z = 9 - 6 = 3 \\ Thus, x = 2, y = 4 and z = 3 . \end{array}$

Q.7

\begin{array}{l} Find the value of a, b, c and d from thee quation : \\ [\begin{array}{l} a - b 2 a + c \\ 2 a - b 3 c + d \end{array}] = [\begin{array}{l} - 1 5 \\ 0 13 \end{array}] \end{array}

Ans.

$\begin{array}{l} [\begin{array}{l} a - b 2 a + c \\ 2 a - b 3 c + d \end{array}] = [\begin{array}{l} - 1 5 \\ 0 13 \end{array}] \\ \Rightarrow a - b = - 1 . .. (i) \\ 2 a + c = 5 . .. (ii) \\ 2 a - b = 0 . .. (iii) \\ 3 c + d = 13 . .. (iv) \\ From equation (i) and equation (iii), we have \\ a = 1 and b = 2 \\ Putting value of a in equation (ii), we get \\ 2 (1) + c = 5 \Rightarrow c = 5 - 2 = 3 \\ Putting value of c in equation (iv), we get \\ 3 (3) + d = 13 \Rightarrow d = 13 - 9 = 4 \end{array}$

Q.8

\begin{array}{l} A = {[a_{ij}]}_{m \times n} is a square matrix, if \\ (A) m < n \\ (B) m > n \\ (C) m = n (D) None of these \end{array}

Ans.

A is a square matrix if m=n.

So, option (C) is correction answer.

Q.9

\begin{array}{l} Which of the given values of x and y make the following \\ pair of matrice sequal \\ [\begin{array}{l} 3 x + 7 5 \\ y + 1 2 - 3 x \end{array}], [\begin{array}{l} 0 y - 2 \\ 8 4 \end{array}] \\ (A) x = - \frac{1}{3}, y = 7 (B) Not possiblet of ind \\ (C) y = 7, x = \frac{- 2}{3} (D) x = - \frac{1}{3}, y = \frac{- 2}{3} \end{array}

Ans.

$\begin{array}{l} If both matrices are equal, then \\ [\begin{array}{l} 3 x +7 5 \\ y +1 2 -3 x \end{array}] = [\begin{array}{l} 0 y -2 \\ 8 4 \end{array}] \\ \Rightarrow 3 x + 7 = 0, y - 2 = 5 \\ and y + 1 = 8, 2 - 3 x = 4 \\ Then, x = - \frac{7}{3}, y = 7 and x = - \frac{2}{3} \\ Since, there are two different values of x, so it is difficult \\ to find the value of x to make matrices equal . \end{array}$

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

(A) 27

(B) 18

(D) 512

Ans.

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

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

So, the option ‘D’ is correct.

Q.11

\begin{array}{l} In the matrix A = [\begin{array}{l} 2 4 \\ 3 2 \end{array}], B = [\begin{array}{l} 1 3 \\ - 2 5 \end{array}], C = [\begin{array}{l} - 2 5 \\ 3 4 \end{array}] \\ Find each of the following : \\ (i) A + B \\ (ii) A - B \\ (iii) 3 A - C \\ (iv) AB (v) BA \end{array}

Ans.

$\begin{array}{l} In the matrix A = [\begin{array}{l} 2 4 \\ 3 2 \end{array}], B = [\begin{array}{l} 1 3 \\ - 2 5 \end{array}], C = [\begin{array}{l} - 2 5 \\ 3 4 \end{array}] \\ (i) A + B = [\begin{array}{l} 2 4 \\ 3 2 \end{array}] + [\begin{array}{l} 1 3 \\ - 2 5 \end{array}] \\ = [\begin{array}{l} 2 + 1 4 + 3 \\ 3 + (- 2) 2 + 5 \end{array}] \\ = [\begin{array}{l} 3 7 \\ 1 7 \end{array}] \\ (ii) A - B = [\begin{array}{l} 2 4 \\ 3 2 \end{array}] - [\begin{array}{l} 1 3 \\ - 2 5 \end{array}] \\ = [\begin{array}{l} 2 - 1 4 - 3 \\ 3 - (- 2) 2 - 5 \end{array}] \\ = [\begin{array}{l} 1 1 \\ 5 - 3 \end{array}] \\ (iii) 3 A - C = 3 [\begin{array}{l} 2 4 \\ 3 2 \end{array}] - [\begin{array}{l} - 2 5 \\ 3 4 \end{array}] \\ = [\begin{array}{l} 6 12 \\ 9 6 \end{array}] - [\begin{array}{l} - 2 5 \\ 3 4 \end{array}] \\ = [\begin{array}{l} 6 - (- 2) 12 - 5 \\ 9 - 3 6 - 4 \end{array}] \\ = [\begin{array}{l} 8 7 \\ 6 2 \end{array}] \\ (iv) AB = [\begin{array}{l} 2 4 \\ 3 2 \end{array}] [\begin{array}{l} 1 3 \\ - 2 5 \end{array}] \\ = [\begin{array}{l} 2 - 8 6 + 20 \\ 3 - 4 9 + 10 \end{array}] \\ = [\begin{array}{l} - 6 26 \\ - 1 19 \end{array}] \\ (v) BA = [\begin{array}{l} 1 3 \\ - 2 5 \end{array}] [\begin{array}{l} 2 4 \\ 3 2 \end{array}] \\ = [\begin{array}{l} 2 + 9 4 + 6 \\ - 4 + 15 - 8 + 10 \end{array}] \\ = [\begin{array}{l} 11 10 \\ 11 2 \end{array}] \end{array}$

Q.12

$\begin{array}{l} Compute the following : \\ (i) [\begin{array}{l} a b \\ - b a \end{array}] + [\begin{array}{l} a b \\ b a \end{array}] \\ (ii) [\begin{array}{l} a^{2} + b^{2} b^{2} + c^{2} \\ a^{2} + c^{2} a^{2} + b^{2} \end{array}] \\ (iii) [\begin{array}{l} - 1 4 - 6 \\ 8 5 16 \\ 2 8 5 \end{array}] + [\begin{array}{l} 12 7 6 \\ 8 0 5 \\ 3 2 4 \end{array}] \\ (iv) [\begin{array}{l} \cos^{2} x \sin^{2} x \\ \sin^{2} x \cos^{2} x \end{array}] + [\begin{array}{l} \sin^{2} x \cos^{2} x \\ \cos^{2} x \sin^{2} x \end{array}] \end{array}$

Ans.

$\begin{array}{l} (i) [\begin{array}{l} a b \\ - b a \end{array}] + [\begin{array}{l} a b \\ b a \end{array}] = [\begin{array}{l} a + a b + b \\ - b + b a + a \end{array}] \\ = [\begin{array}{l} 2 a 2 b \\ 0 2 a \end{array}] \\ (ii) [\begin{array}{l} a^{2} + b^{2} b^{2} + c^{2} \\ a^{2} + c^{2} a^{2} + b^{2} \end{array}] + [\begin{array}{l} 2 ab 2 bc \\ - 2 ac - 2 ab \end{array}] \\ = [\begin{array}{l} a^{2} + b^{2} + 2 ab b^{2} + c^{2} + 2 bc \\ a^{2} + c^{2} - 2 ac a^{2} + b^{2} - 2 ab \end{array}] \\ = [\begin{array}{l} {(a + b)}^{2} {(b + c)}^{2} \\ {(a - c)}^{2} {(a - b)}^{2} \end{array}] \\ (iii) [\begin{array}{l} - 1 4 - 6 \\ 8 5 16 \\ 2 8 5 \end{array}] + [\begin{array}{l} 12 7 6 \\ 8 0 5 \\ 3 2 4 \end{array}] \\ = [\begin{array}{l} - 1 +12 4 + 7 - 6 + 6 \\ 8 + 8 5 + 0 16 + 5 \\ 2 + 3 8 + 2 5 + 4 \end{array}] \\ = [\begin{array}{l} 11 11 0 \\ 16 5 21 \\ 5 10 9 \end{array}] \\ (iv) [\begin{array}{l} \cos^{2} x \sin^{2} x \\ \sin^{2} x \cos^{2} x \end{array}] + [\begin{array}{l} \sin^{2} x \cos^{2} x \\ \cos^{2} x \sin^{2} x \end{array}] \\ = [\begin{array}{l} \cos^{2} x + \sin^{2} x \sin^{2} x + \cos^{2} x \\ \sin^{2} x + \cos^{2} x \cos^{2} x + \sin^{2} x \end{array}] \\ = [\begin{array}{l} 1 1 \\ 1 1 \end{array}] \end{array}$

Q.13

$\begin{array}{l} Compute the indicated products : \\ (i) [\begin{array}{l} a b \\ - b a \end{array}] [\begin{array}{l} a - b \\ b a \end{array}] (ii) [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] [234] \\ (iii) [\begin{array}{l} 1 - 2 \\ 2 3 \end{array}] [\begin{array}{l} 1 2 3 \\ 2 3 1 \end{array}] \\ (iv) [\begin{array}{l} 2 3 4 \\ 3 4 5 \\ 4 5 6 \end{array}] [\begin{array}{l} 1 - 3 5 \\ 0 2 4 \\ 3 0 5 \end{array}] \end{array}$ $\begin{array}{l} (v) [\begin{array}{l} 2 1 \\ 3 2 \\ - 1 1 \end{array}] [\begin{array}{l} 1 0 1 \\ - 1 2 1 \end{array}] \\ (v i) [\begin{array}{l} 3 - 1 3 \\ - 1 0 2 \end{array}] [\begin{array}{l} 2 - 3 \\ 1 0 \\ 3 1 \end{array}] \end{array}$

Ans.

$\begin{array}{l} (i) [\begin{array}{l} ab \\ - ba \end{array}] [\begin{array}{l} a - b \\ b a \end{array}] = [\begin{array}{l} a^{2} + b^{2} - ab + ab \\ - ba + ab b^{2} + a^{2} \end{array}] \\ = [\begin{array}{l} a^{2} + b^{2} - ab + ab \\ - ba + ab b^{2} + a^{2} \end{array}] \\ (ii) [\begin{array}{l} 1 \\ 2 \\ 3 \end{array}] [234] = [\begin{array}{l} 2 3 4 \\ 4 6 8 \\ 6 9 12 \end{array}] \\ (iii) [\begin{array}{l} 1 - 2 \\ 2 3 \end{array}] [\begin{array}{l} 1 2 3 \\ 2 3 1 \end{array}] = [\begin{array}{l} 1 - 42 - 63 - 2 \\ 2 + 64 + 96 + 3 \end{array}] \\ = [\begin{array}{l} - 3 - 41 \\ 8 139 \end{array}] \\ (iv) [\begin{array}{l} 2 3 4 \\ 3 4 5 \\ 4 5 6 \end{array}] [\begin{array}{l} 1 - 3 5 \\ 0 2 4 \\ 3 0 5 \end{array}] \\ = [\begin{array}{l} 2 + 0 + 12 - 6 + 6 + 010 + 12 + 20 \\ 3 + 0 + 15 - 9 + 8 + 015 + 16 + 25 \\ 4 + 0 + 18 - 12 + 10 + 020 + 20 + 30 \end{array}] \\ = [\begin{array}{l} 14042 \\ 18 - 156 \\ 22 - 270 \end{array}] \\ (v) [\begin{array}{l} 21 \\ 32 \\ - 11 \end{array}] [\begin{array}{l} 101 \\ - 121 \end{array}] = [\begin{array}{l} 2 - 10 + 22 + 1 \\ 3 - 20 + 43 + 2 \\ - 1 - 10 + 2 - 1 + 1 \end{array}] \\ = [\begin{array}{l} 123 \\ 145 \\ - 220 \end{array}] \\ (vi) [\begin{array}{l} 3 - 13 \\ - 1 02 \end{array}] [\begin{array}{l} 2 - 3 \\ 1 0 \\ 3 1 \end{array}] = [\begin{array}{l} 6 - 1 + 9 - 9 + 0 + 3 \\ - 2 + 0 + 6 3 + 0 + 2 \end{array}] \\ = [\begin{array}{l} 14 - 6 \\ 45 \end{array}] \end{array}$

Q.14

$\begin{array}{l} If A = [\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 1 1 \end{array}], B = [\begin{array}{l} 3 - 12 \\ 4 25 \\ 2 03 \end{array}] and C = [\begin{array}{l} 4 12 \\ 0 32 \\ 1 - 23 \end{array}], \\ then compute (A + B) and (B - C) . Also, verify that \\ A + (B - C) = (A + B) - C . \end{array}$

Ans.

$\begin{array}{l} A + B = [\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 11 \end{array}] + [\begin{array}{l} 3 - 12 \\ 4 25 \\ 2 03 \end{array}] \\ = [\begin{array}{l} 1 +32+ (- 1) - 3 + 2 \\ 5 + 40 + 22 + 5 \\ 1 + 2 - 1 + 0 1 + 3 \end{array}] \\ = [\begin{array}{l} 4 1 - 1 \\ 9 2 7 \\ 3 - 1 4 \end{array}] \\ B - C = [\begin{array}{l} 3 - 12 \\ 4 25 \\ 2 03 \end{array}] - [\begin{array}{l} 412 \\ 032 \\ 1 - 23 \end{array}] \\ = [\begin{array}{l} 3 - 4 - 1 - 12 - 2 \\ 4 - 0 2 - 35 - 2 \\ 2 - 1 0 - (- 2) 3 - 3 \end{array}] \\ = [\begin{array}{l} - 1 - 20 \\ 4 - 13 \\ 1 20 \end{array}] \\ A + (B - C) = [\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 1 1 \end{array}] + ([\begin{array}{l} 3 - 12 \\ 4 25 \\ 2 03 \end{array}] - [\begin{array}{l} 412 \\ 032 \\ 1 - 23 \end{array}]) \\ = [\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 1 1 \end{array}] + ([\begin{array}{l} 3 - 4 - 1 - 12 - 2 \\ 4 - 0 2 - 35 - 2 \\ 2 - 1 0 + 33 - 3 \end{array}]) \\ = [\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 1 1 \end{array}] + [\begin{array}{l} - 1 - 20 \\ 4 - 13 \\ 1 30 \end{array}] \\ = [\begin{array}{l} 1 - 12 - 2 - 3 + 0 \\ 5 + 40 - 1 2 + 3 \\ 1 + 1 - 1 + 3 1 + 0 \end{array}] \\ = [\begin{array}{l} 00 - 3 \\ 9 - 1 5 \\ 2 2 1 \end{array}] \\ (A + B) - C = ([\begin{array}{l} 12 - 3 \\ 50 2 \\ 1 - 1 1 \end{array}] + [\begin{array}{l} 3 - 12 \\ 4 25 \\ 2 03 \end{array}]) - [\begin{array}{l} 412 \\ 032 \\ 1 - 23 \end{array}] \\ = [\begin{array}{l} 1 + 32 - 1 - 3 + 2 \\ 5 + 40 + 22 + 5 \\ 1 + 2 - 1 + 0 1 + 3 \end{array}] - [\begin{array}{l} 412 \\ 032 \\ 1 - 23 \end{array}] \\ = [\begin{array}{l} 41 - 1 \\ 927 \\ 3 - 14 \end{array}] - [\begin{array}{l} 412 \\ 032 \\ 1 - 23 \end{array}] \\ = [\begin{array}{l} 4 - 41 - 1 - 1 - 2 \\ 9 - 0 2 - 3 7 - 2 \\ 3 - 1 - 1 + 2 4 - 3 \end{array}] \\ = [\begin{array}{l} 00 - 3 \\ 9 - 1 5 \\ 2 1 1 \end{array}] \\ Hence, A + (B - C) = (A + B) - C . \end{array}$

Q.15

$If A= [\begin{array}{l} \frac{2}{3} 1 \frac{5}{3} \\ \frac{1}{3} \frac{2}{3} \frac{4}{3} \\ \frac{7}{3} 2 \frac{2}{3} \end{array}] & B = [\begin{array}{l} \frac{2}{5} \frac{3}{5} 1 \\ \frac{1}{5} \frac{2}{5} \frac{4}{5} \\ \frac{7}{5} \frac{6}{5} \frac{2}{5} \end{array}], then compute 3 A - 5 B .$

Ans.

$\begin{array}{l} Given, A = [\begin{array}{l} \frac{2}{3} 1 \frac{5}{3} \\ \frac{1}{3} \frac{2}{3} \frac{4}{3} \\ \frac{7}{3} 2 \frac{2}{3} \end{array}] and B = [\begin{array}{l} \frac{2}{5} \frac{3}{5} 1 \\ \frac{1}{5} \frac{2}{5} \frac{4}{5} \\ \frac{7}{5} \frac{6}{5} \frac{2}{5} \end{array}] \\ 3 A - 5 B = 3 [\begin{array}{l} \frac{2}{3} 1 \frac{5}{3} \\ \frac{1}{3} \frac{2}{3} \frac{4}{3} \\ \frac{7}{3} 2 \frac{2}{3} \end{array}] - 5 [\begin{array}{l} \frac{2}{5} \frac{3}{5} 1 \\ \frac{1}{5} \frac{2}{5} \frac{4}{5} \\ \frac{7}{5} \frac{6}{5} \frac{2}{5} \end{array}] \\ = [\begin{array}{l} 3 \times \frac{2}{3} 3 \times 13 \times \frac{5}{3} \\ 3 \times \frac{1}{3} 3 \times \frac{2}{3} 3 \times \frac{4}{3} \\ 3 \times \frac{7}{3} 3 \times 23 \times \frac{2}{3} \end{array}] - [\begin{array}{l} 5 \times \frac{2}{5} 5 \times \frac{3}{5} 5 \times 1 \\ 5 \times \frac{1}{5} 5 \times \frac{2}{5} 5 \times \frac{4}{5} \\ 5 \times \frac{7}{5} 5 \times \frac{6}{5} 5 \times \frac{2}{5} \end{array}] \\ = [\begin{array}{l} 2 3 5 \\ 1 2 4 \\ 7 6 2 \end{array}] - [\begin{array}{l} 2 3 5 \\ 1 2 4 \\ 7 6 2 \end{array}] \\ = [\begin{array}{l} 2 - 23 - 35 - 5 \\ 1 - 12 - 24 - 4 \\ 7 - 76 - 62 - 2 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \end{array}$

Q.16

$Simplify cosθ [\begin{array}{l} cosθ sinθ \\ - sinθ cosθ \end{array}] + sinθ [\begin{array}{l} sinθ - cosθ \\ cosθ sinθ \end{array}] .$

Ans.

$\begin{array}{l} cosθ [\begin{array}{l} cosθ sinθ \\ - sinθ cosθ \end{array}] + sinθ [\begin{array}{l} sinθ - cosθ \\ cosθ sinθ \end{array}] \\ = [\begin{array}{l} \cos^{2} θ cosθ sinθ \\ - sinθ cosθ \cos^{2} θ \end{array}] + [\begin{array}{l} \sin^{2} θ - sinθ cosθ \\ sinθ cosθ \sin^{2} θ \end{array}] \\ = [\begin{array}{l} \cos^{2} θ + \sin^{2} θ cosθ sinθ - sinθ cosθ \\ - sinθ cosθ + sinθ cosθ \cos^{2} θ + \sin^{2} θ \end{array}] \\ = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] . \end{array}$

Q.17

$\begin{array}{l} Find X and Y, if \\ (i) X + Y = [\begin{array}{l} 7 0 \\ 2 5 \end{array}] and X - Y = [\begin{array}{l} 3 0 \\ 0 3 \end{array}] \\ (ii) 2 X + 3 Y = [\begin{array}{l} 2 3 \\ 4 0 \end{array}] and 3 X + 2 Y = [\begin{array}{l} 2 - 2 \\ - 1 5 \end{array}] \end{array}$

Ans.

$\begin{array}{l} (i) X + Y = [\begin{array}{l} 7 0 \\ 2 5 \end{array}] . .. (i) \\ and X - Y = [\begin{array}{l} 3 0 \\ 0 3 \end{array}] . .. (ii) \\ Adding both equations, we get \\ X + Y + X - Y = [\begin{array}{l} 7 0 \\ 2 5 \end{array}] + [\begin{array}{l} 3 0 \\ 0 3 \end{array}] \\ 2 X = [\begin{array}{l} 7 + 30 + 0 \\ 2 + 05 + 3 \end{array}] \\ = [\begin{array}{l} 100 \\ 28 \end{array}] \\ X = \frac{1}{2} [\begin{array}{l} 100 \\ 28 \end{array}] \\ = [\begin{array}{l} 50 \\ 14 \end{array}] \\ Putting value of X in equatin (i), we get \\ [\begin{array}{l} 50 \\ 14 \end{array}] + Y = [\begin{array}{l} 70 \\ 25 \end{array}] \\ Y = [\begin{array}{l} 70 \\ 25 \end{array}] - [\begin{array}{l} 50 \\ 14 \end{array}] \\ = [\begin{array}{l} 7 - 50 - 0 \\ 2 - 15 - 4 \end{array}] \\ = [\begin{array}{l} 20 \\ 11 \end{array}] \\ Thus, X = [\begin{array}{l} 50 \\ 14 \end{array}] and Y = [\begin{array}{l} 20 \\ 11 \end{array}] . \\ (ii) 2 X +3 Y = [\begin{array}{l} 23 \\ 40 \end{array}] . .. (i) \\ and 3 X +2 Y = [\begin{array}{l} 2 - 2 \\ - 1 5 \end{array}] . .. (ii) \\ Multiplying equation (i) by 2 and equation (ii) by 3, we get \\ 2 (2 X +3 Y) = 2 [\begin{array}{l} 23 \\ 40 \end{array}] \\ \Rightarrow 4 X +6 Y = [\begin{array}{l} 46 \\ 80 \end{array}] . .. (iii) \\ 3 (3 X +2 Y) = 3 [\begin{array}{l} 2 - 2 \\ - 1 5 \end{array}] \\ \Rightarrow 9 X +6 Y = [\begin{array}{l} 6 - 6 \\ - 3 15 \end{array}] . .. (iv) \\ Subtracting equation (iii) from equation (iv), we get \\ 5 X = [\begin{array}{l} 6 - 6 \\ - 3 15 \end{array}] - [\begin{array}{l} 46 \\ 80 \end{array}] \\ X = \frac{1}{5} [\begin{array}{l} 2 - 12 \\ - 11 15 \end{array}] \\ = [\begin{array}{l} \frac{2}{5} - \frac{12}{5} \\ - \frac{11}{5} 3 \end{array}] \\ Putting value of X in equation (i), we have \\ 2 [\begin{array}{l} \frac{2}{5} - \frac{12}{5} \\ - \frac{11}{5} 3 \end{array}] + 3 Y = [\begin{array}{l} 23 \\ 40 \end{array}] \\ 3 Y = [\begin{array}{l} 23 \\ 40 \end{array}] - [\begin{array}{l} \frac{4}{5} - \frac{24}{5} \\ - \frac{22}{5} 6 \end{array}] \\ = [\begin{array}{l} 2 - \frac{4}{5} 3 - (- \frac{24}{5}) \\ 4 - (- \frac{22}{5}) 0 - 6 \end{array}] \\ Y = \frac{1}{3} [\begin{array}{l} \frac{6}{5} \frac{39}{5} \\ \frac{42}{5} - 6 \end{array}] \\ = [\begin{array}{l} \frac{2}{5} \frac{13}{5} \\ \frac{14}{5} - 2 \end{array}] \\ Thus, X = [\begin{array}{l} \frac{2}{5} - \frac{12}{5} \\ - \frac{11}{5} 3 \end{array}] and Y = [\begin{array}{l} \frac{2}{5} \frac{13}{5} \\ \frac{14}{5} - 2 \end{array}] . \end{array}$

Q.18

$Find X, if Y= [\begin{array}{l} 3 2 \\ 4 0 \end{array}] and 2 X + Y = [\begin{array}{l} 2 - 2 \\ - 1 5 \end{array}] .$

Ans.

$\begin{array}{l} Given, Y = [\begin{array}{l} 3 2 \\ 1 4 \end{array}] \\ and 2 X + Y = [\begin{array}{l} 1 0 \\ - 3 2 \end{array}] . .. (i) \\ Putting value of Y in equation (i), we get \\ 2 X + [\begin{array}{l} 32 \\ 14 \end{array}] = [\begin{array}{l} 1 0 \\ - 3 2 \end{array}] \\ 2 X = [\begin{array}{l} 1 0 \\ - 3 2 \end{array}] - [\begin{array}{l} 3 2 \\ 1 4 \end{array}] \\ = [\begin{array}{l} 1 - 30 - 2 \\ - 3 - 12 - 4 \end{array}] \\ X = \frac{1}{2} [\begin{array}{l} - 2 - 2 \\ - 4 - 2 \end{array}] \\ = [\begin{array}{l} - 1 - 1 \\ - 2 - 1 \end{array}] \end{array}$

Q.19

$If x and y, if 2 [\begin{array}{l} 1 3 \\ 0 x \end{array}] + [\begin{array}{l} y 0 \\ 1 2 \end{array}] = [\begin{array}{l} 5 6 \\ 1 8 \end{array}] .$

Ans.

$\begin{array}{l} Given : 2 [\begin{array}{l} 13 \\ 0 x \end{array}] + [\begin{array}{l} y 0 \\ 1 2 \end{array}] = [\begin{array}{l} 56 \\ 18 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 6 \\ 02 x \end{array}] + [\begin{array}{l} y 0 \\ 1 2 \end{array}] = [\begin{array}{l} 5 6 \\ 1 8 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 + y 6 +0 \\ 0 +12 x +2 \end{array}] = [\begin{array}{l} 56 \\ 18 \end{array}] \\ \Rightarrow 2 + y = 5 and 2 x + 2 = 8 \\ \Rightarrow y = 3 and x = \frac{8 - 2}{2} = 3 \\ Thus, x = 3 and y = 3 . \end{array}$

Q.20

$\begin{array}{l} Solve thee quation for x, y, z and t, if \\ 2 [\begin{array}{l} xz \\ yt \end{array}] + 3 [\begin{array}{l} 1 - 1 \\ 0 2 \end{array}] = 3 [\begin{array}{l} 35 \\ 46 \end{array}] . \end{array}$

Ans.

$\begin{array}{l} Given : \\ 2 [\begin{array}{l} x z \\ y t \end{array}] + 3 [\begin{array}{l} 1 - 1 \\ 0 2 \end{array}] = 3 [\begin{array}{l} 35 \\ 46 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 x 2 z \\ 2 y 2 t \end{array}] + [\begin{array}{l} 3 - 3 \\ 0 6 \end{array}] = [\begin{array}{l} 9 1 5 \\ 12 1 8 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 x + 32 z - 3 \\ 2 y + 02 t + 6 \end{array}] = [\begin{array}{l} 9 1 5 \\ 12 1 8 \end{array}] \\ \Rightarrow 2 x + 3 = 9, 2 y = 12, 2 z - 3 = 15 and 2 t + 6 = 18 \\ \Rightarrow x = 3, y = 6, z = 9 and t = 6 . \end{array}$

Q.21

$If x [\begin{array}{l} 2 \\ 3 \end{array}] + y [\begin{array}{l} - 1 \\ 1 \end{array}] = [\begin{array}{l} 10 \\ 5 \end{array}], find the value of x and y .$

Ans.

$\begin{array}{l} Goven : \\ x [\begin{array}{l} 2 \\ 3 \end{array}] + y [\begin{array}{l} - 1 \\ 1 \end{array}] = [\begin{array}{l} 10 \\ 5 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 x \\ 3 x \end{array}] + [\begin{array}{l} - y \\ y \end{array}] = [\begin{array}{l} 10 \\ 5 \end{array}] \\ \Rightarrow [\begin{array}{l} 2 x - y \\ 3 x + y \end{array}] = [\begin{array}{l} 10 \\ 5 \end{array}] \\ \Rightarrow 2 x - y = 10, 3 x + y = 5 \\ \Rightarrow x = 3 and y = - 4 \end{array}$

Q.22

$\begin{array}{l} Given 3 [\begin{array}{l} xy \\ zw \end{array}] = [\begin{array}{l} x 6 \\ - 12 w \end{array}] + [\begin{array}{l} 4 x + y \\ z + w 3 \end{array}], f in d the \\ value of x, y, z and w . \end{array}$

Ans.

$\begin{array}{l} Given : \\ 3 [\begin{array}{l} xy \\ zw \end{array}] = [\begin{array}{l} x 6 \\ - 12 w \end{array}] + [\begin{array}{l} 4 x + y \\ z + w 3 \end{array}] \\ \Rightarrow [\begin{array}{l} 3 x 3 y \\ 3 z 3 w \end{array}] = [\begin{array}{l} x + 4 6 + x + y \\ - 1 + z + w 2 w + 3 \end{array}] \\ \Rightarrow 3 x = x + 4 \Rightarrow x = 2 \\ 3 y = 6 + x + y \Rightarrow y = 4 \\ 3 w = 2 w + 3 \Rightarrow w = 3 \\ and 3 z = - 1 + z + w \Rightarrow z = 1 \\ Therefore, x = 2, y = 4, z = 1 and w = 3 . \end{array}$

Q.23

$If F(x) = [\begin{array}{l} cosx - sinx 0 \\ sinx cosx 0 \\ 0 0 1 \end{array}], show that F (x) F (y) = F (x + y) .$

Ans.

$\begin{array}{l} Given : F (x) = [\begin{array}{l} cosx - sinx 0 \\ sinx cosx 0 \\ 0 0 1 \end{array}] \\ F (y) = [\begin{array}{l} cosy - siny 0 \\ siny cosy 0 \\ 0 0 1 \end{array}] \\ L . H . S . : \\ F (x) . F (y) = [\begin{array}{l} cosx - sinx 0 \\ sinx cosx 0 \\ 0 0 1 \end{array}] [\begin{array}{l} cosy - siny 0 \\ siny cosy 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} cosx cosy - sinx siny - siny cosx - sinx cosy 0 \\ sinx cosy + cosx siny - sinx siny + cosx cosy 0 \\ 0 01 \end{array}] \\ = [\begin{array}{l} \cos (x + y) - \sin (x + y) 0 \\ \sin (x + y) \cos (x + y) 0 \\ 0 01 \end{array}] \\ = F (x + y) \\ = R . H . S . \end{array}$

Q.24

$\begin{array}{l} Show that \\ (i) [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] [\begin{array}{l} 21 \\ 34 \end{array}] \neq [\begin{array}{l} 21 \\ 34 \end{array}] [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] \\ (ii) [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] \neq [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] \end{array}$

Ans.

$\begin{array}{l} (i) L . H . S . : \\ [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] [\begin{array}{l} 21 \\ 34 \end{array}] = [\begin{array}{l} 10 - 35 - 4 \\ 12 + 216 + 28 \end{array}] \\ = [\begin{array}{l} 7 1 \\ 3334 \end{array}] \\ R . H . S . : \\ [\begin{array}{l} 21 \\ 34 \end{array}] [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] = [\begin{array}{l} 10 + 6 - 2 + 7 \\ 15 + 24 - 3 + 28 \end{array}] \\ = [\begin{array}{l} 165 \\ 3925 \end{array}] \\ Therefore, \\ [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] [\begin{array}{l} 21 \\ 34 \end{array}] \neq [\begin{array}{l} 21 \\ 34 \end{array}] [\begin{array}{l} 5 - 1 \\ 6 7 \end{array}] \\ (ii) [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] \neq [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] \\ L . H . S : \\ [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] = [\begin{array}{l} - 1 + 0 + 61 - 2 + 90 + 2 + 12 \\ 0 + 0 + 00 - 1 + 00 + 1 + 0 \\ - 1 + 0 + 01 - 1 + 0 0 + 1 + 0 \end{array}] \\ = [\begin{array}{l} 5814 \\ 0 - 11 \\ - 10 1 \end{array}] \\ R . H . S : \\ [\begin{array}{l} - 110 \\ 0 - 11 \\ 234 \end{array}] [\begin{array}{l} 123 \\ 010 \\ 110 \end{array}] = [\begin{array}{l} - 1 + 0 + 0 - 2 + 1 + 0 - 3 + 0 + 0 \\ 0 + 0 + 10 - 1 + 10 + 0 + 0 \\ 2 + 0 + 44 + 3 + 46 + 0 + 0 \end{array}] \\ = [\begin{array}{l} - 1 - 1 - 3 \\ 1 0 0 \\ 6 11 6 \end{array}] \\ ∴ L . H . S . \neq R . H . S \end{array}$

Q.25

$F i n d A^{2} - 5 A + 6 I, i f A = [\begin{array}{l} 2 0 1 \\ 2 1 3 \\ 1 - 1 0 \end{array}] .$

Ans.

$\begin{array}{l} A^{2} - 5A+6I, A = [\begin{array}{l} 2 0 1 \\ 2 1 3 \\ 1 -1 0 \end{array}] \\ A^{2} = A \times A \\ = [\begin{array}{l} 2 0 1 \\ 2 1 3 \\ 1 - 1 0 \end{array}] [\begin{array}{l} 2 0 1 \\ 2 1 3 \\ 1 - 1 0 \end{array}] \\ = [\begin{array}{l} 4 + 0 + 1 0 + 0 - 1 2 + 0 + 0 \\ 4 + 2 + 3 0 + 1 - 3 2 + 3 + 0 \\ 2 - 2 + 0 0 - 1 + 0 1 - 3 + 0 \end{array}] \\ = [\begin{array}{l} 5 - 1 2 \\ 9 - 2 5 \\ 0 - 1 - 2 \end{array}] \\ N o w, A^{2} - 5A+6I = [\begin{array}{l} 5 - 1 2 \\ 9 - 2 5 \\ 0 - 1 - 2 \end{array}] - 5 [\begin{array}{l} 2 0 1 \\ 2 1 3 \\ 1 - 1 0 \end{array}] +6 [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ = [\begin{array}{l} 5 - 1 2 \\ 9 - 2 5 \\ 0 - 1 - 2 \end{array}] - [\begin{array}{l} 10 0 5 \\ 10 5 15 \\ 5 - 5 0 \end{array}] + [\begin{array}{l} 6 0 0 \\ 0 6 0 \\ 0 0 6 \end{array}] \\ = [\begin{array}{l} 5 - 10 + 6 - 1 2 - 5 \\ 9 - 10 - 2 - 5 + 6 5 - 15 \\ 0 - 5 - 1 + 5 - 2 + 6 \end{array}] \\ = [\begin{array}{l} 1 - 1 - 3 \\ - 1 - 1 - 10 \\ - 5 4 4 \end{array}] \end{array}$

Q.26

$If A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}], prove that A^{3} - 6 A^{2} + 7 A + 2 I = 0 .$

Ans.

$\begin{array}{l} A = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}], \\ A^{2} = A \times A \\ = [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] \\ = [\begin{array}{l} 1 + 0 + 4 0 + 0 + 02 + 0 + 6 \\ 0 + 0 + 2 0 + 4 + 00 + 2 + 3 \\ 2 + 0 + 6 0 + 0 + 04 + 0 + 9 \end{array}] \\ = [\begin{array}{l} 508 \\ 245 \\ 8013 \end{array}] \\ 6 A^{2} = 6 [\begin{array}{l} 508 \\ 245 \\ 8013 \end{array}] \\ = [\begin{array}{l} 30 048 \\ 122430 \\ 16 078 \end{array}] \\ A^{3} = A^{2} . A \\ = [\begin{array}{l} 508 \\ 245 \\ 8013 \end{array}] [\begin{array}{l} 102 \\ 021 \\ 203 \end{array}] \\ = [\begin{array}{l} 5 + 0 + 160 + 0 + 010 + 0 + 24 \\ 2 + 0 + 100 + 8 + 04 + 4 + 15 \\ 8 + 0 + 260 + 0 + 016 + 0 + 39 \end{array}] \\ = [\begin{array}{l} 21034 \\ 12823 \\ 34055 \end{array}] \\ 7 A = 7 [\begin{array}{l} 102 \\ 021 \\ 203 \end{array}] \\ = [\begin{array}{l} 7014 \\ 0147 \\ 14021 \end{array}] \\ 2 I = 2 [\begin{array}{l} 100 \\ 010 \\ 001 \end{array}] \\ = [\begin{array}{l} 200 \\ 020 \\ 002 \end{array}] \\ L . H . S : \\ A^{3} - 6 A^{2} + 7 A +2 I \\ = [\begin{array}{l} 21034 \\ 12823 \\ 34055 \end{array}] - 6 [\begin{array}{l} 508 \\ 245 \\ 8013 \end{array}] + [\begin{array}{l} 7014 \\ 0147 \\ 14021 \end{array}] + [\begin{array}{l} 200 \\ 020 \\ 002 \end{array}] \\ = [\begin{array}{l} 21034 \\ 12823 \\ 34055 \end{array}] - [\begin{array}{l} 30048 \\ 122430 \\ 48078 \end{array}] + [\begin{array}{l} 7014 \\ 0147 \\ 14021 \end{array}] + [\begin{array}{l} 200 \\ 020 \\ 002 \end{array}] \\ = [\begin{array}{l} 21 - 30 + 7 + 20 - 0 + 0 + 034 - 48 + 14 + 0 \\ 12 - 12 + 0 + 08 - 24 + 14 + 223 - 30 + 7 + 0 \\ 34 - 48 + 14 + 00 + 0 + 0 + 055 - 78 + 21 + 2 \end{array}] \\ = [\begin{array}{l} 000 \\ 000 \\ 000 \end{array}] = 0 \end{array}$

Q.27

$If A= [\begin{array}{l} 3 - 2 \\ 4 - 2 \end{array}], and I = (\begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}), f indk so that A^{2} = kA - 2 I .$

Ans.

$\begin{array}{l} âˆµ A = [\begin{array}{l} 3 - 2 \\ 4 - 2 \end{array}] \\ A^{2} = [\begin{array}{l} 3 - 2 \\ 4 - 2 \end{array}] [\begin{array}{l} 3 - 2 \\ 4 - 2 \end{array}] \\ = [\begin{array}{l} 9 - 8 - 6 + 4 \\ 12 - 8 - 8 + 4 \end{array}] \\ = [\begin{array}{l} 1 - 2 \\ 4 - 4 \end{array}] \\ Now, A^{2} = kA - 2 I \\ [\begin{array}{l} 1 - 2 \\ 4 - 4 \end{array}] = k [\begin{array}{l} 3 - 2 \\ 4 - 2 \end{array}] - 2 [\begin{array}{l} 10 \\ 01 \end{array}] \\ [\begin{array}{l} 1 - 2 \\ 4 - 4 \end{array}] = [\begin{array}{l} 3 k - 2 k \\ 4 k - 2 k \end{array}] - [\begin{array}{l} 20 \\ 02 \end{array}] \\ [\begin{array}{l} 1 - 2 \\ 4 - 4 \end{array}] = [\begin{array}{l} 3 k - 2 - 2 k \\ 4 k - 2 k - 2 \end{array}] \\ \Rightarrow 4 = 4 k \\ \Rightarrow k = 1 \\ Thus, the value of k is 1 . \end{array}$

Q.28

$\begin{array}{l} If A = [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] and I is the identity matrix of order 2, \\ show that I + A = (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \end{array}$

Ans.

$\begin{array}{l} I + A = [\begin{array}{l} 10 \\ 01 \end{array}] + [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] \\ = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] \\ I - A = [\begin{array}{l} 10 \\ 01 \end{array}] - [\begin{array}{l} 0 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 0 \end{array}] \\ = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] \\ R . H . S . = (I - A) [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ = [\begin{array}{l} 1 \tan \frac{α}{2} \\ - \tan \frac{α}{2} 1 \end{array}] [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ = [\begin{array}{l} 1 \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}} \\ - \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}} 1 \end{array}] [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ = [\begin{array}{l} cosα + \frac{sinα . \sin \frac{α}{2}}{\cos \frac{α}{2}} - sinα + cosα \frac{\sin \frac{α}{2}}{\cos \frac{α}{2}} \\ - \frac{cosα . \sin \frac{α}{2}}{\cos \frac{α}{2}} + sinα \frac{sinα . \sin \frac{α}{2}}{\cos \frac{α}{2}} + cosα \end{array}] \\ = [\begin{array}{l} \frac{cosα . \cos \frac{α}{2} + sinα . \sin \frac{α}{2}}{\cos \frac{α}{2}} \frac{- (sinα . \cos \frac{α}{2} - cosαsin \frac{α}{2})}{\cos \frac{α}{2}} \\ \frac{sinαcos \frac{α}{2} - cosα . \sin \frac{α}{2}}{\cos \frac{α}{2}} \frac{cosαcos \frac{α}{2} + sinα . \sin \frac{α}{2}}{\cos \frac{α}{2}} \end{array}] \\ = [\begin{array}{l} \frac{\cos (α - \frac{α}{2})}{\cos \frac{α}{2}} \frac{- \sin (α - \frac{α}{2})}{\cos \frac{α}{2}} \\ \frac{\sin (α - \frac{α}{2})}{\cos \frac{α}{2}} \frac{\cos (α - \frac{α}{2})}{\cos \frac{α}{2}} \end{array}] \\ = [\begin{array}{l} \frac{\cos (\frac{α}{2})}{\cos \frac{α}{2}} \frac{- \sin (\frac{α}{2})}{\cos \frac{α}{2}} \\ \frac{\sin (\frac{α}{2})}{\cos \frac{α}{2}} \frac{\cos (\frac{α}{2})}{\cos \frac{α}{2}} \end{array}] \\ = [\begin{array}{l} 1 - \tan \frac{α}{2} \\ \tan \frac{α}{2} 1 \end{array}] = L . H . S . \end{array}$

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

Ans.

$\begin{array}{l} Total money to invest = 30,000 \\ Money invested in I^{st} type of bond = x \\ Money invested in 2^{nd} type of bond = (30,000- x) \\ Rate of interest on first type of bond =5% \\ Rate of interest on secon type of bond =7% \\ (i) Total obtained interest =1800 \\ (ii) Total obtained interest =2000 \\ (\begin{matrix} I & II \\ x & (30,000- x) \end{matrix}) (\begin{array}{l} 5 % \\ 7 % \end{array}) = (\begin{array}{l} 1800 \\ 2000 \end{array}) \\ (i) 5 % of x +7% (30,000- x) = 1800 \\ \frac{5}{100} x + \frac{7}{100} (30,000- x) = 1800 \\ 5 x +210000-7 x =180000 \\ 210000 -180000=2 x \\ 30000 =2 x \\ x = \frac{30000}{2} \\ = 15000 \\ To get annual interest of 1800, he should invest 15,000 \\ in I type of bond and 15,000 in II type of bond . \\ (ii) 5 % of x +7% (30,000- x) = 2000 \\ \frac{5}{100} x + \frac{7}{100} (30,000- x) = 2000 \\ 5 x +210000-7 x =200000 \\ 210000 -200000=2 x \\ 10000 =2 x \\ x = \frac{10000}{2} \\ = 5000 \\ To get annual interest of 2000 he should invest 5000 & \\ 25000 respectively in I and II type of bond . \end{array}$

Q.30

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

Ans.

$\begin{array}{l} \begin{array}{l} Number of chemistry books on bookshop = 10 dozen \end{array} \\ \begin{array}{l} = 10\times12 \end{array} \\ \begin{array}{l} = 120 \end{array} \\ \begin{array}{l} Number of physics books on bookshop = 8 dozen \end{array} \\ \begin{array}{l} = 8\times12 \end{array} \\ = \begin{array}{l} 96 \end{array} \\ \begin{array}{l} Number of economics books on bookshop = 10 dozen \end{array} \\ \begin{array}{l} = 10\times12 \end{array} \\ \begin{array}{l} = 120 \end{array} \\ \begin{array}{l} Selling price of one chemistry book = \end{array} 80 \\ \begin{array}{l} Selling price of one physics book = \end{array} 60 \\ \begin{array}{l} Selling price of one economics book = \end{array} 40 \\ \begin{array}{l} The total amount the bookshop will receive \end{array} \\ \begin{array}{l} = [12096120] [\begin{array}{l} 80 \\ 60 \\ 40 \end{array}] \end{array} \\ \begin{array}{l} = [9600 +5760+4800] \end{array} \\ \begin{array}{l} = [20160] \end{array} \\ \begin{array}{l} Thus, the total amount the bookshop will receive from selling \end{array} \begin{array}{l} all the books is \end{array} 20, 160 . \end{array}$

Q.31

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

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

(A) k = 3, p = n

(B) k is arbitrary, p = 2

(D) k = 2, p = 3

Ans.

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

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

= ( p×k )×( 3×k )+( n×3 )×( 3×k )

= ( p×k )+( n×k )

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

Thus, option ( A ) is correct.

Q.32

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

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

(A) P × 2

(B) 2 × n

(D) p × n

Ans.

$\begin{array}{l} The order of matrix X is 2 \times n \\ The order of matrix Z is 2 \times p \\ 7 X - 5 Z is defined when X and Z an of the same order . \\ \Rightarrow n = p (Given) \\ Thus the order of 7 X - 5 Z is 2 \times n . \\ Thus, option (B) is correct answer . \end{array}$

Q.33

$\begin{array}{l} Find the t ranspose of each of the following matrices : \\ (i) [\begin{array}{l} 5 \\ \frac{1}{2} \\ - 1 \end{array}] (ii) [\begin{array}{l} 1 - 1 \\ 2 3 \end{array}] (iii) [\begin{array}{l} - 15 6 \\ \sqrt{3} 5 6 \\ 23 - 1 \end{array}] \end{array}$

Ans.

$\begin{matrix} (i) Transpose of [\begin{array}{l} 5 \\ \frac{1}{2} \\ - 1 \end{array}] = [5 \frac{1}{2} - 1] \\ (ii) Transpose of [\begin{array}{l} 1 - 1 \\ 2 3 \end{array}] = [\begin{array}{l} 12 \\ - 13 \end{array}] \\ (iii) Transpose of [\begin{array}{l} - 15 6 \\ \sqrt{3} 5 6 \\ 23 - 1 \end{array}] = [\begin{array}{l} - 1 \sqrt{3} 2 \\ 5 5 3 \\ 6 6 - 1 \end{array}] \end{matrix}$

Q.34

$\begin{array}{l} If A = [\begin{array}{l} - 123 \\ 579 \\ - 211 \end{array}] andB = [\begin{array}{l} - 41 - 5 \\ 12 0 \\ 13 1 \end{array}], then verify that \\ (i) (A + B) ‘ = A ‘ + B ‘ \\ (ii) (A - B) ‘ = A ‘ - B ‘ \end{array}$

Ans.

$\begin{array}{l} (i) A + B = [\begin{array}{l} - 123 \\ 579 \\ - 211 \end{array}] + [\begin{array}{l} - 41 - 5 \\ 12 0 \\ 13 1 \end{array}] \\ = [\begin{array}{l} - 1 - 42 + 13 - 5 \\ 5 + 17 + 29 + 0 \\ - 2 + 11 + 31 + 1 \end{array}] \\ = [\begin{array}{l} - 53 - 2 \\ 69 9 \\ - 14 2 \end{array}] \\ (A + B) ‘ = {[\begin{array}{l} - 53 - 2 \\ 69 9 \\ - 14 2 \end{array}]}^{‘} \\ = [\begin{array}{l} - 56 - 1 \\ 39 4 \\ - 29 2 \end{array}] \\ A ‘ + B ‘ = {[\begin{array}{l} - 123 \\ 579 \\ - 211 \end{array}]}^{‘} + {[\begin{array}{l} - 41 - 5 \\ 12 0 \\ 13 1 \end{array}]}^{‘} \\ = [\begin{array}{l} - 15 - 2 \\ 27 1 \\ 39 1 \end{array}] + [\begin{array}{l} - 411 \\ 123 \\ - 501 \end{array}] \\ = [\begin{array}{l} - 1 - 45 + 1 - 2 + 1 \\ 2 + 17 + 2 1 + 3 \\ 3 - 59 + 0 1 + 1 \end{array}] \\ = [\begin{array}{l} - 56 - 1 \\ 39 4 \\ - 29 2 \end{array}] \\ Thus, (A + B) ‘ = A ‘ + B ‘ . \\ (ii) A - B = [\begin{array}{l} - 123 \\ 579 \\ - 211 \end{array}] - [\begin{array}{l} - 41 - 5 \\ 12 0 \\ 13 1 \end{array}] \\ = [\begin{array}{l} - 1 - (- 4) 2 - 13 - (- 5) \\ 5 - 17 - 29 - 0 \\ - 2 - 11 - 31 - 1 \end{array}] \\ = [\begin{array}{l} 318 \\ 459 \\ - 3 - 2 0 \end{array}] \\ (A - B) ‘ = {[\begin{array}{l} 318 \\ 459 \\ - 3 - 2 0 \end{array}]}^{‘} \\ = [\begin{array}{l} 34 - 3 \\ 15 - 2 \\ 89 0 \end{array}] \\ A ‘ - B ‘ = {[\begin{array}{l} - 123 \\ 579 \\ - 211 \end{array}]}^{‘} - {[\begin{array}{l} - 41 - 5 \\ 12 0 \\ 13 1 \end{array}]}^{‘} \\ = [\begin{array}{l} - 15 - 2 \\ 27 1 \\ 39 1 \end{array}] - [\begin{array}{l} - 411 \\ 123 \\ - 501 \end{array}] \\ = [\begin{array}{l} - 1 + 45 - 1 - 2 - 1 \\ 2 - 17 - 2 1 - 3 \\ 3 + 59 - 0 1 - 1 \end{array}] \\ = [\begin{array}{l} 34 - 3 \\ 15 - 2 \\ 89 0 \end{array}] \\ Thus, (A - B) ‘ = A ‘ - B ‘ . \end{array}$

Q.35

$\begin{array}{l} If A ‘ = [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] and B = [\begin{array}{l} - 121 \\ 123 \end{array}], then verify that \\ (i) (A + B) ‘ = A ‘ + B ‘ \\ (ii) (A - B) ‘ = A ‘ - B ‘ \end{array}$

Ans.

$\begin{array}{l} Since, A ‘= [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] \Rightarrow A = [\begin{array}{l} 3 - 10 \\ 4 21 \end{array}] \\ and B = [\begin{array}{l} - 121 \\ 123 \end{array}] \\ So, A + B = [\begin{array}{l} 3 - 10 \\ 4 21 \end{array}] + [\begin{array}{l} - 121 \\ 123 \end{array}] \\ = [\begin{array}{l} 3 - 1 - 1 + 20 + 1 \\ 4 + 1 2 + 21 + 3 \end{array}] \\ = [\begin{array}{l} 211 \\ 5 43 \end{array}] \\ ∴ (A + B) ‘ = {[\begin{array}{l} 211 \\ 5 44 \end{array}]}^{‘} \\ = [\begin{array}{l} 25 \\ 14 \\ 14 \end{array}] \\ A ‘ + B ‘ = {[\begin{array}{l} 3 - 10 \\ 4 21 \end{array}]}^{‘} + {[\begin{array}{l} - 121 \\ 123 \end{array}]}^{‘} \\ = [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] + [\begin{array}{l} - 11 \\ 22 \\ 13 \end{array}] \\ = [\begin{array}{l} 3 - 14 + 1 \\ - 1 + 22 + 2 \\ 0 + 11 + 3 \end{array}] \\ = [\begin{array}{l} 25 \\ 14 \\ 14 \end{array}] \\ Thus, (A + B) ‘ = A ‘ + B ‘ . \\ (ii) Since, A ‘= [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] \Rightarrow A = [\begin{array}{l} 3 - 10 \\ 4 21 \end{array}] \\ and B = [\begin{array}{l} - 121 \\ 123 \end{array}] \\ So, A - B = [\begin{array}{l} 3 - 10 \\ 4 21 \end{array}] - [\begin{array}{l} - 121 \\ 123 \end{array}] \\ = [\begin{array}{l} 3 + 1 - 1 - 20 - 1 \\ 4 - 1 2 - 21 - 3 \end{array}] \\ = [\begin{array}{l} 4 - 3 - 1 \\ 3 0 - 2 \end{array}] \\ ∴ (A - B) ‘ = {[\begin{array}{l} 4 - 3 - 1 \\ 3 0 - 2 \end{array}]}^{‘} \\ = [\begin{array}{l} 4 3 \\ - 3 0 \\ - 1 - 2 \end{array}] \\ A ‘ - B ‘ = [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] - {[\begin{array}{l} - 121 \\ 123 \end{array}]}^{‘} \\ = [\begin{array}{l} 34 \\ - 12 \\ 01 \end{array}] - [\begin{array}{l} - 11 \\ 22 \\ 13 \end{array}] \\ = [\begin{array}{l} 3 + 14 - 1 \\ - 1 - 22 - 2 \\ 0 - 11 - 3 \end{array}] \\ = [\begin{array}{l} 4 3 \\ - 3 0 \\ - 1 - 2 \end{array}] \\ Thus, (A - B) ‘ = A ‘ - B ‘ . \end{array}$

Q.36

$If A’= [\begin{array}{l} - 23 \\ 12 \end{array}] and B = [\begin{array}{l} - 10 \\ 12 \end{array}], then find (A + 2 B) ‘ .$

Ans.

$\begin{array}{l} Since, A ‘ = [\begin{array}{l} - 23 \\ 12 \end{array}] \Rightarrow A = [\begin{array}{l} - 21 \\ 32 \end{array}] \\ and B = [\begin{array}{l} - 10 \\ 12 \end{array}] \\ A + 2 B = [\begin{array}{l} - 21 \\ 32 \end{array}] + 2 [\begin{array}{l} - 10 \\ 12 \end{array}] \\ = [\begin{array}{l} - 21 \\ 32 \end{array}] + [\begin{array}{l} - 20 \\ 24 \end{array}] \\ = [\begin{array}{l} - 41 \\ 56 \end{array}] \\ = [\begin{array}{l} - 45 \\ 16 \end{array}] \\ ∴ (A + 2 B) ‘ = {[\begin{array}{l} - 45 \\ 16 \end{array}]}^{‘} \\ = [\begin{array}{l} - 41 \\ 56 \end{array}] \end{array}$

Q.37

$\begin{array}{l} For the matrices A and B, verify that (AB) ‘ = B ‘ A ‘, where \\ (i) A = [\begin{array}{l} 1 \\ - 4 \\ 3 \end{array}], B = [- 121] \\ (ii) A = [\begin{array}{l} 0 \\ 1 \\ 2 \end{array}], B = [157] \end{array}$

Ans.

$\begin{array}{l} For the matrices A and B, verify that (AB) ‘ = B ‘ A ‘, where \\ (i) A = [\begin{array}{l} 1 \\ - 4 \\ 3 \end{array}], B = [- 1 2 1] \\ AB = [\begin{array}{l} 1 \\ - 4 \\ 3 \end{array}] [- 1 2 1] \\ = [\begin{array}{l} - 1 2 1 \\ 4 - 8 - 4 \\ - 3 6 3 \end{array}] \\ (AB) ‘ = {[\begin{array}{l} - 1 2 1 \\ 4 - 8 - 4 \\ - 3 6 3 \end{array}]}^{‘} \\ = [\begin{array}{l} - 1 4 - 3 \\ 2 - 8 6 \\ 1 - 4 3 \end{array}] \\ A ‘ = {[\begin{array}{l} 1 \\ - 4 \\ 3 \end{array}]}^{‘} = [1 - 43] \\ B ‘ = {[- 121]}^{‘} \\ = [\begin{array}{l} - 1 \\ 2 \\ 1 \end{array}] \\ B ‘ A ‘ = [\begin{array}{l} - 1 \\ 2 \\ 1 \end{array}] [1 - 43] \\ = [\begin{array}{l} - 1 4 - 3 \\ 2 - 8 6 \\ 1 - 4 3 \end{array}] \\ Thus, (AB) ‘ = B ‘ A ‘ . \\ (ii) A = [\begin{array}{l} 0 \\ 1 \\ 2 \end{array}], B = [1 5 7] \\ AB = [\begin{array}{l} 0 \\ 1 \\ 2 \end{array}] [1 5 7] \\ = [\begin{array}{l} 0 0 0 \\ 1 5 7 \\ 21 0 14 \end{array}] \\ (AB) ‘ = {[\begin{array}{l} 0 0 0 \\ 1 5 7 \\ 21 0 14 \end{array}]}^{‘} \\ = [\begin{array}{l} 0 1 2 \\ 0 5 10 \\ 0 7 14 \end{array}] \\ A ‘ = {[\begin{array}{l} 0 \\ 1 \\ 2 \end{array}]}^{‘} \\ = [012] \\ B ‘ = {[157]}^{‘} \\ = [\begin{array}{l} 1 \\ 5 \\ 7 \end{array}] \\ B ‘ A ‘ = [\begin{array}{l} 1 \\ 5 \\ 7 \end{array}] [012] \\ = [\begin{array}{l} 01 2 \\ 0510 \\ 0714 \end{array}] \\ Thus, (AB) ‘ = B ‘ A ‘ \end{array}$

Q.38

$\begin{array}{l} If (i) A = [\begin{array}{l} cosα sinα \\ - sinαcosα \end{array}], the nver if y AA ‘ = I \\ (ii) A = [\begin{array}{l} sinα cosα \\ - cosα sinα \end{array}], the nver if y AA ‘ = I \end{array}$

Ans.

$\begin{array}{l} (i) Given A = [\begin{array}{l} cosα sinα \\ - sinαcosα \end{array}] \\ A ‘ = {[\begin{array}{l} cosα sinα \\ - sinαcosα \end{array}]}^{‘} \\ = [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ L . H . S : AA ‘ = [\begin{array}{l} cosα sinα \\ - sinαcosα \end{array}] [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}] \\ = [\begin{array}{l} \cos^{2} α + \sin^{2} α - cosαsinα + sinαcosα \\ - sinαcosα + {cosαsinαsin}^{2} α + \cos^{2} α \end{array}] \\ = [\begin{array}{l} 10 \\ 01 \end{array}] \\ = I = R . H . S . \\ (ii) A = [\begin{array}{l} sinα cosα \\ - cosα sinα \end{array}] \\ A ‘ = {[\begin{array}{l} sinα cosα \\ - cosα sinα \end{array}]}^{‘} \\ = [\begin{array}{l} sinα - cosα \\ cosα sinα \end{array}] \\ AA ‘ = [\begin{array}{l} sinα cosα \\ - cosα sinα \end{array}] [\begin{array}{l} sinα - cosα \\ cosα sinα \end{array}] \\ = [\begin{array}{l} \sin^{2} α + \cos^{2} α - sinαcosα + cosαsinα \\ - cosαsinα + {sinαcosαcos}^{2} α + \sin^{2} α \end{array}] \\ = [\begin{array}{l} 10 \\ 01 \end{array}] \\ = I \\ Thus, AA ‘ = I . \end{array}$

Q.39

$\begin{array}{l} (i) Show tha t the matrix A= [\begin{array}{l} 1 - 15 \\ 1 21 \\ 5 13 \end{array}] i s a symmetric matrix. \\ (i i) Show tha tthe matrix A= [\begin{array}{l} 0 1 - 1 \\ - 1 0 1 \\ 1 - 1 0 \end{array}] is a skew \\ symmetric matrix . \end{array}$

Ans.

$\begin{array}{l} (i) A = [\begin{array}{l} 1 - 15 \\ - 1 21 \\ 5 13 \end{array}] then A ‘ = {[\begin{array}{l} 1 - 15 \\ - 1 21 \\ 5 13 \end{array}]}^{} \\ = [\begin{array}{l} 1 - 15 \\ - 1 21 \\ 5 13 \end{array}] = A \\ Since, A ‘ = A, so A is symmetric matrix . \\ (ii) A = [\begin{array}{l} 0 1 - 1 \\ - 1 0 1 \\ 1 - 1 0 \end{array}] \\ A ‘ = {[\begin{array}{l} 0 1 - 1 \\ - 1 0 1 \\ 1 - 1 0 \end{array}]}^{‘} \\ = [\begin{array}{l} 0 - 1 1 \\ 1 0 - 1 \\ - 1 - 1 0 \end{array}] \\ = - [\begin{array}{l} 0 1 - 1 \\ - 1 0 1 \\ 1 - 1 0 \end{array}] \\ = - A \\ Since, A ‘ = - A, so A is skew symmetric matrix . \end{array}$

Q.40

$\begin{array}{l} For the matrix A = [\begin{array}{l} 15 \\ 67 \end{array}], verify that \\ (i) (A + A ‘) is a symmetric matrix . \\ (ii) (A - A ‘) is as kew symmetric matrix . \end{array}$

Ans.

$\begin{array}{l} Given : A = [\begin{array}{l} 15 \\ 67 \end{array}] \Rightarrow A ‘ = [\begin{array}{l} 16 \\ 57 \end{array}] \\ (i) A + A ‘ = [\begin{array}{l} 15 \\ 67 \end{array}] + [\begin{array}{l} 16 \\ 57 \end{array}] \\ = [\begin{array}{l} 1 + 15 + 6 \\ 6 + 57 + 7 \end{array}] \\ = [\begin{array}{l} 211 \\ 1114 \end{array}] \\ (A + A ‘) ‘ = {[\begin{array}{l} 211 \\ 1114 \end{array}]}^{‘} \\ = [\begin{array}{l} 211 \\ 1114 \end{array}] \\ = A + A ‘ \\ Therefore, (A + A ‘) is symmetric matrix . \\ (ii) A - A ‘ = [\begin{array}{l} 15 \\ 67 \end{array}] - [\begin{array}{l} 16 \\ 57 \end{array}] \\ = [\begin{array}{l} 1 - 15 - 6 \\ 6 - 57 - 7 \end{array}] \\ = [\begin{array}{l} 0 - 1 \\ 10 \end{array}] \\ (A - A ‘) ‘ = {[\begin{array}{l} 0 - 1 \\ 10 \end{array}]}^{‘} \\ = [\begin{array}{l} 01 \\ - 10 \end{array}] = - [\begin{array}{l} 0 - 1 \\ 10 \end{array}] \\ = - (A - A ‘) \\ Therefore, (A - A ‘) is a skew symmetric matrix . \end{array}$

Q.41

$Find \frac{1}{2} (A + A ‘) and \frac{1}{2} (A - A ‘), when A = [\begin{array}{l} 0 a b \\ - a 0 c \\ - b - c 0 \end{array}] .$

Ans.

$\begin{array}{l} Given, A = [\begin{array}{l} 0 a b \\ - a 0 c \\ - b - c 0 \end{array}]] \\ so, A ‘ = [{\begin{array}{l} 0 a b \\ - a 0 c \\ - b - c 0 \end{array}}^{‘}] \\ = [\begin{array}{l} 0 - a - b \\ a 0 - c \\ b c 0 \end{array}] \\ (i) A + A ‘ = [\begin{array}{l} 0 b \\ - a 0 c \\ - b - c 0 \end{array}] + [\begin{array}{l} 0 - a - b \\ a 0 - c \\ b c 0 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 000 \\ 0 0 0 \end{array}] \\ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 0 0 0 \\ 000 \\ 0 0 0 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 000 \\ 0 0 0 \end{array}] \\ (ii) A - A ‘ = [\begin{array}{l} 0 ab \\ - a 0 c \\ - b - c 0 \end{array}] - [\begin{array}{l} 0 - a - b \\ a 0 - c \\ b c 0 \end{array}] \\ = [\begin{array}{l} 0 2 a 2 b \\ - 2 a 0 2 c \\ - 2 b - 2 c 0 \end{array}] \\ \frac{1}{2} (A - A ‘) = \frac{1}{2} [\begin{array}{l} 0 2 a 2 b \\ - 2 a 0 2 c \\ - 2 b - 2 c 0 \end{array}] \\ = [\begin{array}{l} 0 a b \\ - a 0 c \\ - b - c 0 \end{array}] \end{array}$

Q.42

$\begin{array}{l} Express the following matrices as the sum of a symmetric \\ and as kew symmetric matrix : \\ (i) [\begin{array}{l} 3 5 \\ 1 - 1 \end{array}] \\ (ii) [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] \\ (iii) [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] \\ (iv) [\begin{array}{l} 15 \\ - 12 \end{array}] \end{array}$

Ans.

$\begin{array}{l} (i) Let A = [\begin{array}{l} 3 5 \\ 1 - 1 \end{array}] \\ ∴ A ‘ = [\begin{array}{l} 3 1 \\ 5 - 1 \end{array}] \\ \frac{1}{2} (A + A ‘) = \frac{1}{2} {[\begin{array}{l} 3 5 \\ 1 - 1 \end{array}] + [\begin{array}{l} 3 1 \\ 5 - 1 \end{array}]} \\ = \frac{1}{2} [\begin{array}{l} 6 6 \\ 6 - 2 \end{array}] \\ = [\begin{array}{l} 3 3 \\ 3 - 1 \end{array}] = B (Let) \\ B ‘ = {[\begin{array}{l} 3 3 \\ 3 - 1 \end{array}]}^{‘} \\ = [\begin{array}{l} 3 3 \\ 3 - 1 \end{array}] \\ So, \frac{1}{2} (A + A ‘) is symmetric matrix . \\ \frac{1}{2} (A - A ‘) = \frac{1}{2} {[\begin{array}{l} 3 5 \\ 1 - 1 \end{array}] - [\begin{array}{l} 3 1 \\ 5 - 1 \end{array}]} \\ = \frac{1}{2} [\begin{array}{l} 0 4 \\ - 4 0 \end{array}] \\ = [\begin{array}{l} 0 2 \\ - 2 0 \end{array}] = C (Let) \\ C ‘ = {[\begin{array}{l} 0 2 \\ - 2 0 \end{array}]}^{‘} \\ = [\begin{array}{l} 0 - 2 \\ 2 0 \end{array}] \\ = - [\begin{array}{l} 0 2 \\ - 2 0 \end{array}] = - C \\ So, \frac{1}{2} (A - A ‘) is skew symmetric matrix . \\ Now, B + C = [\begin{array}{l} 3 3 \\ 3 - 1 \end{array}] + [\begin{array}{l} 0 2 \\ - 2 0 \end{array}] \\ = [\begin{array}{l} 3 5 \\ 1 - 1 \end{array}] = A \\ (ii) \\ Let A = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] \\ ∴ A ‘ = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] \\ A + A ‘ = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] + [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] \\ = [\begin{array}{l} 12 - 4 4 \\ - 4 6 - 2 \\ 4 - 2 6 \end{array}] \\ ∴ \frac{1}{2} (A + A ‘) = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] = B (Let) \\ A - A ‘ = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] - [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] \\ = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ ∴ \frac{1}{2} (A - A ‘) = [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] = C (Let) \\ C is skew symmetric matrix . \\ ∴ B + C = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] + [\begin{array}{l} 0 0 0 \\ 0 0 0 \\ 0 0 0 \end{array}] \\ = [\begin{array}{l} 6 - 2 2 \\ - 2 3 - 1 \\ 2 - 1 3 \end{array}] = A \\ (iii) Let A = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] \\ ∴ A ‘ = [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ A + A ‘ = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] + [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ = [\begin{array}{l} 6 1 - 5 \\ 1 - 4 - 4 \\ - 5 - 4 4 \end{array}] \\ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 6 1 - 5 \\ 1 - 4 - 4 \\ - 5 - 4 4 \end{array}] \\ = [\begin{array}{l} 3 \frac{1}{2} - \frac{5}{2} \\ \frac{1}{2} - 2 - 2 \\ - \frac{5}{2} - 2 2 \end{array}] = B (Let) \\ B ‘ = [\begin{array}{l} 3 \frac{1}{2} - \frac{5}{2} \\ \frac{1}{2} - 2 - 2 \\ - \frac{5}{2} - 2 2 \end{array}] = B \\ So, \frac{1}{2} (A + A ‘) is symmetric matrix . \\ A - A ‘ = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] - [\begin{array}{l} 3 - 2 - 4 \\ 3 - 2 - 5 \\ - 1 1 2 \end{array}] \\ = [\begin{array}{l} 0 53 \\ - 5 06 \\ - 3 - 60 \end{array}] \\ \frac{1}{2} (A - A ‘) = \frac{1}{2} [\begin{array}{l} 0 53 \\ - 5 06 \\ - 3 - 60 \end{array}] \\ = [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} 03 \\ - \frac{3}{2} - 30 \end{array}] = C (Let) \\ C ‘ = {[\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} 03 \\ - \frac{3}{2} - 30 \end{array}]}^{} = - C \\ So, \frac{1}{2} (A - A ‘) is skew symmetric matrix . \\ B + C = [\begin{array}{l} 3 \frac{1}{2} - \frac{5}{2} \\ \frac{1}{2} - 2 - 2 \\ - \frac{5}{2} - 2 2 \end{array}] + [\begin{array}{l} 0 \frac{5}{2} \frac{3}{2} \\ - \frac{5}{2} 03 \\ - \frac{3}{2} - 30 \end{array}] \\ = [\begin{array}{l} 3 3 - 1 \\ - 2 - 2 1 \\ - 4 - 5 2 \end{array}] = A \\ (iv) \\ Let A = [\begin{array}{l} 15 \\ - 12 \end{array}], \\ ∴ A ‘ = [\begin{array}{l} 1 - 1 \\ 5 2 \end{array}] \\ A + A ‘ = [\begin{array}{l} 15 \\ - 12 \end{array}] + [\begin{array}{l} 1 - 1 \\ 5 2 \end{array}] \\ = [\begin{array}{l} 1 + 15 - 1 \\ - 1 + 52 + 2 \end{array}] \\ = [\begin{array}{l} 24 \\ 44 \end{array}] \\ ∴ \frac{1}{2} (A + A ‘) = \frac{1}{2} [\begin{array}{l} 24 \\ 44 \end{array}] \\ = [\begin{array}{l} 12 \\ 22 \end{array}] = B \\ B ‘ = {[\begin{array}{l} 12 \\ 22 \end{array}]}^{‘} \\ = [\begin{array}{l} 12 \\ 22 \end{array}] = B \\ Therefore, \frac{1}{2} (A + A ‘) is symmetric matrix . \\ A - A ‘ = [\begin{array}{l} 15 \\ - 12 \end{array}] - [\begin{array}{l} 1 - 1 \\ 5 2 \end{array}] \\ = [\begin{array}{l} 1 - 15 + 1 \\ - 1 - 52 - 2 \end{array}] \\ = [\begin{array}{l} 06 \\ - 60 \end{array}] \\ ∴ \frac{1}{2} (A - A ‘) \\ = \frac{1}{2} [\begin{array}{l} 06 \\ - 60 \end{array}] \\ = [\begin{array}{l} 03 \\ - 30 \end{array}] = C \\ C ‘ = {[\begin{array}{l} 03 \\ - 30 \end{array}]}^{‘} \\ = [\begin{array}{l} 0 - 3 \\ 3 0 \end{array}] = C \\ Therefore, \frac{1}{2} (A - A ‘) is a skew symmetric matrix . \\ Now, \\ B + C = [\begin{array}{l} 12 \\ 22 \end{array}] + [\begin{array}{l} 03 \\ - 30 \end{array}] \\ = [\begin{array}{l} 1 + 02 + 3 \\ 2 - 32 + 0 \end{array}] \\ = [\begin{array}{l} 15 \\ - 12 \end{array}] = A \end{array}$

Q.43

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

(A) Skew symmetric matrix

(B) Symmetric matrix

(D) Identity matrix

Ans.

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

Now, (AB – BA)’= (AB)’ – (BA)’

= B’A’ – A’B’

= BA – AB

= – (AB–BA)

Thus, AB – BA is a skew symmetric matrix.

Therefore, option (A) is correct solution.

Q.44

$\begin{array}{l} If A = [\begin{array}{l} \cos α - \sin α \\ \sin α \cos α \end{array}], thenA + A ‘ = I, if the value of α is \\ (A) \frac{π}{6} \\ (B) \frac{π}{3} \\ (C) π \\ (D) \frac{3 π}{2} \end{array}$

Ans.

$\begin{array}{l} Given : \\ A = [\begin{array}{l} cosα - sinα \\ sinα cosα \end{array}], A ‘ = [\begin{array}{l} cosαsinα \\ - sinαcosα \end{array}] \\ A + A ‘ = [\begin{array}{l} cosα + cosα - sinα + sinα \\ sinα + (- sinα) cosα + cosα \end{array}] \\ = [\begin{array}{l} 2 cosα 0 \\ 02 cosα \end{array}] = [\begin{array}{l} 10 \\ 01 \end{array}] \\ \Rightarrow 2 cosα = 1 \\ \Rightarrow cosα = \frac{1}{2} = \cos \frac{π}{3} \\ \Rightarrow α = \frac{π}{3} \\ The correct option is (B) . \end{array}$

Q.45

$\begin{array}{l} Using elementary transformations, find the inverse \\ of each of the matrices, [\begin{array}{l} 1 -1 \\ 23 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 1 - 1 \\ 2 3 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 1 - 1 \\ 2 3 \end{array}] = [\begin{array}{l} 10 \\ 01 \end{array}] A \\ Apply R_{2} \to R_{2} - 2 R_{1} \\ [\begin{array}{l} 1 - 1 \\ 0 5 \end{array}] = [\begin{array}{l} 10 \\ - 21 \end{array}] A \\ R_{2} \to \frac{1}{5} R_{2} \\ ∴ [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] = [\begin{array}{l} 1 0 \\ - \frac{2}{5} \frac{1}{5} \end{array}] A \\ Apply R_{1} \to R_{1} + R_{2} \\ [\begin{array}{l} 10 \\ 01 \end{array}] = [\begin{array}{l} 1 0 \\ - \frac{2}{5} \frac{1}{5} \end{array}] A \\ A^{- 1} = [\begin{array}{l} 1 0 \\ - \frac{2}{5} \frac{1}{5} \end{array}] \end{array}$

Q.46

$\begin{array}{l} Using elementary transformations, find the inverse \\ of each of the matrices [\begin{array}{l} 2 1 \\ 1 1 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 1 \\ 1 1 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 1 \\ 1 1 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 0 \\ 1 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] A \\ R_{2} \to R_{2} - R_{1} \\ ∴ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ - 1 2 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 1 - 1 \\ - 1 2 \end{array}] \end{array}$

Q.47

$\begin{array}{l} Using elementary transformations, find the inverse \\ of each of the matrices [\begin{array}{l} 13 \\ 27 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 1 3 \\ 2 7 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 1 3 \\ 2 7 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - 2 R_{1} \\ [\begin{array}{l} 13 \\ 0 1 \end{array}] = [\begin{array}{l} 1 0 \\ - 21 \end{array}] A \\ R_{1} \to R_{1} - 3 R_{2} \\ ∴ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 7 - 3 \\ - 2 1 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 7 - 3 \\ - 2 1 \end{array}] \end{array}$

Q.48

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 23 \\ 57 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 3 \\ 5 7 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 3 \\ 5 7 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \leftrightarrow R_{2} \\ [\begin{array}{l} 5 7 \\ 2 3 \end{array}] = [\begin{array}{l} 0 1 \\ 1 0 \end{array}] A \\ Apply R_{1} \to R_{1} - 2 R_{2} \\ ∴ [\begin{array}{l} 1 1 \\ 2 3 \end{array}] = [\begin{array}{l} - 2 1 \\ 1 0 \end{array}] A \\ Apply R_{2} \to R_{2} - 2 R_{1} \\ [\begin{array}{l} 1 1 \\ 0 1 \end{array}] = [\begin{array}{l} - 2 1 \\ 5 - 2 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} - 7 3 \\ 5 - 2 \end{array}] A \\ A^{- 1} = [\begin{array}{l} - 7 3 \\ 5 - 2 \end{array}] \end{array}$

Q.49

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 21 \\ 74 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 1 \\ 7 4 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 1 \\ 7 4 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - 3 R_{1} \\ ∴ [\begin{array}{l} 2 1 \\ 1 1 \end{array}] = [\begin{array}{l} 1 0 \\ - 3 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 0 \\ 1 1 \end{array}] = [\begin{array}{l} 4 - 1 \\ - 3 1 \end{array}] A \\ Apply R_{2} \to R_{2} - R_{1} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 4 - 1 \\ - 7 2 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 4 - 1 \\ - 7 2 \end{array}] \end{array}$

Q.50

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 25 \\ 13 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 5 \\ 1 3 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 5 \\ 1 3 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ ∴ [\begin{array}{l} 1 2 \\ 1 3 \end{array}] = [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - R_{1} \\ [\begin{array}{l} 1 2 \\ 0 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ - 1 2 \end{array}] A \\ Apply R_{1} \to R_{1} - 2 R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 3 - 5 \\ - 1 2 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 3 - 5 \\ - 1 2 \end{array}] \end{array}$

Q.51

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 31 \\ 52 \end{array}] \end{array}$

`
Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 3 1 \\ 5 2 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 3 1 \\ 5 2 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to 2 R_{1} - R_{2} \\ ∴ [\begin{array}{l} 1 0 \\ 5 2 \end{array}] = [\begin{array}{l} 2 - 1 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - 5 R_{1} \\ [\begin{array}{l} 1 0 \\ 0 2 \end{array}] = [\begin{array}{l} 2 - 1 \\ - 10 6 \end{array}] A \\ Apply R_{2} \to \frac{1}{2} R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 2 - 1 \\ - 5 3 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 2 - 1 \\ - 5 3 \end{array}] \end{array}$

Q.52

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 45 \\ 34 \end{array}] \end{array}$

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 4 5 \\ 3 4 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 4 5 \\ 3 4 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ ∴ [\begin{array}{l} 1 1 \\ 3 4 \end{array}] = [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - 3 R_{1} \\ [\begin{array}{l} 1 1 \\ 0 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ - 3 4 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 4 - 5 \\ - 3 4 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 4 - 5 \\ - 3 4 \end{array}] \end{array}$

Q.53

$\begin{array}{l} Using elementary transformations, find the inverse of \\ each of the matrices [\begin{array}{l} 310 \\ 27 \end{array}] \end{array}$

Ans.

$\begin{matrix} Let A = [\begin{array}{l} 3 10 \\ 2 7 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 3 10 \\ 2 7 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ S 0, [\begin{array}{l} 13 \\ 27 \end{array}] = [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - 2 R_{1} \\ [\begin{array}{l} 1 3 \\ 0 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ - 2 3 \end{array}] A \\ Apply R_{1} \to R_{1} - 3 R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 7 - 10 \\ - 2 3 \end{array}] A \\ A^{- 1} = [\begin{array}{l} 7 - 10 \\ - 2 3 \end{array}] \end{matrix}$

Q.54

$\begin{array}{l} Using elementary transformations, find the inverse \\ of each of the matrices [\begin{array}{l} 3 -1 \\ - 42 \end{array}] \end{array}$

Ans.

$\begin{matrix} Let A = [\begin{array}{l} 3 - 1 \\ - 4 2 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 3 - 1 \\ - 4 2 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \leftrightarrow R_{2} \\ ∴ [\begin{array}{l} - 4 2 \\ 3 - 1 \end{array}] = [\begin{array}{l} 0 1 \\ 1 0 \end{array}] A \\ Apply R_{1} \to (- 1) R_{1} \\ [\begin{array}{l} 4 - 2 \\ 3 - 1 \end{array}] = [\begin{array}{l} 0 - 1 \\ 1 0 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 - 1 \\ 3 - 1 \end{array}] = [\begin{array}{l} - 1 - 1 \\ 1 0 \end{array}] A \\ Apply R_{2} \to R_{2} - 3 R_{1} \\ [\begin{array}{l} 1 - 1 \\ 0 2 \end{array}] = [\begin{array}{l} - 1 - 1 \\ 4 3 \end{array}] A \\ Apply R_{2} \to \frac{1}{2} R_{2} \\ [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] = [\begin{array}{l} - 1 - 1 \\ 2 \frac{3}{2} \end{array}] A \\ Apply R_{1} \to R_{1} + R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 1 \frac{1}{2} \\ 2 \frac{3}{2} \end{array}] A \Rightarrow A^{- 1} = [\begin{array}{l} 1 \frac{1}{2} \\ 2 \frac{3}{2} \end{array}] \end{matrix}$

Q.55

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

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 - 6 \\ 1 - 2 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 - 6 \\ 1 - 2 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 - 4 \\ 1 - 2 \end{array}] = [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] A \\ Apply R_{2} \to R_{2} - R_{1} \\ [\begin{array}{l} 1 - 4 \\ 0 2 \end{array}] = [\begin{array}{l} 1 - 1 \\ - 1 2 \end{array}] A \\ Apply R_{2} \to \frac{1}{2} R_{2} \\ [\begin{array}{l} 1 - 4 \\ 0 1 \end{array}] = [\begin{array}{l} 1 - 1 \\ - \frac{1}{2} 1 \end{array}] A \\ Apply R_{1} \to R_{1} + 4 R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} - 1 3 \\ - \frac{1}{2} 1 \end{array}] A \\ ∴ A^{- 1} = [\begin{array}{l} - 1 3 \\ - \frac{1}{2} 1 \end{array}] \end{array}$

Q.56

$\begin{array}{r} Using elementary transformations, find the \\ inverse of each of the matrices [\begin{array}{l} 6 -3 \\ - 21 \end{array}] \end{array}$

Ans.

$\begin{array}{l} L e t A = [\begin{array}{l} 6 - 3 \\ - 2 1 \end{array}] \\ S i n c e, A = IA \\ [\begin{array}{l} 6 - 3 \\ - 2 1 \end{array}] = [\begin{array}{l} 10 \\ 0 1 \end{array}] A \\ A p p l y R_{1} \leftrightarrow R_{2} \\ [\begin{array}{l} - 2 1 \\ 6 - 3 \end{array}] = [\begin{array}{l} 0 1 \\ 10 \end{array}] A \\ A p p l y R_{2} \to R_{2} + 3 R_{1} \\ [\begin{array}{l} 1 - 4 \\ 0 0 \end{array}] = [\begin{array}{l} 01 \\ 1 3 \end{array}] A \\ In second row of L .H .S ., all elements are zero . \\ ∴ A^{- 1} does not exist . \end{array}$

Q.57

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

Ans.

$\begin{array}{l} L e t A = [\begin{array}{l} 2 - 3 \\ - 1 2 \end{array}] \\ S i n c e, A = IA \\ [\begin{array}{l} 2 - 3 \\ - 1 2 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ A p p l y R_{1} \to R_{1} + R_{2} \\ [\begin{array}{l} 1 - 1 \\ - 1 2 \end{array}] = [\begin{array}{l} 1 1 \\ 0 1 \end{array}] A \\ A p p l y R_{2} \to R_{2} + R_{1} \\ [\begin{array}{l} 1 - 1 \\ 0 1 \end{array}] = [\begin{array}{l} 1 1 \\ 1 2 \end{array}] A \\ A p p l y R_{1} \to R_{1} + R_{2} \\ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} 2 3 \\ 1 2 \end{array}] A \\ ∴ A^{- 1} = [\begin{array}{l} 2 3 \\ 1 2 \end{array}] \end{array}$

Q.58

$\begin{array}{l} Using elementary transformations, find the \\ inverse of each of the matrices [\begin{array}{l} 21 \\ 42 \end{array}] \end{array}$

Ans.

$\begin{array}{l} L e t A = [\begin{array}{l} 2 1 \\ 4 2 \end{array}] \\ S i n c e, A = IA \\ [\begin{array}{l} 2 1 \\ 4 2 \end{array}] = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] A \\ A p p l y R_{2} \to R_{2} - 2 R_{1} \\ [\begin{array}{l} 2 1 \\ 00 \end{array}] = [\begin{array}{l} 1 0 \\ - 2 1 \end{array}] A \\ S i n c e, in second row of L .H .S ., all elements are zero . \\ {So, A}^{-1} does not exist . \end{array}$

Q.59

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

Ans.

$\begin{array}{l} Let A = [\begin{array}{l} 2 - 3 3 \\ 2 2 3 \\ 3 - 2 2 \end{array}] \\ Since, A = IA \\ [\begin{array}{l} 2 - 3 3 \\ 2 2 3 \\ 3 - 2 2 \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A \\ Apply R_{1} \leftrightarrow R_{3} \\ [\begin{array}{l} 3 - 2 2 \\ 2 2 3 \\ 2 - 3 3 \end{array}] = [\begin{array}{l} 0 0 1 \\ 0 1 0 \\ 1 0 0 \end{array}] A \\ Apply R_{1} \to R_{1} - R_{2} \\ [\begin{array}{l} 1 - 4 - 1 \\ 2 2 3 \\ 2 - 3 3 \end{array}] = [\begin{array}{l} 0 - 1 1 \\ 0 1 0 \\ 1 0 0 \end{array}] A \\ Apply R_{2} \to R_{2} - 2 R_{1} and R_{3} \to R_{3} - 2 R_{1} \\ [\begin{array}{l} 1 - 4 - 1 \\ 0 10 5 \\ 0 5 5 \end{array}] = [\begin{array}{l} 0 - 1 1 \\ 0 3 - 2 \\ 1 2 - 2 \end{array}] A \\ Apply R_{2} \to R_{2} - R_{3} \\ [\begin{array}{l} 1 - 4 - 1 \\ 0 5 0 \\ 0 5 5 \end{array}] = [\begin{array}{l} 0 - 1 1 \\ - 1 1 0 \\ 1 2 - 2 \end{array}] A \\ Apply R_{2} \to \frac{1}{5} R_{2} and R_{3} \to \frac{1}{5} R_{3} \\ [\begin{array}{l} 1 - 4 - 1 \\ 0 1 0 \\ 0 1 1 \end{array}] = [\begin{array}{l} 0 - 1 1 \\ - \frac{1}{5} \frac{1}{5} 0 \\ \frac{1}{5} \frac{2}{5} - \frac{2}{5} \end{array}] A \\ Apply R_{3} \to R_{3} - R_{2} \\ [\begin{array}{l} 1 - 4 - 1 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} 0 - 1 1 \\ - \frac{1}{5} \frac{1}{5} 0 \\ \frac{2}{5} \frac{1}{5} - \frac{2}{5} \end{array}] A \\ Apply R_{1} \to R_{1} + 4 R_{2} \\ [\begin{array}{l} 1 0 - 1 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} - \frac{4}{5} - \frac{1}{5} 1 \\ - \frac{1}{5} \frac{1}{5} 0 \\ \frac{2}{5} \frac{1}{5} - \frac{2}{5} \end{array}] A \\ Apply R_{1} \to R_{1} + R_{3} \\ [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} - \frac{2}{5} 0 \frac{3}{5} \\ - \frac{1}{5} \frac{1}{5} 0 \\ \frac{2}{5} \frac{1}{5} - \frac{2}{5} \end{array}] A \\ ∴ A^{- 1} = [\begin{array}{l} - \frac{2}{5} 0 \frac{3}{5} \\ - \frac{1}{5} \frac{1}{5} 0 \\ \frac{2}{5} \frac{1}{5} - \frac{2}{5} \end{array}] \end{array}$

Q.60

$Using elementary transformations, find the inverse of each of the matrices [\begin{array}{l} 1 3 - 2 \\ - 30 - 5 \\ 2 5 0 \end{array}]$

Ans.

$\begin{matrix} L e t A = [\begin{array}{l} 1 3 - 2 \\ - 3 0 - 5 \\ 2 5 0 \end{array}] \\ S i n c e, A = IA \\ [\begin{array}{l} 1 3 - 2 \\ - 3 0 - 5 \\ 2 5 0 \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A \\ A p p l y R_{2} \to R_{2} + 3 R_{1} a n d R_{3} \to R_{3} - 2 R_{1}, we get \\ [\begin{array}{l} 1 3 - 2 \\ 0 9 - 11 \\ 0 - 1 4 \end{array}] = [\begin{array}{l} 1 0 0 \\ 3 1 0 \\ - 2 0 1 \end{array}] A \\ A p p l y R_{2} \to R_{2} + 8 R_{3} \\ [\begin{array}{l} 1 3 - 2 \\ 0 1 21 \\ 0 - 1 4 \end{array}] = [\begin{array}{l} 1 0 0 \\ - 13 1 8 \\ - 2 0 1 \end{array}] A \\ A p p l y R_{1} \to R_{1} - 3 R_{2} a n d R_{3} \to R_{3} + R_{2} \\ [\begin{array}{l} 1 0 - 65 \\ 0 1 21 \\ 0 0 25 \end{array}] = [\begin{array}{l} 40 - 3 - 24 \\ - 13 1 8 \\ - 15 1 9 \end{array}] A \\ A p p l y R_{3} \to \frac{1}{25} R_{3} \\ [\begin{array}{l} 1 0 - 65 \\ 0 1 21 \\ 0 0 1 \end{array}] = [\begin{array}{l} 40 - 3 - 24 \\ - 13 1 8 \\ - \frac{15}{25} \frac{1}{25} \frac{9}{25} \end{array}] A \\ = \frac{1}{25} [\begin{array}{l} 40 - 3 - 24 \\ - 13 1 8 \\ - 15 1 9 \end{array}] A \\ A p p l y R_{1} \to R_{1} + 65 R_{3} a n d R_{2} \to R_{2} + 21 R_{3}, w e g e t \\ [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = \frac{1}{25} [\begin{array}{l} 25 - 10 - 15 \\ - 10 4 11 \\ - 15 1 9 \end{array}] A \\ = [\begin{array}{l} 1 - \frac{10}{25} - \frac{15}{25} \\ - \frac{10}{25} \frac{4}{25} \frac{11}{25} \\ - \frac{15}{25} \frac{1}{25} \frac{9}{25} \end{array}] A \\ = [\begin{array}{l} 1 - \frac{2}{5} - \frac{3}{5} \\ - \frac{2}{5} \frac{4}{25} \frac{11}{25} \\ - \frac{3}{5} \frac{1}{25} \frac{9}{25} \end{array}] A \\ T h e r e f o r e, A^{- 1} = [\begin{array}{l} 1 - \frac{2}{5} - \frac{3}{5} \\ - \frac{2}{5} \frac{4}{25} \frac{11}{25} \\ - \frac{3}{5} \frac{1}{25} \frac{9}{25} \end{array}] \end{matrix}$

Q.61

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

Ans.

$\begin{array}{l} L e t A = [\begin{array}{l} 2 0 - 1 \\ 5 1 0 \\ 0 1 3 \end{array}] \\ S i n c e, A = IA \\ [\begin{array}{l} 2 0 - 1 \\ 5 1 0 \\ 0 1 3 \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] A \\ A p p l y in g R_{1} \leftrightarrow R_{2}, we get \\ [\begin{array}{l} 5 1 0 \\ 2 0 - 1 \\ 0 1 3 \end{array}] = [\begin{array}{l} 0 1 0 \\ 1 0 0 \\ 0 0 1 \end{array}] A \\ Applying R_{1} \to R_{1} - 2 R_{2}, w e g e t \\ [\begin{array}{l} 1 1 2 \\ 2 0 - 1 \\ 0 1 3 \end{array}] = [\begin{array}{l} - 2 1 0 \\ 1 0 0 \\ 0 0 1 \end{array}] A \\ Applying R_{2} \to R_{2} - 2 R_{1}, w e g e t \\ [\begin{array}{l} 1 1 2 \\ 0 - 2 - 5 \\ 0 1 3 \end{array}] = [\begin{array}{l} - 2 1 0 \\ 5 - 2 0 \\ 0 0 1 \end{array}] A \\ Applying R_{2} \to (- 1) R_{2}, w e g e t \\ [\begin{array}{l} 1 1 2 \\ 0 2 5 \\ 0 1 3 \end{array}] = [\begin{array}{l} - 2 1 0 \\ - 5 2 0 \\ 0 0 1 \end{array}] A \\ Applying R_{2} \to R_{2} - R_{3}, w e g e t \\ [\begin{array}{l} 1 1 2 \\ 0 1 2 \\ 0 1 3 \end{array}] = [\begin{array}{l} - 2 1 0 \\ - 5 2 - 1 \\ 0 0 1 \end{array}] A \\ Applying R_{1} \to R_{1} - R_{2}, w e g e t \\ [\begin{array}{l} 1 0 0 \\ 0 1 2 \\ 0 1 3 \end{array}] = [\begin{array}{l} 3 - 1 1 \\ - 5 2 - 1 \\ 0 0 1 \end{array}] A \\ Applying R_{3} \to R_{3} - R_{2}, w e g e t \\ [\begin{array}{l} 1 0 0 \\ 0 1 2 \\ 0 0 1 \end{array}] = [\begin{array}{l} 3 - 1 1 \\ - 5 2 - 1 \\ 5 - 2 2 \end{array}] A \\ Applying R_{2} \to R_{2} - 2 R_{3}, w e g e t \\ [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] = [\begin{array}{l} 3 - 1 1 \\ - 15 6 - 5 \\ 5 - 2 2 \end{array}] A \Rightarrow A^{- 1} = [\begin{array}{l} 3 - 1 1 \\ - 15 6 - 5 \\ 5 - 2 2 \end{array}] \end{array}$

Q.62

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

(A) AB = BA

(B) AB = BA = 0

(D) AB = BA = I

Ans.

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

Thus, option (D) is correct.

Q.63

$\begin{array}{l} Let A = [\begin{array}{l} 0 1 \\ 0 0 \end{array}], show that {(aI + bA)}^{n} = a^{n} I + {na}^{n - 1}, \\ where is the identity matrix of order 2 and n \in N . \end{array}$

Ans.

$\begin{array}{l} L . H . S . = {(a I + b A)}^{n} \\ = {(a [\begin{array}{l} 1 0 \\ 01 \end{array}] + b [\begin{array}{l} 0 1 \\ 0 0 \end{array}])}^{n} \\ = {[\begin{array}{l} a b \\ 0 a \end{array}]}^{n} \\ R . H . S . = a^{n} I + n a^{n - 1} b A \\ = a^{n} [\begin{array}{l} 1 0 \\ 01 \end{array}] + n a^{n - 1} b [\begin{array}{l} 01 \\ 00 \end{array}] \\ = [\begin{array}{l} a^{n} 0 \\ 0 a^{n} \end{array}] + [\begin{array}{l} 0 n a^{n - 1} b \\ 00 \end{array}] \\ = [\begin{array}{l} a^{n} n a^{n - 1} b \\ 0 a^{n} \end{array}] \\ S o, {[\begin{array}{l} a b \\ 0 a \end{array}]}^{n} = [\begin{array}{l} a^{n} n a^{n - 1} b \\ 0 a^{n} \end{array}] \\ Apply principle of Mathematical Induction \\ P (n) : {[\begin{array}{l} a b \\ 0 a \end{array}]}^{n} = [\begin{array}{l} a^{n} n a^{n - 1} b \\ 0 a^{n} \end{array}] \\ F o r n = 1, \\ L . H . S . = {[\begin{array}{l} a b \\ 0 a \end{array}]}^{1} = [\begin{array}{l} a b \\ 0 a \end{array}] \\ R . H . S = [\begin{array}{l} a^{1} 1. a^{1 - 1} b \\ 0 a^{1} \end{array}] = [\begin{array}{l} a a b \\ 0 a \end{array}] \\ S o, \\ [\begin{array}{l} a b \\ 0 a \end{array}] = [\begin{array}{l} a a b \\ 0 a \end{array}] \\ S o, \\ [\begin{array}{l} a b \\ 0 a \end{array}] = [\begin{array}{l} a a b \\ 0 a \end{array}] \\ ∴ P (n) is true for n = 1. \\ Let P (n) be true for n = k . \\ ∴ P (k) : {[\begin{array}{l} a b \\ 0 a \end{array}]}^{k} = [\begin{array}{l} a^{k} k a^{k - 1} b \\ 0 a^{k} \end{array}] … (i) \\ Multiple both sides by [\begin{array}{l} a b \\ 0 a \end{array}] \\ {[\begin{array}{l} a b \\ 0 a \end{array}]}^{k} \times [\begin{array}{l} a b \\ 0 a \end{array}] = [\begin{array}{l} a^{k} k a^{k - 1} b \\ 0 a^{k} \end{array}] \times [\begin{array}{l} a b \\ 0 a \end{array}] \\ L . H . S . = {[\begin{array}{l} a b \\ 0 a \end{array}]}^{k} \times [\begin{array}{l} a b \\ 0 a \end{array}] \\ = [\begin{array}{l} a^{k} k a^{k - 1} b \\ 0 a^{k} \end{array}] \times [\begin{array}{l} a b \\ 0 a \end{array}] \\ [F r o m equation (i), w e g e t] \\ = [\begin{array}{l} a^{k + 1} a^{k} b + k a^{k} b \\ 0 a^{k + 1} \end{array}] \\ = [\begin{array}{l} a^{k + 1} (k + 1) a^{k} b \\ 0 a^{k + 1} \end{array}] \\ L . H . S . = [\begin{array}{l} a^{k} k a^{k - 1} b \\ 0 a^{k} \end{array}] \times [\begin{array}{l} a b \\ 0 a \end{array}] \\ = [\begin{array}{l} a^{k + 1} b a^{k} + k a^{k} b \\ 0 a^{k + 1} \end{array}] \\ = [\begin{array}{l} a^{k + 1} (k + 1) a^{k} b \\ 0 a^{k + 1} \end{array}] \\ This shows P(n) is true for n = k + 1 then by principle of mathematical \\ induction, P(n) is true for all positive integral values of n . \end{array}$

Q.64

$If A= [\begin{array}{l} 1 1 1 \\ 1 1 1 \\ 1 1 1 \end{array}], prove that A^{n} = [\begin{array}{l} 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \\ 3^{n - 1} 3^{n - 1} 3^{n - 1} \end{array}], n \in N .$

Ans.

$\begin{matrix} Let P (n) : A^{n} = [\begin{array}{l} 3^{n -1} 3^{n -1} 3^{n -1} \\ 3^{n -1} 3^{n -1} 3^{n -1} \\ 3^{n -1} 3^{n -1} 3^{n -1} \end{array}] and A = [\begin{array}{l} 1 1 1 \\ 1 1 1 \\ 1 1 1 \end{array}] \\ For n = 1 \\ L . H . S . = A = [\begin{array}{l} 1 1 1 \\ 1 1 1 \\ 1 1 1 \end{array}] \\ R . H . S . = [\begin{array}{l} 3^{1 -1} 3^{1 -1} 3^{1 -1} \\ 3^{1 -1} 3^{1 -1} 3^{1 -1} \\ 3^{1 -1} 3^{1 -1} 3^{1 -1} \end{array}] \\ = [\begin{array}{l} 3^{0} 3^{0} 3^{0} \\ 3^{0} 3^{0} 3^{0} \\ 3^{0} 3^{0} 3^{0} \end{array}] \\ = [\begin{array}{l} 1 1 1 \\ 1 1 1 \\ 1 1 1 \end{array}] \\ ∴ P (n) is true for n = 1 . \\ Let P (n) be true for n = k, then \\ P (k) : A^{k} = [\begin{array}{l} 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \end{array}] . .. (i) \\ Multiplying both sides by A, we get \\ A^{k} . A = [\begin{array}{l} 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \end{array}] . A \\ A^{k +1} = [\begin{array}{l} 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \end{array}] . [\begin{array}{l} 1 1 1 \\ 1 1 1 \\ 1 1 1 \end{array}] \\ R . H . S . = [\begin{array}{l} 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \\ 3^{k -1} 3^{k -1} 3^{k -1} \end{array}] . [\begin{array}{l} 111 \\ 111 \\ 111 \end{array}] \\ = [\begin{array}{l} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \\ 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \\ 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} 3^{k - 1} + 3^{k - 1} + 3^{k - 1} \end{array}] \\ = [\begin{array}{l} 3 . 3^{k - 1} 3 . 3^{k - 1} 3 . 3^{k - 1} \\ 3 . 3^{k - 1} 3 . 3^{k - 1} 3 . 3^{k - 1} \\ 3 . 3^{k - 1} 3 . 3^{k - 1} 3 . 3^{k - 1} \end{array}] \\ = [\begin{array}{l} 3^{(k - 1) + 1} 3^{(k - 1) + 1} 3^{(k - 1) + 1} \\ 3^{(k - 1) + 1} 3^{(k - 1) + 1} 3^{(k - 1) + 1} \\ 3^{(k - 1) + 1} 3^{(k - 1) + 1} 3^{(k - 1) + 1} \end{array}] \\ ∴ P (n) is true forn = k + 1 \\ By principle of ma the matical induction that P (n) is true f oral \ln \in N . \end{matrix}$

Q.65

$\begin{array}{l} If A = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}], then prove tha t A^{n} = [\begin{array}{l} 1 + 2 n - 4 n \\ n 1 - 2 n \end{array}], \\ where n is any positive integer . \end{array}$

Ans.

$\begin{array}{l} Let P (n) : A^{n} = [\begin{array}{l} 1 +2 n - 4 n \\ n 1 - 2 n \end{array}], where A = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}] \\ Put n = 1, \\ A = [\begin{array}{l} 1 +2 (1) - 4 (1) \\ (1) 1 - 2 (1) \end{array}] = [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}] \\ ∴ P (n) is true for n = 1. \\ Let p (n) be true for n = k \\ ∴ P (k) : A^{k} = [\begin{array}{l} 1 +2 k - 4 k \\ k 1 - 2 k \end{array}] \\ Multiply both sides by A, we get \\ A^{k} . A = A^{k +1} = [\begin{array}{l} 1 +2 (k +1) - 4 (k +1) \\ (k +1) 1 - 2 (k +1) \end{array}] \\ L . H . S . = A^{k} . A \\ = [\begin{array}{l} 1 +2 k - 4 k \\ k 1 - 2 k \end{array}] . [\begin{array}{l} 3 - 4 \\ 1 - 1 \end{array}] \\ = [\begin{array}{l} 3 (1 + 2 k) - 4 k - 4 (1 + 2 k) + 4 k \\ 3 k + 1 - 2 k - 4 k - 1 (1 - 2 k) \end{array}] \\ = [\begin{array}{l} 3 + 6 k - 4 k - 4 - 8 k + 4 k \\ 3 k + 1 - 2 k - 4 k - 1 + 2 k \end{array}] \\ = [\begin{array}{l} 3 + 2 k - 4 - 4 k \\ k + 1 - 2 k - 1 \end{array}] \\ = [\begin{array}{l} 1 + 2 (1 + k) - 4 (1 + k) \\ (k + 1) 1 - 2 (k + 1) \end{array}] = R . H . S . \\ Therefore, P (n) is true for n = k +1 . \\ Henc e, P (n) is true for all n \in N . \end{array}$

Q.66

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

Ans.

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

Now, (AB – BA)’= (AB)’ – (BA)’

= B’A’ – A’B’

= BA – AB

= – (AB – BA)

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

Q.67

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

Ans.

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

Then,

(B’AB)’ = B’(B’A)’

= B’A’(B’)’

= B’A’B

= B’AB [Since, A’ = A]

Therefore, B’AB is symmetric matrix.

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

Then,

(B’AB)’ = B’(B’A)’

= B’A’(B’)’

= B’A’B

= B’(– A)B [Since, A’ = A]

= –(B’AB)

Therefore, B’AB is skew symmetric matrix.

Q.68

$\begin{array}{l} Find the values of x, y, z if the matrix A = [\begin{array}{l} 0 2 y z \\ x y - z \\ x - y z \end{array}] \\ satisfiy the equation A ‘ A = I . \end{array}$

Ans.

$\begin{array}{l} A = [\begin{array}{l} 0 2 y z \\ x y - z \\ x - y z \end{array}] \\ A ‘ = [\begin{array}{l} 0 x x \\ 2 y y - y \\ z - z z \end{array}] \\ A ‘ A = [\begin{array}{l} 0 x x \\ 2 y y - y \\ z - z z \end{array}] [\begin{array}{l} 0 2 y z \\ x y - z \\ x - y z \end{array}] \\ = [\begin{array}{l} 0 + x^{2} + x^{2} 0 + xy - xy 0 - xz + xz \\ 0 + xy - xy 4 y^{2} + y^{2} + y^{2} 2 yz - yz - yz \\ 0 - xz + xz 2 yz - zy - yz z^{2} + z^{2} + z^{2} \end{array}] \\ = [\begin{matrix} 2 x^{2} & 0 & 0 \\ 0 & 6 y^{2} & 0 \\ 0 & 0 & 3 z^{2} \end{matrix}] \\ Since, A ‘ A = I \\ So, \\ [\begin{array}{l} 2 x^{2} 0 0 \\ 0 6 y^{2} 0 \\ 0 0 3 z^{2} \end{array}] = [\begin{array}{l} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] \\ \Rightarrow 2 x^{2} = 1, 6 y^{2} = 1 and 3 z^{2} = 1 \\ \Rightarrow x = \pm \frac{1}{\sqrt{2}}, y = \pm \frac{1}{\sqrt{6}} and z = \pm \frac{1}{\sqrt{3}} \end{array}$

Q.69

$F o r w h a t v a l u e o f x : [1 2 1] [\begin{array}{l} 1 2 0 \\ 2 0 1 \\ 1 0 2 \end{array}] [\begin{array}{l} 0 \\ 2 \\ x \end{array}] = 0$

Ans.

$\begin{array}{l} [1 2 1] [\begin{array}{l} 1 2 0 \\ 2 0 1 \\ 1 0 2 \end{array}] [\begin{array}{l} 0 \\ 2 \\ x \end{array}] = 0 \\ [1 + 4 + 1 2 + 0 + 0 0 + 2 + 2] [\begin{array}{l} 0 \\ 2 \\ x \end{array}] = 0 \\ [6 2 4] [\begin{array}{l} 0 \\ 2 \\ x \end{array}] = 0 \\ [0 + 4 + 4 x] = 0 \\ 4 + 4 x = 0 \\ \Rightarrow x = - 1 \end{array}$

Q.70

$If A = [\begin{array}{l} 31 \\ - 12 \end{array}], show that A^{2} - 5 A + 7 I = 0 .$

Ans.

$\begin{array}{l} Since, A = [\begin{array}{l} 3 1 \\ - 1 2 \end{array}], then \\ A^{2} = [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] \\ = [\begin{array}{l} 9 - 1 3 + 2 \\ - 3 - 2 - 1 + 4 \end{array}] \\ = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] \\ L . H . S . = A^{2} - 5 A + 7 I \\ = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] - 5 [\begin{array}{l} 3 1 \\ - 1 2 \end{array}] + 7 [\begin{array}{l} 1 0 \\ 0 1 \end{array}] \\ = [\begin{array}{l} 8 5 \\ - 5 3 \end{array}] - [\begin{array}{l} 15 5 \\ - 5 10 \end{array}] + [\begin{array}{l} 7 0 \\ 0 7 \end{array}] \\ = [\begin{array}{l} 8 - 15 + 7 5 - 5 + 0 \\ - 5 + 5 + 0 3 - 10 + 7 \end{array}] \\ = [\begin{array}{l} 0 0 \\ 0 0 \end{array}] \\ = 0 = R . H . S . \end{array}$

Q.71

$Find x, if [x-5-1] [\begin{array}{l} 1 0 2 \\ 0 2 1 \\ 2 0 3 \end{array}] [\begin{array}{l} x \\ 4 \\ 1 \end{array}] = 0 .$

Ans.

$\begin{array}{l} [x - 5 - 1] [\begin{array}{l} 102 \\ 021 \\ 203 \end{array}] [\begin{array}{l} x \\ 4 \\ 1 \end{array}] = 0 \\ \Rightarrow [x + 0 - 20 - 10 + 02 x - 5 - 3] [\begin{array}{l} x \\ 4 \\ 1 \end{array}] = 0 \\ \Rightarrow [x - 2 - 102 x - 8] [\begin{array}{l} x \\ 4 \\ 1 \end{array}] = 0 \\ \Rightarrow [x (x - 2) - 40 + 2 x - 8] = 0 \\ \Rightarrow x^{2} - 2 x - 40 + 2 x - 8 = 0 \\ \Rightarrow x^{2} - 48 = 0 \\ \Rightarrow x = \pm \sqrt{48} \\ = \pm 4 \sqrt{3} \end{array}$

Q.72

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

Market Products

I 10,000 2,000 18,000

II 6,000 20,000 8,000

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

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

Ans.

$\begin{array}{l} (a) Matrix corresponding to sale price of each product = [\begin{array}{l} 2.50 \\ 1.50 \\ 1.00 \end{array}] \\ The total revenue collected by each market is given by \\ [\begin{array}{l} 10,0002,00018,000 \\ 6,00020,0008,000 \end{array}] [\begin{array}{l} 2.50 \\ 1.50 \\ 1.00 \end{array}] = [\begin{array}{l} 25,000+3,000+18,000 \\ 15,000+30,000+8,000 \end{array}] \\ = [\begin{array}{l} 46,000 \\ 53,000 \end{array}] \\ Total cost prices of the commodities each market I and II are \\ ‘ 46,000, ‘ 53,000 . \\ Total revnue by both markets = 46,000+ 53,000 \\ = 99,000 \\ (b) The cost price of commodities x, y and z are respectively ‘ 2 . 00, \\ ‘ 1.00 and ‘ 0 . 50 cost price of each market is given below \\ [\begin{array}{l} 10,0002,00018,000 \\ 6,00020,0008,000 \end{array}] [\begin{array}{l} 2.00 \\ 1.00 \\ 0.50 \end{array}] = [\begin{array}{l} 20,000+2,000+9,000 \\ 12,000+20,000+4,000 \end{array}] \\ = [\begin{array}{l} 31000 \\ 36000 \end{array}] \\ Total cost prices of the commodities each market I and II are 31,000, \\ ZZ 36,000 . \\ Total revnue by both markets = 31,000+ 36,000 \\ = 67,000 \\ Gross profit = 99,000- 67,000 \\ = 32,000 \end{array}$

Q.73

$Find the matrix X so that X [\begin{array}{l} 1 2 3 \\ 4 5 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}]$

Ans.

$\begin{array}{l} X [\begin{array}{l} 12 3 \\ 45 6 \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ Let order of matrix X be m \times n . \\ Since, the order of [\begin{array}{l} 1 2 3 \\ 45 6 \end{array}] is 2 \times 3 and order of matrix X \\ is m \times n . \\ So, n = 2 . \\ Now, the order of [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] is 2 \times 3 and order of matrix X is m \times n . \\ So, m = 2 . \\ Thus, order of matrix X is 2 \times 2 . \\ Let X = [\begin{array}{l} a b \\ c d \end{array}], then \\ [\begin{array}{l} a b \\ c d \end{array}] [\begin{array}{l} 1 23 \\ 4 56 \end{array}] [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] = \\ [\begin{array}{l} a + 4 b 2 a + 5 b 3 a + 6 b \\ c + 4 d 2 c + 5 d 3 c + 6 d \end{array}] = [\begin{array}{l} - 7 - 8 - 9 \\ 2 4 6 \end{array}] \\ \Rightarrow a + 4 b = - 7 . .. (i) \\ 2 a + 5 b = - 8 . .. (ii) \\ c + 4 d = 2 . .. (iii) \\ 2 c + 5 d = 4 . . . (iv) \\ 3 c + 6 d = 6 . .. (v) \\ Solving equation (i) and equation (ii), we get \\ a = 1, b = - 2 \\ value of a and b satis fies 3 a +6 b = - 9 . \\ Solving equation (iii) and equation (iv), we get \\ c = 2, d = 0 \\ value of c and d satisfies 3 c + 6 d = 6 . \\ Hence, X = [\begin{array}{l} 1 - 2 \\ 2 0 \end{array}] \end{array}$

Q.74

$\begin{array}{l} If A and B are square matric e so f thesame that AB = BA, \\ then prove by induction that {AB}^{n} = {BA}^{n} . Further, prove \\ that {(AB)}^{n} = A^{n} B^{n} forall n \in N . \end{array}$

Ans.

$\begin{array}{l} (i) Let P (n) : {AB}^{n} = B^{n} A \\ For n = 1 \\ P (1) : AB = BA [Given] \\ ∴ P (n) is true for n = 1. \\ Let P (n) be true for n = k . \\ P (k) : {AB}^{k} = B^{k} A . .. (i) \\ For n = k + 1 \\ {AB}^{k +1} = B^{k +1} A . .. (ii) \\ L . H . S . = {AB}^{k +1} \\ = ({AB}^{k}) B \\ = B ({AB}^{k}) \\ = B (B^{k} A) [From equation (i)] \\ = B^{k +1} A = R . H . S . \\ So, P (n) is true for n = k + 1 . \\ Therefore, by the Principle of mathematical induction, \\ P (n) is true for all n \in N . \\ (ii) Let P (n) : {(AB)}^{n} = A^{n} B^{n} \\ For n = 1 \\ L . H . S . = AB \\ R . H . S . = AB \\ ∴ P (n) is true for n = 1. \\ Let P (n) be true for n = k . \\ P (k) : {(AB)}^{k} = A^{k} B^{k} . .. (i) \\ Multiply both sides by AB, we get \\ {(AB)}^{k} (AB) = A^{k} B^{k} (AB) \\ L . H . S . = {(AB)}^{k + 1} \\ R . H . S . = A^{k} B^{k} (AB) \\ = A^{k} B^{k} (BA) [AB = BA] \\ = A^{k} (B^{k} . B) A \\ = A^{k} (B^{k + 1} A) \\ = A^{k} ({AB}^{k + 1}) [âˆµ {AB}^{n} = B^{n} A] \\ = A^{k + 1} B^{k + 1} \\ Therefore, P (n) is true for n = k + 1 . \\ By the principle of mathematical inducation, \\ P (n) is true for n \in N . \end{array}$

Q.75

$\begin{array}{l} If A = [\begin{array}{l} α β \\ γ - α \end{array}] is such that A^{2} = I, then \\ (A) 1 + α^{2} + βγ = 0 \\ (B) 1 - α^{2} + βγ = 0 \\ (C) 1 - α^{2} - βγ = 0 \\ (D) 1 + α^{2} - βγ = 0 \end{array}$

Ans.

$\begin{array}{l} A^{2} = [\begin{array}{l} α β \\ γ - α \end{array}] [\begin{array}{l} α β \\ γ - α \end{array}] \\ = [\begin{array}{l} α^{2} + βγ αβ - αβ \\ αγ - αγ βγ + α^{2} \end{array}] \\ âˆµ A^{2} = [\begin{array}{l} 1 0 \\ 0 1 \end{array}] \\ ∴ [\begin{array}{l} 1 0 \\ 0 1 \end{array}] = [\begin{array}{l} α^{2} + βγ 0 \\ 0 β γ + α^{2} \end{array}] \\ \Rightarrow α^{2} + βγ = 1 \\ \Rightarrow 1 - α^{2} - βγ = 0 \\ ∴ Option C is the correct answer . \end{array}$

Q.76

$\begin{array}{l} If the matrix A is both symmetric ands kew symmetric, then \\ (A) A is a diagonal matrix \\ (B) A is a zero matrix \\ (C) A is a square matrix \\ (D) None of these \end{array}$

Ans.

A is symmetric matrix then a_ij=a_ji

A is skew symmetric matrix then

a_ij =– a_ji

This implies that a_ij = a_ji = – a_ji

2a_ji = 0

Or a_ji = 0

Therefore, A is a zero matrix.

Option (B) is correct.

Q.77

$\begin{array}{l} If A is square matrix such that A^{2} = A, then {(I + A)}^{3} = 7 A \\ is equal to \\ (A) A \\ (B) I - A \\ (C) I \\ (D) 3 A \end{array}$

Ans.

$\begin{array}{l} Given, A^{2} = A, \\ {(I + A)}^{3} - 7 A = I^{3} + 3 I^{2} A + 3 {IA}^{2} + A^{3} - 7 A [\begin{array}{l} {(a + b)}^{3} = a^{3} + 3 a^{2} b \\ + 3 {ab}^{2} + b^{3} \end{array}] \\ = I + 3 A + 3 A + A^{2} A - 7 A [âˆµ I^{3} = I] \\ = I + 6 A + A . A - 7 A [âˆµ A^{2} = A] \\ = I + A^{2} - A \\ = I + A - A [âˆµ A^{2} = A] \\ = I \\ Hence, option (C) is correct option . \end{array}$

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Chapter 4 - Determinants

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Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

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Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

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Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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Related Chapters

Chapter 1 - Relations and Functions

Chapter 2 - Inverse Trigonometric Functions

Chapter 4 - Determinants

Chapter 5 - Continuity and Differentiability

Chapter 6 - Application of Derivatives

Chapter 7 - Integrals

Chapter 8 - Application of Integrals

Chapter 9 - Differential Equations

Chapter 10 - Vector Algebra

Chapter 11 - Three Dimensional Geometry

Chapter 12 - Linear Programming

Chapter 13 - Probability

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1. Why should we study from NCERT Solutions Class 12 Mathematics Chapter 3?

2. What is the importance of NCERT Solutions Class 12 Mathematics Chapter 3?

3. Shall I practise all the questions from NCERT Solutions Class 12 Mathematics Chapter 3?

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