# NCERT Solutions Class 12 Mathematics Chapter 3 – Matrices

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 3.2 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 73525 2123  15 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

$\begin{array}{l}\mathrm{Constructa}2×2\mathrm{matrix},\mathrm{A}=\left[{\mathrm{a}}_{\mathrm{ij}}\right],\mathrm{whose}\mathrm{}\mathrm{elementsare}\\ \mathrm{givenby}:\mathrm{ }\left(\mathrm{i}\right)\mathrm{}{\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+\mathrm{j}\right)}^{2}}{2}\left(\mathrm{ii}\right){\mathrm{a}}_{\mathrm{ij}}=\frac{\mathrm{i}}{\mathrm{j}}\left(\mathrm{iii}\right){\mathrm{a}}_{\mathrm{ij}}=\frac{{\left(\mathrm{i}+2\mathrm{j}\right)}^{2}}{2}\end{array}$

Ans.

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

Q.5

$\begin{array}{l}\mathbf{Construct}\mathrm{}\mathbf{a}\mathrm{}\mathbf{3}\mathrm{}×\mathrm{}\mathbf{4}\mathrm{}\mathbf{matrix},\mathrm{}\mathbf{whose}\mathrm{}\mathbf{elements}\mathrm{}\mathbf{are} \mathbf{given}\mathrm{}\mathbf{by}:\\ \left(\mathrm{i}\right)\mathrm{ }{\mathrm{a}}_{\mathrm{ij}}=\frac{1}{2}|-3\mathrm{i}+\mathrm{j}|\left(\mathrm{ii}\right){\mathrm{a}}_{\mathrm{ij}}=2\mathrm{i}-\mathrm{j}\end{array}$

Ans.

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

Q.6

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{from}\mathrm{the}\mathrm{following} \mathrm{equations}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l}43\\ \mathrm{x}5\end{array}\right]=\left[\begin{array}{l}\mathrm{y}\mathrm{z}\\ 15\end{array}\right] \\ \left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}3\\ 5+\mathrm{z}\mathrm{xy}\end{array}\right]=\left[\begin{array}{l}62\\ 58\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}\mathrm{x}+\mathrm{y}+\mathrm{z}\\ \mathrm{x}+\mathrm{z}\\ \mathrm{y}+\mathrm{z}\end{array}\right]=\left[\begin{array}{l}9\\ 5\\ 7\end{array}\right]\end{array}$

Ans.

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

Q.7

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{ }\mathrm{a},\mathrm{b},\mathrm{c} \mathrm{and}\mathrm{ }\mathrm{d}\mathrm{from}\mathrm{thee}\mathrm{quation}:\\ \left[\begin{array}{l}\mathrm{a}–\mathrm{b}2\mathrm{a}+\mathrm{c}\\ 2\mathrm{a}–\mathrm{b}3\mathrm{c}+\mathrm{d}\end{array}\right]=\left[\begin{array}{l}-1 5\\ 013\end{array}\right]\end{array}$

Ans.

$\begin{array}{l}\left[\begin{array}{l}\mathrm{a}-\mathrm{b}2\mathrm{a}+\mathrm{c}\\ 2\mathrm{a}-\mathrm{b}3\mathrm{c}+\mathrm{d}\end{array}\right]=\left[\begin{array}{l}-1 5\\ 013\end{array}\right]\\ ⇒\mathrm{ }\mathrm{a}-\mathrm{b}=-1...\left(\mathrm{i}\right)\mathrm{ }\\ \mathrm{ }2\mathrm{a}+\mathrm{c}=5\mathrm{ }...\left(\mathrm{ii}\right)\\ \mathrm{ }2\mathrm{a}-\mathrm{b}=0\mathrm{ }...\left(\mathrm{iii}\right)\mathrm{ }\\ 3\mathrm{c}+\mathrm{d}=\mathrm{13}...\left(\mathrm{iv}\right)\\ \mathrm{From}\mathrm{equation}\left(\mathrm{i}\right) \mathrm{and}\mathrm{ }\mathrm{equation}\left(\mathrm{iii}\right),\mathrm{we}\mathrm{have}\\ \mathrm{a}=1\mathrm{and}\mathrm{b}=\mathrm{2}\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{in}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \mathrm{2}\left(1\right)+\mathrm{c}=5⇒\mathrm{c}=5-2=3\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{c}\mathrm{in}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ \mathrm{3}\left(3\right)+\mathrm{d}=13⇒\mathrm{d}=13-9=4\end{array}$

Q.8

$\begin{array}{l}\mathrm{A}={\left[{\mathrm{a}}_{\mathrm{ij}}\right]}_{\mathrm{m}×\mathrm{n}}\mathrm{is}\mathrm{a}\mathrm{square}\mathrm{matrix},\mathrm{if}\\ \left(\mathrm{A}\right)\mathrm{m}<\mathrm{n}\\ \left(\mathrm{B}\right)\mathrm{m}>\mathrm{n}\\ \left(\mathrm{C}\right)\mathrm{m}=\mathrm{n}\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)\mathrm{None}\mathrm{of}\mathrm{these}\end{array}$

Ans.

A is a square matrix if m=n.

So, option (C) is correction answer.

Q.9

$\begin{array}{l}\mathrm{Which}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{values}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}\mathrm{make}\mathrm{the}\mathrm{following}\\ \mathrm{pair}\mathrm{of}\mathrm{matrice}\mathrm{sequal}\\ \left[\begin{array}{l}3\mathrm{x}+75\\ \mathrm{y}+12–3\mathrm{x}\end{array}\right],\mathrm{ }\left[\begin{array}{l}0\mathrm{y}–2\\ 84\end{array}\right]\\ \left(\mathrm{A}\right)\mathrm{x}=-\frac{1}{3}, \mathrm{y}=7\left(\mathrm{B}\right)\mathrm{Not}\mathrm{possiblet}\mathrm{of}\mathrm{ind}\\ \left(\mathrm{C}\right)\mathrm{y}=7,\mathrm{x}=\frac{-2}{3} \left(\mathrm{D}\right)\mathrm{x}=-\frac{1}{3},\mathrm{ }\mathrm{y}=\frac{-2}{3}\end{array}$

Ans.

$\begin{array}{l}\mathrm{If}\mathrm{both}\mathrm{matrices}\mathrm{are}\mathrm{equal},\mathrm{then}\\ \left[\begin{array}{l}3\mathrm{x}+7\mathrm{5}\\ \mathrm{y}+12-3\mathrm{x}\end{array}\right]= \left[\begin{array}{l}0\mathrm{y}-2\\ 8\mathrm{4}\end{array}\right]\\ ⇒ \mathrm{ }3\mathrm{x}+7=0, \mathrm{ }\mathrm{y}-\mathrm{2}=5\\ \mathrm{and} \mathrm{y}+1=8, \mathrm{ }2-3\mathrm{x}=4\\ \mathrm{Then},\mathrm{ }\mathrm{x}=-\frac{7}{3},\mathrm{ }\mathrm{y}=7\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{x}=-\frac{2}{3}\\ \mathrm{Since},\mathrm{there}\mathrm{are}\mathrm{two}\mathrm{different}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{so}\mathrm{it}\mathrm{is}\mathrm{difficult}\\ \mathrm{to}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{to}\mathrm{make}\mathrm{matrices}\mathrm{equal}\mathrm{.}\end{array}$

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

(A) 27

(B) 18

(C) 81

(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

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}24\\ 32\end{array}\right],\mathrm{ }\mathrm{B}=\left[\begin{array}{l} \mathrm{ }13\\ -25\end{array}\right],\mathrm{ }\mathrm{C}=\left[\begin{array}{l}-25\\ \mathrm{ }34\end{array}\right]\\ \mathrm{Find}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{i}\right)\mathrm{A}+\mathrm{B}\\ \left(\mathrm{ii}\right)\mathrm{A}-\mathrm{B}\\ \left(\mathrm{iii}\right)3\mathrm{A}-\mathrm{C}\\ \left(\mathrm{iv}\right)\mathrm{AB}\left(\mathrm{v}\right)\mathrm{BA}\end{array}$

Ans.

$\begin{array}{l}\mathrm{In}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right], \mathrm{B}=\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right], \mathrm{C}=\left[\begin{array}{l}-\mathrm{2}\mathrm{5}\\ 3\mathrm{4}\end{array}\right]\\ \left(\mathrm{i}\right)\mathrm{A}+\mathrm{B}=\left[\begin{array}{l}\mathrm{2}\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]+\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}\mathrm{2}+\mathrm{1}\mathrm{4}+\mathrm{3}\\ \mathrm{3}+\left(-2\right)\mathrm{2}+\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}37\\ 17\end{array}\right]\\ \left(\mathrm{ii}\right)\mathrm{A}-\mathrm{B}=\left[\begin{array}{l}\mathrm{2}\mathrm{4}\\ 3\mathrm{2}\end{array}\right]-\left[\begin{array}{l} \mathrm{ }1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}\mathrm{2}-1\mathrm{4}-\mathrm{3}\\ \mathrm{3}-\left(-2\right)\mathrm{2}-5\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1 \mathrm{ }1\\ 5-3\end{array}\right]\\ \left(\mathrm{iii}\right)3\mathrm{A}-\mathrm{C}=3\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]-\left[\begin{array}{l}-25\\ \mathrm{ }34\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{6}\mathrm{12}\\ 9 \mathrm{ }6\end{array}\right]-\left[\begin{array}{l}-25\\ \mathrm{ }34\end{array}\right]\\ =\left[\begin{array}{l}6-\left(-2\right)12-5\\ 9-36-4\end{array}\right]\\ =\left[\begin{array}{l}87\\ 62\end{array}\right]\\ \left(\mathrm{iv}\right) \mathrm{AB}=\left[\begin{array}{l}2\mathrm{4}\\ \mathrm{3}\mathrm{2}\end{array}\right]\left[\begin{array}{l} 1\mathrm{3}\\ -2\mathrm{5}\end{array}\right]\\ =\left[\begin{array}{l}2-86+20\\ 3-49+10\end{array}\right]\\ =\left[\begin{array}{l}-626\\ -119\end{array}\right]\\ \left(\mathrm{v}\right) \mathrm{BA}=\left[\begin{array}{l} 1\mathrm{3}\\ -\mathrm{2}\mathrm{5}\end{array}\right]\left[\begin{array}{l}2\mathrm{4}\\ 3\mathrm{2}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }2+9\mathrm{ } 4+6\\ -4+15-8+10\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1110\\ 11 2\end{array}\right]\end{array}$

Q.12

$\begin{array}{l}\mathrm{Compute}\mathrm{the}\mathrm{following}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ –\mathrm{b}\mathrm{a}\end{array}\right]+\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{b}\mathrm{a}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}{\mathrm{a}}^{2}+{\mathrm{b}}^{2}{\mathrm{b}}^{2}+{\mathrm{c}}^{2}\\ {\mathrm{a}}^{2}+{\mathrm{c}}^{2}{\mathrm{a}}^{2}+{\mathrm{b}}^{2}\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}–14–6\\ 8516\\ 28\mathrm{ } 5\end{array}\right]+\left[\begin{array}{l}1276\\ 805\\ 324\end{array}\right]\mathrm{ }\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}{\mathrm{cos}}^{2}\mathrm{x}{\mathrm{sin}}^{2}\mathrm{x}\\ {\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\end{array}\right]+\left[\begin{array}{l}{\mathrm{sin}}^{2}\mathrm{x}{\mathrm{cos}}^{2}\mathrm{x}\\ {\mathrm{cos}}^{2}\mathrm{x}{\mathrm{sin}}^{2}\mathrm{x}\end{array}\right]\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ -\mathrm{b}\mathrm{a}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{b}\mathrm{a}\end{array}\right]=\left[\begin{array}{l} \mathrm{a}+\mathrm{a}\mathrm{b}+\mathrm{b}\\ -\mathrm{ }\mathrm{b}+\mathrm{b}\mathrm{a}+\mathrm{a}\end{array}\right]\\ =\left[\begin{array}{l}2\mathrm{a}2\mathrm{b}\\ 02\mathrm{a}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{2}{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}\\ {\mathrm{a}}^{\mathrm{2}}+{\mathrm{c}}^{2}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}\end{array}\right]+\left[\begin{array}{l} \mathrm{ }2\mathrm{ab}2\mathrm{bc}\\ -2\mathrm{ac} -2\mathrm{ab}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}+2\mathrm{ab}{\mathrm{b}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}+2\mathrm{bc}\\ {\mathrm{a}}^{\mathrm{2}}+{\mathrm{c}}^{\mathrm{2}}-2\mathrm{ac}{\mathrm{a}}^{\mathrm{2}}+{\mathrm{b}}^{\mathrm{2}}-2\mathrm{ab}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\left(\mathrm{a}+\mathrm{b}\right)}^{2}{\left(\mathrm{b}+\mathrm{c}\right)}^{2}\\ {\left(\mathrm{a}-\mathrm{c}\right)}^{2}{\left(\mathrm{a}-\mathrm{b}\right)}^{2}\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}-1\mathrm{4}- 6\\ \mathrm{8}5\mathrm{16}\\ 28 \mathrm{ }5\end{array}\right]\mathrm{+}\left[\begin{array}{l}127\mathrm{6}\\ 80\mathrm{5}\\ 32\mathrm{4}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-1+12\mathrm{4}+7 - 6+6\\ \mathrm{8}+\mathrm{8}\mathrm{5}+0\mathrm{16}+5\\ \mathrm{2}+3\mathrm{8}+2 \mathrm{5}+4\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}11110\\ 16\mathrm{ } 521\\ 510 9\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}+{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}+{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}\\ {\mathrm{sin}}^{\mathrm{2}}\mathrm{x}+{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}{\mathrm{cos}}^{\mathrm{2}}\mathrm{x}+{\mathrm{sin}}^{\mathrm{2}}\mathrm{x}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}11\\ 11\end{array}\right]\end{array}$

Q.13

$\begin{array}{l}\mathrm{Compute}\mathrm{the}\mathrm{indicated}\mathrm{products}:\\ \left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{a}\mathrm{b}\\ -\mathrm{b}\mathrm{a}\end{array}\right]\left[\begin{array}{l}\mathrm{a}–\mathrm{b}\\ \mathrm{b} \mathrm{ }\mathrm{a}\end{array}\right] \left(\mathrm{ii}\right)\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\left[234\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}1-2\\ 2 \mathrm{ }3\end{array}\right]\left[\begin{array}{l}123\\ 23\mathrm{}1\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}234\\ 345\\ 456\end{array}\right]\left[\begin{array}{l}1\mathrm{}–35\\ 0 \mathrm{ }24\\ 3 05\end{array}\right]\end{array}$ $\begin{array}{l}\left(v\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\text{}1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}2\\ -1\text{}1\end{array}\right]\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{}0\text{}1\\ -1\text{}2\text{}1\end{array}\right]\\ \left(vi\right)\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{}-1\text{}3\\ -1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{}2\end{array}\right]\left[\begin{array}{l}2\text{}-3\\ 1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 3\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\end{array}\right]\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right) \left[\begin{array}{l} \mathrm{ab}\\ -\mathrm{ba}\end{array}\right]\left[\begin{array}{l}\mathrm{a}-\mathrm{b}\\ \mathrm{b} \mathrm{ }\mathrm{a}\end{array}\right]=\left[\begin{array}{l} \mathrm{ }{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-\mathrm{ab}+\mathrm{ab}\\ -\mathrm{ba}+\mathrm{ab} \mathrm{ }{\mathrm{b}}^{2}+{\mathrm{a}}^{2}\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }{\mathrm{a}}^{2}+{\mathrm{b}}^{2}-\mathrm{ab}+\mathrm{ab}\\ -\mathrm{ba}+\mathrm{ab} \mathrm{ }{\mathrm{b}}^{2}+{\mathrm{a}}^{2}\end{array}\right]\\ \left(\mathrm{ii}\right)\left[\begin{array}{l}\mathrm{1}\\ \mathrm{2}\\ \mathrm{3}\end{array}\right]\left[234\right]=\left[\begin{array}{l}234\\ 468\\ 6912\end{array}\right]\\ \left(\mathrm{iii}\right)\left[\begin{array}{l}\mathrm{1}-\mathrm{2}\\ 2 \mathrm{3}\end{array}\right]\left[\begin{array}{l}12 3\\ 23 1\end{array}\right]=\left[\begin{array}{l}1-42-63-2\\ 2+64+96+3\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-3-41\\ \mathrm{ }8\mathrm{ }139\end{array}\right]\\ \left(\mathrm{iv}\right)\left[\begin{array}{l}234\\ 345\\ 456\end{array}\right]\left[\begin{array}{l}\mathrm{1}-35\\ 0 24\\ 3 05\end{array}\right]\\ =\left[\begin{array}{l}2+0+12-6+6+010+12+20\\ 3+0+15-9+8+015+16+25\\ 4+0+18-12+10+020+20+30\end{array}\right]\\ =\left[\begin{array}{l}14042\\ 18-156\\ 22-270\end{array}\right]\\ \left(\mathrm{v}\right) \left[\begin{array}{l} \mathrm{ }21\\ \mathrm{ }32\\ -11\end{array}\right]\left[\begin{array}{l} \mathrm{ }101\\ -121\end{array}\right]=\left[\begin{array}{l}2-10+22+1\\ 3-20+43+2\\ -1-10+2-1+1\end{array}\right]\\ =\left[\begin{array}{l}123\\ 145\\ -220\end{array}\right]\\ \left(\mathrm{vi}\right)\left[\begin{array}{l} 3-13\\ -1 02\end{array}\right]\left[\begin{array}{l}\mathrm{2}-\mathrm{3}\\ 1 0\\ 3 1\end{array}\right]=\left[\begin{array}{l} \mathrm{ }6-1+9-9+0+3\\ -2+0+6 \mathrm{ }3+0+2\end{array}\right]\\ =\left[\begin{array}{l}14-6\\ 45\end{array}\right]\end{array}$

Q.14

$\begin{array}{l}\mathrm{If}\mathrm{A}=\left[\begin{array}{l}12-3\\ 50\mathrm{ } 2\\ 1 -1 1\end{array}\right],\mathrm{ }\mathrm{B}=\left[\begin{array}{l}3-12\\ 4\mathrm{ } 25\\ 2 03\end{array}\right]\mathrm{and}\mathrm{C}=\left[\begin{array}{l}412\\ 032\\ 1-23\end{array}\right],\\ \mathrm{then}\mathrm{compute}\left(\mathrm{A}+\mathrm{B}\right)\mathrm{and}\left(\mathrm{B}-\mathrm{C}\right).\mathrm{ }\mathrm{Also},\mathrm{verify}\mathrm{that}\\ \mathrm{A}+\left(\mathrm{B}-\mathrm{C}\right)=\left(\mathrm{A}+\mathrm{B}\right)-\mathrm{C}.\end{array}$

Ans.

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

Q.15

$\text{If\hspace{0.17em} A=}\text{[}\begin{array}{l}\frac{2}{3}1\frac{5}{3}\\ \frac{1}{3}\frac{2}{3}\frac{4}{3}\\ \frac{7}{3}2\frac{2}{3}\end{array}\right]&\mathrm{B}=\left[\begin{array}{l}\frac{2}{5}\frac{3}{5}1\\ \frac{1}{5}\frac{2}{5}\frac{4}{5}\\ \frac{7}{5}\frac{6}{5}\frac{2}{5}\end{array}\right],\mathrm{then}\mathrm{compute}3\mathrm{A}-5\mathrm{B}.$

Ans.

$\begin{array}{l}\mathrm{Given},\mathrm{A}=\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{3}}\mathrm{1}\frac{5}{\mathrm{3}}\\ \frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{2}}{\mathrm{3}}\frac{\mathrm{4}}{\mathrm{3}}\\ \frac{\mathrm{7}}{\mathrm{3}}2\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{3}}{\mathrm{5}}\mathrm{1}\\ \frac{\mathrm{1}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{4}}{\mathrm{5}}\\ \frac{\mathrm{7}}{\mathrm{5}}\frac{\mathrm{6}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\\ 3\mathrm{A}-5\mathrm{B}=3\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{3}}1\frac{\mathrm{5}}{\mathrm{3}}\\ \frac{\mathrm{1}}{\mathrm{3}}\frac{\mathrm{2}}{\mathrm{3}}\frac{\mathrm{4}}{\mathrm{3}}\\ \frac{\mathrm{7}}{\mathrm{3}}2\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]-5\left[\begin{array}{l}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{3}}{\mathrm{5}}\mathrm{1}\\ \frac{\mathrm{1}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\frac{\mathrm{4}}{\mathrm{5}}\\ \frac{\mathrm{7}}{\mathrm{5}}\frac{\mathrm{6}}{\mathrm{5}}\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}3×\frac{\mathrm{2}}{\mathrm{3}}3×13×\frac{\mathrm{5}}{\mathrm{3}}\\ 3×\frac{\mathrm{1}}{\mathrm{3}}3×\frac{\mathrm{2}}{\mathrm{3}}3×\frac{\mathrm{4}}{\mathrm{3}}\\ 3×\frac{\mathrm{7}}{\mathrm{3}}3×23×\frac{\mathrm{2}}{\mathrm{3}}\end{array}\right]-\left[\begin{array}{l}5×\frac{\mathrm{2}}{\mathrm{5}}5×\frac{\mathrm{3}}{\mathrm{5}}5×\mathrm{1}\\ 5×\frac{\mathrm{1}}{\mathrm{5}}5×\frac{\mathrm{2}}{\mathrm{5}}5×\frac{\mathrm{4}}{\mathrm{5}}\\ 5×\frac{\mathrm{7}}{\mathrm{5}}5×\frac{\mathrm{6}}{\mathrm{5}}5×\frac{\mathrm{2}}{\mathrm{5}}\end{array}\right]\mathrm{ }\\ =\left[\begin{array}{l}235\\ 124\\ 762\end{array}\right]-\left[\begin{array}{l}235\\ 124\\ 762\end{array}\right]\\ =\left[\begin{array}{l}2-23-35-5\\ 1-12-24-4\\ 7-76-62-2\end{array}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]\end{array}$

Q.16

$\text{Simplify\hspace{0.17em}cosθ}\text{[}\begin{array}{l}\mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }\end{array}\right]+\mathrm{sin\theta }\left[\begin{array}{l}\mathrm{sin\theta } -\mathrm{cos\theta }\\ \mathrm{cos\theta }\mathrm{sin\theta }\end{array}\right].$

Ans.

$\begin{array}{l}\mathrm{cos\theta }\left[\begin{array}{l}\mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }\end{array}\right]+\mathrm{sin\theta }\left[\begin{array}{l}\mathrm{sin\theta } -\mathrm{cos\theta }\\ \mathrm{cos\theta }\mathrm{sin\theta }\end{array}\right]\\ =\left[\begin{array}{l} {\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta } \mathrm{cos\theta }\mathrm{sin\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\mathrm{+}\left[\begin{array}{l}{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta } -\mathrm{sin\theta }\mathrm{cos\theta }\\ \mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\\ =\left[\begin{array}{l} {\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }+{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta } \mathrm{cos\theta }\mathrm{sin\theta }-\mathrm{sin\theta }\mathrm{cos\theta }\\ -\mathrm{sin\theta }\mathrm{cos\theta }+\mathrm{sin\theta }\mathrm{cos\theta }{\mathrm{cos}}^{\mathrm{2}}\mathrm{\theta }+{\mathrm{sin}}^{\mathrm{2}}\mathrm{\theta }\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{ }10\\ 0\mathrm{1}\end{array}\right].\end{array}$

Q.17

$\begin{array}{l}\mathrm{Find}\mathrm{X}\mathrm{and}\mathrm{Y},\mathrm{if}\\ \left(\mathrm{i}\right)\mathrm{ }\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right] \mathrm{and} \mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}30\\ 03\end{array}\right]\\ \left(\mathrm{ii}\right)2\mathrm{X}+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right] \mathrm{and} 3\mathrm{X}+2\mathrm{Y}=\left[\begin{array}{l} \mathrm{ }2-2\\ -1 \mathrm{ }5\end{array}\right]\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right) \mathrm{ }\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }...\left(\mathrm{i}\right) \\ \mathrm{and} \mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}30\\ 03\end{array}\right]\mathrm{ }...\left(\mathrm{ii}\right) \\ \mathrm{Adding}\mathrm{both}\mathrm{equations},\mathrm{we}\mathrm{get}\\ \mathrm{X}+\mathrm{Y}+\mathrm{X}-\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]+\left[\begin{array}{l}30\\ 03\end{array}\right]\\ 2\mathrm{X}=\left[\begin{array}{l}\mathrm{7}+30+\mathrm{0}\\ \mathrm{2}+05+\mathrm{3}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}100\\ 28\end{array}\right]\\ \mathrm{X}=\frac{1}{2}\left[\begin{array}{l}100\\ 28\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}50\\ 14\end{array}\right]\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{X}\mathrm{in}\mathrm{equatin}\left(\mathrm{i}\right), \mathrm{we}\mathrm{ }\mathrm{get}\\ \left[\begin{array}{l}50\\ 14\end{array}\right]+\mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }\\ \mathrm{Y}=\left[\begin{array}{l}70\\ 25\end{array}\right]\mathrm{ }-\left[\begin{array}{l}50\\ 14\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{7}-50-0\\ \mathrm{2}-15-4\end{array}\right]\\ =\mathrm{ }\left[\begin{array}{l}20\\ 11\end{array}\right]\\ \mathrm{Thus},\mathrm{ }\mathrm{X}=\left[\begin{array}{l}50\\ 14\end{array}\right]\mathrm{and}\mathrm{Y}=\left[\begin{array}{l}20\\ 11\end{array}\right].\\ \left(\mathrm{ii}\right)2\mathrm{X}+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]\mathrm{ }...\left(\mathrm{i}\right)\mathrm{ }\\ \mathrm{and} 3\mathrm{X}+2\mathrm{Y}=\left[\begin{array}{l} 2-\mathrm{2}\\ -1 \mathrm{5}\end{array}\right]\mathrm{ }...\left(\mathrm{ii}\right)\mathrm{ }\\ \mathrm{Multiplying}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{by}2\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right)\mathrm{by}3,\mathrm{we}\mathrm{get}\\ 2\left(2\mathrm{X}+3\mathrm{Y}\right)=2\left[\begin{array}{l}23\\ 40\end{array}\right]\\ ⇒ 4\mathrm{X}+6\mathrm{Y}=\left[\begin{array}{l}46\\ 80\end{array}\right]...\left(\mathrm{iii}\right)\\ 3\left(3\mathrm{X}+2\mathrm{Y}\right)=3\left[\begin{array}{l} 2-\mathrm{2}\\ -1 \mathrm{5}\end{array}\right]\\ ⇒9\mathrm{X}+6\mathrm{Y}=\left[\begin{array}{l} 6-\mathrm{6}\\ -3 \mathrm{15}\end{array}\right]...\left(\mathrm{iv}\right)\\ \mathrm{Subtracting}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{from}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ 5\mathrm{X}=\left[\begin{array}{l} 6-\mathrm{6}\\ -3 \mathrm{15}\end{array}\right]-\left[\begin{array}{l}46\\ 80\end{array}\right]\\ \mathrm{ }\mathrm{X}=\frac{1}{5}\left[\begin{array}{l} 2-\mathrm{12}\\ -11 \mathrm{15}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{X}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{we}\mathrm{have}\\ \mathrm{2}\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]+3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]\\ \mathrm{ }3\mathrm{Y}=\left[\begin{array}{l}23\\ 40\end{array}\right]-\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{4}}{5}-\frac{\mathrm{24}}{5}\\ -\frac{\mathrm{22}}{5} 6\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{2}-\frac{\mathrm{4}}{5}\mathrm{3}-\left(-\frac{\mathrm{24}}{5}\right)\\ \mathrm{4}-\left(-\frac{\mathrm{22}}{5}\right)\mathrm{0}-6\end{array}\right]\\ \mathrm{ }\mathrm{Y}=\frac{1}{3}\left[\begin{array}{l}\frac{\mathrm{6}}{5}\frac{\mathrm{39}}{5}\\ \frac{\mathrm{42}}{5}-6\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{2}}{5}\frac{\mathrm{13}}{5}\\ \frac{\mathrm{14}}{5}-2\end{array}\right]\\ \mathrm{Thus},\mathrm{X}=\left[\begin{array}{l} \mathrm{ }\frac{\mathrm{2}}{5}-\frac{\mathrm{12}}{5}\\ -\frac{\mathrm{11}}{5} 3\end{array}\right]\mathrm{and} \mathrm{Y}=\left[\begin{array}{l}\frac{\mathrm{2}}{5}\frac{\mathrm{13}}{5}\\ \frac{\mathrm{14}}{5}-2\end{array}\right].\end{array}$

Q.18

$\text{Find X, if Y=}\text{[}\begin{array}{l}32\\ 40\end{array}\right]\mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l} 2-2\\ -1 \mathrm{ }5\end{array}\right].$

Ans.

$\begin{array}{l}\mathrm{Given},\mathrm{Y}=\left[\begin{array}{l}32\\ 14\end{array}\right]\mathrm{}\\ \mathrm{and}2\mathrm{X}+\mathrm{Y}=\left[\begin{array}{l} 10\\ -32\end{array}\right]...\left(\mathrm{i}\right)\\ \mathrm{Putting}\mathrm{value}\mathrm{of}\mathrm{Y}\mathrm{in}\mathrm{equation}\left(\mathrm{i}\right),\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ 2\mathrm{X}+\left[\begin{array}{l}32\\ 14\end{array}\right]=\left[\begin{array}{l} 10\\ -32\end{array}\right]\\ \mathrm{ }2\mathrm{X}=\left[\begin{array}{l} 10\\ -32\end{array}\right]-\left[\begin{array}{l}32\\ 14\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }1-30-2\\ -3-12-4\end{array}\right]\\ \mathrm{X}\mathrm{ }=\frac{1}{2}\left[\begin{array}{l}-2-2\\ -4-2\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-1-1\\ -2 -1\end{array}\right]\end{array}$

Q.19

$\text{If\hspace{0.17em}\hspace{0.17em}x\hspace{0.17em}\hspace{0.17em}and y, if 2}\text{[}\begin{array}{l}13\\ 0\mathrm{x}\end{array}\right]+\left[\begin{array}{l}\mathrm{y}0\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right].$

Ans.

$\begin{array}{l}\mathrm{Given}: 2\left[\begin{array}{l}13\\ \mathrm{0}\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{y}\mathrm{0}\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒\left[\begin{array}{l}26\\ 02\mathrm{x}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{y}\mathrm{0}\\ 12\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{l}2+\mathrm{y} \mathrm{ }6+0\\ 0+12\mathrm{x}+2\end{array}\right]=\left[\begin{array}{l}56\\ 18\end{array}\right]\\ ⇒ 2+\mathrm{y}=5 \mathrm{and} 2\mathrm{x}+2=8\\ ⇒ \mathrm{ }\mathrm{y}=3 \mathrm{and}\mathrm{ }\mathrm{x}=\frac{8-2}{2}=3\\ \mathrm{Thus},\mathrm{x}=3\mathrm{and}\mathrm{y}=3.\\ \end{array}$

Q.20

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

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{2}\left[\begin{array}{l}\mathrm{x}\mathrm{z}\\ \mathrm{y}\mathrm{t}\end{array}\right]+3\left[\begin{array}{l}1-\mathrm{1}\\ \mathrm{0} 2\end{array}\right]=\mathrm{3}\left[\begin{array}{l}35\\ 46\end{array}\right]\\ ⇒ \left[\begin{array}{l}2\mathrm{x}2\mathrm{z}\\ 2\mathrm{y}2\mathrm{t}\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{3}-\mathrm{3}\\ \mathrm{0} 6\end{array}\right]=\left[\begin{array}{l}91 5\\ 121 8\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{l}2\mathrm{x}+32\mathrm{z}-\mathrm{3}\\ 2\mathrm{y}+02\mathrm{t}+\mathrm{6}\end{array}\right]=\left[\begin{array}{l}91 5\\ 121 8\end{array}\right]\\ ⇒2\mathrm{x}+\mathrm{3}=9, 2\mathrm{y}=12, 2\mathrm{z}-\mathrm{3}=15 \mathrm{and} 2\mathrm{t}+\mathrm{6}=18\\ ⇒\mathrm{x}=3,\mathrm{y}=6, \mathrm{z}=9\mathrm{}\mathrm{and}\mathrm{t}=6.\end{array}$

Q.21

$\text{If\hspace{0.17em} x}\text{[}\begin{array}{l}2\\ 3\end{array}\right]+\mathrm{y}\left[\begin{array}{l}-1\\ \mathrm{ }1\end{array}\right]=\left[\begin{array}{l}10\\ 5\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{and}\mathrm{y}.$

Ans.

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

Q.22

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

Ans.

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

Q.23

$\text{If F(x) =}\left[\begin{array}{l}\mathrm{cosx}-\mathrm{sinx}0\\ \mathrm{sinx}\mathrm{cosx}0\\ 0 \mathrm{ }01\end{array}\right], \mathrm{show}\mathrm{that}\mathrm{F}\left(\mathrm{x}\right)\mathrm{F}\left(\mathrm{y}\right)=\mathrm{F}\left(\mathrm{x}+\mathrm{y}\right).$

Ans.

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

Q.24

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

Ans.

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

Q.25

$Find\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{2}-5A+6I,ifA=\left[\begin{array}{l}2\text{}0\text{}1\\ 2\text{}1\text{}3\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1\text{}0\end{array}\right].$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}-\text{5A+6I, A}=\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{-1}\text{}\text{0}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}=A×A\\ \text{}=\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\\ \text{}=\left[\begin{array}{l}4+0+1\text{}0+0-1\text{}2+0+0\\ 4+2+3\text{}0+1-3\text{}2+3+0\\ 2-2+0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0-1+0\text{}1-3+0\end{array}\right]\\ \text{}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]\\ Now,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{2}}-\text{5A+6I}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]-\text{5}\left[\begin{array}{l}\text{2}\text{}\text{0}\text{}\text{1}\\ \text{2}\text{}\text{1}\text{}\text{3}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{1}\text{}\text{0}\end{array}\right]\text{+6}\left[\begin{array}{l}1\text{}0\text{}0\\ 0\text{}1\text{}0\\ 0\text{}0\text{}1\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}5\text{}-1\text{}2\\ 9\text{}-2\text{}5\\ 0\text{}-1-2\end{array}\right]-\left[\begin{array}{l}\text{10}\text{}\text{0}\text{}\text{5}\\ \text{10}\text{}\text{5}\text{}\text{15}\\ \text{5}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\text{5}\text{}\text{0}\end{array}\right]\text{+}\left[\begin{array}{l}6\text{}0\text{}0\\ 0\text{}6\text{}0\\ 0\text{}0\text{}6\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}5-10+6\text{}\text{}-1\text{}\text{}2-5\\ 9-10\text{}-2-5+6\text{}5-15\\ 0-5\text{}\text{}-1+5\text{}\text{}-2+6\end{array}\right]\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}1\text{}-1\text{}-3\\ -1\text{}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}-10\\ -5\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\text{}\text{\hspace{0.17em}}4\end{array}\right]\end{array}$

Q.26

$\text{If A =}\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\mathrm{}, \mathrm{prove}\mathrm{that} {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+7\mathrm{A}+2\mathrm{I}=0.$

Ans.

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right], \\ \mathrm{ }{\mathrm{A}}^{\mathrm{2}}\mathrm{}=\mathrm{A}×\mathrm{A}\\ =\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\left[\begin{array}{l}102\\ 021\\ 2 03\end{array}\right]\\ =\left[\begin{array}{l}1+0+4 0+0+02+0+6\\ 0+0+2 \mathrm{ }0+4+00+2+3\\ 2+0+6 \mathrm{ }0+0+04+0+9\end{array}\right]\\ =\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\\ 6{\mathrm{A}}^{2}=6\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\\ =\left[\begin{array}{l}30 \mathrm{ }048\\ 122430\\ 16 \mathrm{ }078\end{array}\right]\\ \\ {\mathrm{A}}^{3}={\mathrm{A}}^{2}.\mathrm{A}\\ =\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]\left[\begin{array}{l}102\\ 021\\ 203\end{array}\right]\\ =\left[\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}\right]\\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]\\ 7\mathrm{A}=7\left[\begin{array}{l}102\\ 021\\ 203\end{array}\right]\\ =\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]\\ 2\mathrm{I}=2\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]\\ =\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}:\\ {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{\mathrm{2}}+7\mathrm{A}+2\mathrm{I}\\ \\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]-6\left[\begin{array}{l}508\\ 245\\ 8013\end{array}\right]+\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]+\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ =\left[\begin{array}{l}21034\\ 12823\\ 34055\end{array}\right]-\left[\begin{array}{l}30048\\ 122430\\ 48078\end{array}\right]+\left[\begin{array}{l}7014\\ 0147\\ 14021\end{array}\right]+\left[\begin{array}{l}200\\ 020\\ 002\end{array}\right]\\ =\left[\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}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]=0\end{array}$

Q.27

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

Ans.

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

Q.28

$\begin{array}{l}\mathrm{If} \mathrm{A}=\left[\begin{array}{l}0-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}0\end{array}\right]\mathrm{and}\mathrm{I}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\\ \mathrm{show}\mathrm{that}\mathrm{I}+\mathrm{A}=\left(\mathrm{I}-\mathrm{A}\right)\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{ }\mathrm{sin\alpha }\\ \mathrm{sin\alpha }\mathrm{cos\alpha }\end{array}\right]\end{array}$

Ans.

$\begin{array}{l}\mathrm{I}+\mathrm{ }\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]\mathrm{+}\left[\begin{array}{l}\mathrm{0}-\mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\mathrm{0}\end{array}\right]\\ =\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \mathrm{I}-\mathrm{ }\mathrm{A}=\left[\begin{array}{l}10\\ 01\end{array}\right]-\left[\begin{array}{l}\mathrm{0}-\mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{\mathrm{2}}\mathrm{0}\end{array}\right]\\ =\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}.=\left(\mathrm{I}-\mathrm{A}\right)\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}1\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ -\mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}1\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ -\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}1\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{cos\alpha }+\frac{\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}-\mathrm{sin\alpha }+\mathrm{cos\alpha }\frac{\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ -\frac{\mathrm{cos\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}+\mathrm{sin\alpha }\frac{\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}+\mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos\alpha }.\mathrm{cos}\frac{\mathrm{\alpha }}{2}+\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\left(\mathrm{sin\alpha }.\mathrm{cos}\frac{\mathrm{\alpha }}{2}-\mathrm{cos\alpha sin}\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin\alpha cos}\frac{\mathrm{\alpha }}{2}-\mathrm{cos\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos\alpha cos}\frac{\mathrm{\alpha }}{2}+\mathrm{sin\alpha }.\mathrm{sin}\frac{\mathrm{\alpha }}{2}}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\mathrm{sin}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos}\left(\mathrm{\alpha }-\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}\frac{\mathrm{cos}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{-\mathrm{sin}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\\ \frac{\mathrm{sin}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\frac{\mathrm{cos}\left(\frac{\mathrm{\alpha }}{2}\right)}{\mathrm{cos}\frac{\mathrm{\alpha }}{2}}\end{array}\right]\\ =\left[\begin{array}{l}1-\mathrm{tan}\frac{\mathrm{\alpha }}{2}\\ \mathrm{tan}\frac{\mathrm{\alpha }}{2}1\end{array}\right]=\mathrm{L}.\mathrm{H}.\mathrm{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} \mathrm{Total}\mathrm{money}\mathrm{to}\mathrm{invest}=\mathrm{} 30,000\\ \mathrm{Money}\mathrm{invested}\mathrm{in}{\mathrm{I}}^{\mathrm{st}}\mathrm{type}\mathrm{of}\mathrm{bond}= \mathrm{x}\\ \mathrm{Money}\mathrm{invested}\mathrm{in}{2}^{\mathrm{nd}}\mathrm{type}\mathrm{of}\mathrm{bond}=\left(30,000-\mathrm{x}\right)\\ \mathrm{Rate}\mathrm{of}\mathrm{interest}\mathrm{on}\mathrm{first}\mathrm{type}\mathrm{of}\mathrm{bond}=5%\\ \mathrm{Rate}\mathrm{of}\mathrm{interest}\mathrm{on}\mathrm{secon}\mathrm{type}\mathrm{of}\mathrm{bond}=7%\\ \left(\mathrm{i}\right) \mathrm{Total}\mathrm{obtained}\mathrm{interest}=1800\\ \left(\mathrm{ii}\right) \mathrm{Total}\mathrm{obtained}\mathrm{interest}=2000\\ \mathrm{ }\left(\begin{array}{cc}\mathrm{I}& \mathrm{II}\\ \mathrm{x}& \left(30,000-\mathrm{x}\right)\end{array}\right)\left(\begin{array}{l}5%\\ 7%\end{array}\right)\mathrm{=}\left(\begin{array}{l}\mathrm{1800}\\ \mathrm{2000}\end{array}\right)\\ \left(\mathrm{i}\right)5% \mathrm{of}\mathrm{ }\mathrm{x}+7%\left(30,000-\mathrm{x}\right)=1800\\ \frac{\mathrm{5}}{\mathrm{100}}\mathrm{x}+\frac{\mathrm{7}}{\mathrm{100}}\left(30,000-\mathrm{x}\right)=1800\\ \mathrm{ }5\mathrm{x}+210000-7\mathrm{x}=180000\\ 210000-180000=2\mathrm{x}\\ 30000=2\mathrm{x}\\ \mathrm{ }\mathrm{x}=\frac{\mathrm{30000}}{\mathrm{2}}\\ \mathrm{ }=15000\\ \mathrm{To}\mathrm{get}\mathrm{annual}\mathrm{interest}\mathrm{of}1800,\mathrm{he}\mathrm{should}\mathrm{invest}\mathrm{}15,000\\ \mathrm{in}\mathrm{I}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{and}\mathrm{}15,000\mathrm{in}\mathrm{II}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{.}\\ \left(\mathrm{ii}\right)5% \mathrm{of}\mathrm{ }\mathrm{x}+7%\left(30,000-\mathrm{x}\right)=2000\\ \frac{\mathrm{5}}{\mathrm{100}}\mathrm{x}+\frac{\mathrm{7}}{\mathrm{100}}\left(30,000-\mathrm{x}\right)=2000\\ \mathrm{ }5\mathrm{x}+210000-7\mathrm{x}=200000\\ 210000-200000=2\mathrm{x}\\ 10000=2\mathrm{x}\\ \mathrm{ }\mathrm{x}=\frac{\mathrm{10000}}{\mathrm{2}}\\ = 5000\\ \mathrm{To}\mathrm{get}\mathrm{annual}\mathrm{interest}\mathrm{of}\mathrm{}2000\mathrm{he}\mathrm{should}\mathrm{invest}\mathrm{}5000&\\ 25000\mathrm{respectively}\mathrm{in}\mathrm{I}\mathrm{and}\mathrm{II}\mathrm{type}\mathrm{of}\mathrm{bond}\mathrm{.}\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.

$\begin{array}{l}\begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{chemistry}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 10\mathrm{dozen}\end{array}\\ \begin{array}{l}=10×12\end{array}\\ \begin{array}{l}=120\end{array}\\ \begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{physics}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 8\mathrm{dozen}\end{array}\\ \begin{array}{l}=8×12\end{array}\\ =\mathrm{}\begin{array}{l}\mathrm{96}\end{array}\\ \begin{array}{l}\mathrm{Number}\mathrm{of}\mathrm{economics}\mathrm{books}\mathrm{on}\mathrm{bookshop}= 10\mathrm{dozen}\end{array}\\ \begin{array}{l}=10×12\end{array}\\ \begin{array}{l}=120\end{array}\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{chemistry}\mathrm{book}=\end{array}80\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{physics}\mathrm{book}=\end{array}60\\ \begin{array}{l}\mathrm{Selling}\mathrm{price}\mathrm{of}\mathrm{one}\mathrm{economics}\mathrm{book}=\end{array}40\\ \begin{array}{l}\mathrm{The}\mathrm{total}\mathrm{amount}\mathrm{the}\mathrm{bookshop}\mathrm{will}\mathrm{receive}\end{array}\\ \begin{array}{l}=\left[\mathrm{12096120}\right]\left[\begin{array}{l}\mathrm{80}\\ \mathrm{60}\\ \mathrm{40}\end{array}\right]\end{array}\\ \begin{array}{l}=\left[9600+5760+4800\right]\end{array}\\ \begin{array}{l}=\left[\mathrm{20160}\right]\end{array}\\ \begin{array}{l}\mathrm{Thus},\mathrm{the}\mathrm{total}\mathrm{amount}\mathrm{the}\mathrm{bookshop}\mathrm{will}\mathrm{receive}\mathrm{from}\mathrm{selling}\end{array}\mathrm{}\begin{array}{l}\mathrm{all}\mathrm{the}\mathrm{books}\mathrm{ }\mathrm{is}\end{array}\mathrm{}20,160\mathrm{.}\end{array}$

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

(C) p is arbitrary, k = 3

(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

(C) n × 3

(D) p × n

Ans.

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

Q.33

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

Ans.

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

Q.34

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

Ans.

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

Q.35

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

Ans.

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

Q.36

$\text{If\hspace{0.17em}A’=}\text{[}\begin{array}{l}-23\\ 12\end{array}\right]\mathrm{ }\mathrm{and}\mathrm{B}=\left[\begin{array}{l}-10\\ \mathrm{ }12\end{array}\right],\mathrm{then}\mathrm{find}\left(\mathrm{A}+2\mathrm{B}\right)‘.$

Ans.

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

Q.37

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

Ans.

$\begin{array}{l}\mathrm{For}\mathrm{the}\mathrm{matrices}\mathrm{A}\mathrm{and}\mathrm{B},\mathrm{verify}\mathrm{that}\left(\mathrm{AB}\right)‘=\mathrm{B}‘\mathrm{A}‘,\mathrm{ }\mathrm{where}\\ \left(\mathrm{i}\right)\mathrm{A}=\left[\begin{array}{l} 1\\ -\mathrm{4}\\ 3\end{array}\right], \mathrm{ }\mathrm{B}=\left[-12 1\right]\\ \mathrm{ }\mathrm{AB}=\left[\begin{array}{l} 1\\ -\mathrm{4}\\ 3\end{array}\right]\left[-12 1\right]\\ =\left[\begin{array}{l}-1 2 \mathrm{1}\\ 4-8-4\\ -3 \mathrm{ }6 \mathrm{ }3\end{array}\right]\\ \left(\mathrm{AB}\right)‘={\left[\begin{array}{l}-1 2 \mathrm{1}\\ 4-8-4\\ -3 \mathrm{ }6 \mathrm{ }3\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}-1 \mathrm{ }4-3\\ \mathrm{ }2-8 6\\ \mathrm{ }1-4 3\end{array}\right]\\ \\ \mathrm{A}‘={\left[\begin{array}{l} 1\\ -\mathrm{4}\\ 3\end{array}\right]}^{‘}=\left[1-43\right]\\ \mathrm{B}‘={\left[-121\right]}^{‘}\\ =\left[\begin{array}{l}-1\\ \mathrm{ }2\\ \mathrm{ }1\end{array}\right]\\ \mathrm{B}‘\mathrm{A}‘\mathrm{ }=\left[\begin{array}{l}-1\\ \mathrm{ }2\\ \mathrm{ }1\end{array}\right]\left[1-43\right]\\ =\left[\begin{array}{l}-1 \mathrm{ }4-3\\ \mathrm{ }2-8 6\\ \mathrm{ }1-4 \mathrm{ }3\end{array}\right]\\ \mathrm{Thus},\mathrm{ }\left(\mathrm{AB}\right)‘=\mathrm{B}‘\mathrm{A}‘.\\ \left(\mathrm{ii}\right)\mathrm{A}=\left[\begin{array}{l}\mathrm{0}\\ \mathrm{1}\\ \mathrm{2}\end{array}\right], \mathrm{B}=\left[15 7\right]\\ \mathrm{ }\mathrm{AB}=\left[\begin{array}{l}\mathrm{0}\\ \mathrm{1}\\ \mathrm{2}\end{array}\right]\left[15 7\right]\\ =\left[\begin{array}{l}000\\ 157\\ 21014\end{array}\right]\\ \\ \left(\mathrm{AB}\right)‘={\left[\begin{array}{l}000\\ 157\\ 21014\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}012\\ 0510\\ 0714\end{array}\right]\\ \mathrm{ }\mathrm{A}‘={\left[\begin{array}{l}\mathrm{0}\\ \mathrm{1}\\ \mathrm{2}\end{array}\right]}^{‘}\\ =\left[012\right]\\ \mathrm{ }\mathrm{B}‘={\left[157\right]}^{‘}\\ =\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]\\ \mathrm{B}‘\mathrm{A}‘=\left[\begin{array}{l}1\\ 5\\ 7\end{array}\right]\left[012\right]\\ =\left[\begin{array}{l}01 2\\ 0510\\ 0714\end{array}\right]\\ \mathrm{Thus},\mathrm{ }\left(\mathrm{AB}\right)‘=\mathrm{B}‘\mathrm{A}‘\\ \end{array}$

Q.38

$\begin{array}{l}\mathrm{If} \mathrm{ }\left(\mathrm{i}\right)\mathrm{A}=\left[\begin{array}{l} \mathrm{ }\mathrm{cos\alpha } \mathrm{sin\alpha }\\ -\mathrm{sin\alpha cos\alpha }\end{array}\right],\mathrm{the}\mathrm{nver}\mathrm{if}\mathrm{y}\mathrm{AA}‘=\mathrm{I}\\ \mathrm{ }\left(\mathrm{ii}\right)\mathrm{A}=\left[\begin{array}{l} \mathrm{ }\mathrm{sin\alpha } \mathrm{cos\alpha }\\ -\mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right],\mathrm{the}\mathrm{nver}\mathrm{if}\mathrm{y}\mathrm{AA}‘=\mathrm{I}\end{array}$

Ans.

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Given} \mathrm{A}=\left[\begin{array}{l} \mathrm{ }\mathrm{cos\alpha } \mathrm{sin\alpha }\\ -\mathrm{sin\alpha cos\alpha }\end{array}\right]\\ \mathrm{A}‘={\left[\begin{array}{l} \mathrm{ }\mathrm{cos\alpha } \mathrm{sin\alpha }\\ -\mathrm{sin\alpha cos\alpha }\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}: \mathrm{AA}‘=\left[\begin{array}{l} \mathrm{ }\mathrm{cos\alpha } \mathrm{sin\alpha }\\ -\mathrm{sin\alpha cos\alpha }\end{array}\right]\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}{\mathrm{cos}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\alpha }-\mathrm{cos\alpha sin\alpha }+\mathrm{sin\alpha cos\alpha }\\ -\mathrm{sin\alpha cos\alpha }+{\mathrm{cos\alpha sin\alpha sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }\end{array}\right]\\ =\left[\begin{array}{l}10\\ 01\end{array}\right]\\ =\mathrm{I}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{ }\left(\mathrm{ii}\right)\mathrm{ }\mathrm{A}=\left[\begin{array}{l} \mathrm{ }\mathrm{sin\alpha } \mathrm{cos\alpha }\\ -\mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right]\\ \mathrm{A}‘={\left[\begin{array}{l} \mathrm{ }\mathrm{sin\alpha } \mathrm{cos\alpha }\\ -\mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}\mathrm{sin\alpha }-\mathrm{cos\alpha }\\ \mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right]\\ \mathrm{ }\mathrm{AA}‘=\left[\begin{array}{l} \mathrm{ }\mathrm{sin\alpha } \mathrm{cos\alpha }\\ -\mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right]\left[\begin{array}{l}\mathrm{sin\alpha }-\mathrm{cos\alpha }\\ \mathrm{cos\alpha } \mathrm{ }\mathrm{sin\alpha }\end{array}\right]\\ =\left[\begin{array}{l}{\mathrm{sin}}^{2}\mathrm{\alpha }+{\mathrm{cos}}^{2}\mathrm{\alpha }-\mathrm{sin\alpha cos\alpha }+\mathrm{cos\alpha sin\alpha }\\ -\mathrm{cos\alpha sin\alpha }+{\mathrm{sin\alpha cos\alpha cos}}^{2}\mathrm{\alpha }+{\mathrm{sin}}^{2}\mathrm{\alpha }\end{array}\right]\\ =\left[\begin{array}{l}10\\ 01\end{array}\right]\\ =\mathrm{I}\\ \mathrm{Thus},\mathrm{AA}‘=\mathrm{I}.\end{array}$

Q.39

$\begin{array}{l}\left(\text{i}\right)\mathit{}\mathit{\text{Show tha t the matrix A=}}\left[\begin{array}{l}1-15\\ 1\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}21\\ 5\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}13\end{array}\right]\mathit{}\mathit{i}\mathit{\text{s a symmetric matrix.}}\\ \\ \left(ii\right)\mathit{\text{Show tha tthe matrix A=}}\left[\begin{array}{l}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}\hspace{0.17em}}1-1\\ -1\text{\hspace{0.17em}\hspace{0.17em}}0\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1\\ \text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}1-1\text{\hspace{0.17em}\hspace{0.17em}}0\end{array}\right]\mathit{}\mathit{\text{is a skew}}\\ \mathit{\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}symmetric matrix}}\mathit{.}\end{array}$

Ans.

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

Q.40

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

Ans.

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

Q.41

$\text{Find \hspace{0.17em}\hspace{0.17em}}\frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)\mathrm{}\mathrm{and}\frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right),\mathrm{when}\mathrm{A}=\left[\begin{array}{l} 0 \mathrm{a}\mathrm{b}\\ -\mathrm{a} 0\mathrm{c}\\ -\mathrm{b}-\mathrm{c}0\end{array}\right].$

Ans.

$\begin{array}{l}\mathrm{Given}, \mathrm{A}=\left[\begin{array}{l} 0 \mathrm{a}\mathrm{b}\\ -\mathrm{a} 0\mathrm{c}\\ -\mathrm{b}-\mathrm{c}0\end{array}\right]\right]\\ \mathrm{so}, \mathrm{A}‘=\left[{\begin{array}{l} 0 \mathrm{a}\mathrm{b}\\ -\mathrm{a} 0\mathrm{c}\\ -\mathrm{b}-\mathrm{c}0\end{array}}^{‘}\right]\\ \mathrm{ }=\left[\begin{array}{l}\mathrm{0}- \mathrm{a}-\mathrm{b}\\ \mathrm{a} 0-\mathrm{c}\\ \mathrm{b} \mathrm{c} 0\end{array}\right]\\ \left(\mathrm{i}\right) \mathrm{A}+\mathrm{A}‘=\left[\begin{array}{l} 0 \mathrm{b}\\ -\mathrm{a} 0\mathrm{c}\\ -\mathrm{b}-\mathrm{c}0\end{array}\right]+\left[\begin{array}{l}\mathrm{0}- \mathrm{a}-\mathrm{b}\\ \mathrm{a} 0-\mathrm{c}\\ \mathrm{b} \mathrm{c} 0\end{array}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 00 0\end{array}\right]\\ \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)=\frac{1}{2}\left[\begin{array}{l}000\\ 000\\ 00 0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}000\\ 000\\ 00 0\end{array}\right]\\ \left(\mathrm{ii}\right) \mathrm{A}-\mathrm{A}‘=\left[\begin{array}{l} 0 \mathrm{ }\mathrm{ab}\\ -\mathrm{a} \mathrm{0}\mathrm{c}\\ -\mathrm{b}-\mathrm{c}\mathrm{0}\end{array}\right]-\left[\begin{array}{l}\mathrm{0}-\mathrm{ }\mathrm{a}-\mathrm{b}\\ \mathrm{a} \mathrm{0}-\mathrm{c}\\ \mathrm{b} \mathrm{c} 0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} 02\mathrm{a}2\mathrm{b}\\ -2\mathrm{a} 02\mathrm{c}\\ -2\mathrm{b} -2\mathrm{c} 0\end{array}\right]\\ \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)=\frac{1}{2}\left[\begin{array}{l} 02\mathrm{a}2\mathrm{b}\\ -2\mathrm{a}02\mathrm{c}\\ -2\mathrm{b} -2\mathrm{c} 0\end{array}\right]\\ =\left[\begin{array}{l} 0 \mathrm{a}\mathrm{b}\\ -\mathrm{a} 0\mathrm{c}\\ -\mathrm{b}-\mathrm{c}0\end{array}\right]\\ \end{array}$

Q.42

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

Ans.

$\begin{array}{l}\left(\mathrm{i}\right) \mathrm{Let} \mathrm{A}=\left[\begin{array}{l}3 \mathrm{ }5\\ 1-1\end{array}\right]\\ \mathrm{ }\therefore \mathrm{ }\mathrm{A}‘=\left[\begin{array}{l}3 \mathrm{ }1\\ 5-1\end{array}\right]\\ \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)=\frac{1}{2}\left\{\left[\begin{array}{l}3 \mathrm{ }5\\ 1-1\end{array}\right]+\left[\begin{array}{l}3 \mathrm{ }1\\ 5-1\end{array}\right]\right\}\\ =\frac{1}{2}\left[\begin{array}{l}6 \mathrm{ }6\\ 6-2\end{array}\right]\\ =\left[\begin{array}{l}3 \mathrm{ }3\\ 3-1\end{array}\right]=\mathrm{B}\left(\mathrm{Let}\right)\\ \mathrm{B}‘={\left[\begin{array}{l}3 \mathrm{ }3\\ 3-1\end{array}\right]}^{‘}\\ \mathrm{ }=\left[\begin{array}{l}3 \mathrm{ }3\\ 3-1\end{array}\right]\\ \mathrm{So}, \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)\mathrm{is}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)=\frac{1}{2}\left\{\left[\begin{array}{l}3 \mathrm{ }5\\ 1-1\end{array}\right]-\left[\begin{array}{l}3 \mathrm{ }1\\ 5-1\end{array}\right]\right\}\\ =\frac{1}{2}\left[\begin{array}{l} \mathrm{ }0 \mathrm{ }4\\ -4 \mathrm{ }0\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }0 \mathrm{ }2\\ -2 \mathrm{ }0\end{array}\right]=\mathrm{C}\left(\mathrm{Let}\right)\\ \mathrm{C}‘={\left[\begin{array}{l} \mathrm{ }0 \mathrm{ }2\\ -2 \mathrm{ }0\end{array}\right]}^{‘}\\ \mathrm{ }=\left[\begin{array}{l}0\mathrm{ }-2\\ 2 0\end{array}\right]\\ \mathrm{ }=-\left[\begin{array}{l} \mathrm{ }0 \mathrm{ }2\\ -2 \mathrm{ }0\end{array}\right]=-\mathrm{C}\\ \mathrm{So}, \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)\mathrm{is}\mathrm{skew} \mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \mathrm{Now}, \mathrm{B}+\mathrm{C}=\left[\begin{array}{l}3 \mathrm{ }3\\ 3-1\end{array}\right]+\left[\begin{array}{l} \mathrm{ }0 \mathrm{ }2\\ -2 \mathrm{ }0\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}3 \mathrm{ }5\\ 1-1\end{array}\right]=\mathrm{A}\\ \left(\mathbf{ii}\right)\\ \mathrm{Let} \mathrm{ }\mathrm{A}=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]\\ \therefore \mathrm{ }\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]\\ \mathrm{ }\mathrm{A}+\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]+\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{ }12-4 \mathrm{ }4\\ -4 \mathrm{ }6-2\\ 4-2 \mathrm{ }6\end{array}\right]\\ \therefore \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]=\mathrm{B}\left(\mathrm{Let}\right)\\ \mathrm{A}-\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]-\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]\\ =\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]\\ \therefore \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)=\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]=\mathrm{C}\left(\mathrm{Let}\right)\\ \mathrm{C}\mathrm{is}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \therefore \mathrm{B}+\mathrm{C}=\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]+\left[\begin{array}{l}000\\ 000\\ 000\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }6-2 \mathrm{ }2\\ -2 \mathrm{ }3-1\\ 2-1 \mathrm{ }3\end{array}\right]=\mathrm{A}\\ \left(\mathbf{iii}\right) \mathrm{Let} \mathrm{A}=\left[\begin{array}{l} 3 \mathrm{ }3-1\\ -2-2 \mathrm{ }1\\ -4-5 \mathrm{ }2\end{array}\right]\\ \therefore \mathrm{ }\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }3-2-4\\ \mathrm{ }3-2-5\\ -1 \mathrm{ }1 \mathrm{ }2\end{array}\right]\\ \mathrm{A}+\mathrm{A}‘=\left[\begin{array}{l} 3 \mathrm{ }3-1\\ -2-2 \mathrm{ }1\\ -4-5 \mathrm{ }2\end{array}\right]+\left[\begin{array}{l} \mathrm{ }3-2-4\\ \mathrm{ }3-2-5\\ -1 \mathrm{ }1 \mathrm{ }2\end{array}\right]\\ =\left[\begin{array}{l} 6 \mathrm{ }1-5\\ \mathrm{ }1-4-4\\ -5-4 \mathrm{ }4\end{array}\right]\\ \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)=\frac{1}{2}\left[\begin{array}{l} 6 \mathrm{ }1-5\\ \mathrm{ }1-4-4\\ -5-4 \mathrm{ }4\end{array}\right]\\ =\left[\begin{array}{l} 3 \mathrm{ }\frac{1}{2}-\frac{5}{2}\\ \mathrm{ }\frac{1}{2}-2-2\\ -\frac{5}{2}-2 \mathrm{ }2\end{array}\right]=\mathrm{B} \mathrm{ }\left(\mathrm{Let}\right)\\ \mathrm{B}‘=\left[\begin{array}{l} 3 \mathrm{ }\frac{1}{2}-\frac{5}{2}\\ \mathrm{ }\frac{1}{2}-2-2\\ -\frac{5}{2}-2 \mathrm{ }2\end{array}\right]=\mathrm{B}\\ \mathrm{So},\mathrm{ }\frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)\mathrm{is}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \mathrm{A}-\mathrm{A}‘=\left[\begin{array}{l} 3 \mathrm{ }3-1\\ -2-2 \mathrm{ }1\\ -4-5 \mathrm{ }2\end{array}\right]-\left[\begin{array}{l} \mathrm{ }3-2-4\\ \mathrm{ }3-2-5\\ -1 \mathrm{ }1 \mathrm{ }2\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} 0 \mathrm{ }53\\ -5 \mathrm{ }06\\ -3-60\end{array}\right]\\ \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)=\frac{1}{2}\left[\begin{array}{l} 0 \mathrm{ }53\\ -5 \mathrm{ }06\\ -3-60\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} 0 \mathrm{ }\frac{5}{2}\frac{3}{2}\\ -\frac{5}{2} \mathrm{ }03\\ -\frac{3}{2}-30\end{array}\right]=\mathrm{C} \mathrm{ }\left(\mathrm{Let}\right)\\ \mathrm{C}‘={\left[\begin{array}{l} 0 \mathrm{ }\frac{5}{2}\frac{3}{2}\\ -\frac{5}{2} \mathrm{ }03\\ -\frac{3}{2}-30\end{array}\right]}^{}=-\mathrm{C}\\ \mathrm{So},\mathrm{ }\frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)\mathrm{is}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \mathrm{B}+\mathrm{C}=\left[\begin{array}{l} 3 \mathrm{ }\frac{1}{2}-\frac{5}{2}\\ \mathrm{ }\frac{1}{2}-2-2\\ -\frac{5}{2}-2 \mathrm{ }2\end{array}\right]+\left[\begin{array}{l} 0 \mathrm{ }\frac{5}{2}\frac{3}{2}\\ -\frac{5}{2} \mathrm{ }03\\ -\frac{3}{2}-30\end{array}\right]\\ =\left[\begin{array}{l} 3 \mathrm{ }3-1\\ -2-2 \mathrm{ }1\\ -4-5 \mathrm{ }2\end{array}\right]=\mathrm{A}\\ \left(\mathrm{iv}\right)\\ \mathrm{Let}\mathrm{A}=\left[\begin{array}{l} \mathrm{ }15\\ -12\end{array}\right],\\ \therefore \mathrm{ }\mathrm{A}‘=\left[\begin{array}{l}1-1\\ 5 2\end{array}\right]\\ \mathrm{A}+\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }15\\ -12\end{array}\right]+\left[\begin{array}{l}1-1\\ 5 2\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }1+15-1\\ -1+52+2\end{array}\right]\\ =\left[\begin{array}{l}24\\ 44\end{array}\right]\\ \therefore \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right)\mathrm{ }=\frac{1}{2}\left[\begin{array}{l}24\\ 44\end{array}\right]\\ =\left[\begin{array}{l}12\\ 22\end{array}\right]=\mathrm{B}\\ \mathrm{B}‘={\left[\begin{array}{l}12\\ 22\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}12\\ 22\end{array}\right]=\mathrm{B}\\ \mathrm{Therefore}, \frac{1}{2}\left(\mathrm{A}+\mathrm{A}‘\right) \mathrm{is}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \mathrm{A}-\mathrm{A}‘=\left[\begin{array}{l} \mathrm{ }15\\ -12\end{array}\right]-\left[\begin{array}{l}1-1\\ 5 2\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }1-15+1\\ -1-52-2\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }06\\ -60\end{array}\right]\\ \therefore \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right)\\ =\frac{1}{2}\left[\begin{array}{l} \mathrm{ }06\\ -60\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }03\\ -30\end{array}\right]=\mathrm{C}\\ \mathrm{C}‘={\left[\begin{array}{l} \mathrm{ }03\\ -30\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}0-3\\ 3 0\end{array}\right]=\mathrm{C}\\ \mathrm{Therefore}, \frac{1}{2}\left(\mathrm{A}-\mathrm{A}‘\right) \mathrm{is}\mathrm{a}\mathrm{skew}\mathrm{symmetric}\mathrm{matrix}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ \mathrm{B}+\mathrm{C}=\left[\begin{array}{l}12\\ 22\end{array}\right]+\left[\begin{array}{l} \mathrm{ }03\\ -30\end{array}\right]\\ =\left[\begin{array}{l}1+02+3\\ 2-32+0\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }15\\ -12\end{array}\right]=\mathrm{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

(C) Zero 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

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

Ans.

$\begin{array}{l}\mathrm{Given}:\\ \mathrm{A}=\left[\begin{array}{l}\mathrm{cos\alpha }-\mathrm{sin\alpha }\\ \mathrm{sin\alpha } \mathrm{cos\alpha }\end{array}\right],\mathrm{ } \mathrm{A}‘=\left[\begin{array}{l} \mathrm{cos\alpha sin\alpha }\\ -\mathrm{sin\alpha cos\alpha }\end{array}\right]\\ \mathrm{A}+\mathrm{A}‘=\left[\begin{array}{l}\mathrm{cos\alpha }+\mathrm{cos\alpha }-\mathrm{sin\alpha }+\mathrm{sin\alpha }\\ \mathrm{sin\alpha }+\left(-\mathrm{sin\alpha }\right) \mathrm{ }\mathrm{cos\alpha }+\mathrm{cos\alpha }\end{array}\right]\\ =\left[\begin{array}{l}2\mathrm{cos\alpha }0\\ 02\mathrm{cos\alpha }\end{array}\right]=\left[\begin{array}{l}10\\ 01\end{array}\right]\\ ⇒2\mathrm{cos\alpha }=1\\ ⇒ \mathrm{ }\mathrm{cos\alpha }=\frac{1}{2}=\mathrm{cos}\frac{\mathrm{\pi }}{3}\\ ⇒\mathrm{\alpha }=\frac{\mathrm{\pi }}{3}\\ \mathrm{The}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{B}\right).\end{array}$

Q.45

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

Ans.

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

Q.46

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}21\\ 11\end{array}\right]\end{array}$

Ans.

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

Q.47

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

Ans.

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

Q.48

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

Ans.

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

Q.49

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

Ans.

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

Q.50

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

Ans.

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

Q.51

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

`
Ans.

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

Q.52

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

Ans.

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

Q.53

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

Ans.

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

Q.54

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

Ans.

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

Q.55

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

Ans.

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

Q.56

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

Ans.

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

Q.57

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

Ans.

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

Q.58

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

Ans.

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

Q.59

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\\ \mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}2-33\\ 223\\ 3-22\end{array}\right]\end{array}$

Ans.

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

Q.60

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

Ans.

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

Q.61

$\begin{array}{l}\mathrm{Using}\mathrm{elementary}\mathrm{transformations},\mathrm{find}\mathrm{the}\mathrm{inverse}\\ \mathrm{of}\mathrm{each}\mathrm{of}\mathrm{the}\mathrm{matrices}\left[\begin{array}{l}20-1\\ 510\\ 003\end{array}\right]\end{array}$

Ans.

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

Q.62

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

(A) AB = BA

(B) AB = BA = 0

(C) AB = 0, BA = I

(D) AB = BA = I

Ans.

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

Thus, option (D) is correct.

Q.63

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{A}=\left[\begin{array}{l}01\\ 00\end{array}\right],\mathrm{ }\mathrm{show}\mathrm{that}{\left(\mathrm{aI}+\mathrm{bA}\right)}^{\mathrm{n}}={\mathrm{a}}^{\mathrm{n}}\mathrm{I}+{\mathrm{na}}^{\mathrm{n}–1},\\ \mathrm{where}\mathrm{is}\mathrm{the}\mathrm{identity}\mathrm{matrix}\mathrm{of}\mathrm{order}2\mathrm{and}\mathrm{n}\in \mathrm{N}.\end{array}$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}L.H.S.={\left(aI+bA\right)}^{n}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left(a\left[\begin{array}{l}1\text{}0\\ 01\end{array}\right]+b\left[\begin{array}{l}0\text{}1\\ 0\text{}0\end{array}\right]\right)}^{n}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}={\left[\begin{array}{l}ab\\ 0a\end{array}\right]}^{n}\\ R.H.S.={a}^{n}I+n{a}^{n-1}bA\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}={a}^{n}\left[\begin{array}{l}1\text{}0\\ 01\end{array}\right]+n{a}^{n-1}b\left[\begin{array}{l}01\\ 00\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}{a}^{n}0\\ 0\text{}{a}^{n}\end{array}\right]+\left[\begin{array}{l}0n{a}^{n-1}b\\ 00\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}{a}^{n}n{a}^{n-1}b\\ 0\text{}{a}^{n}\end{array}\right]\\ So,\text{\hspace{0.17em}}{\left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]}^{n}=\left[\begin{array}{l}{a}^{n}n{a}^{n-1}b\\ 0{a}^{n}\end{array}\right]\\ \text{Apply principle of Mathematical Induction}\\ \text{P}\left(n\right):{\left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]}^{n}=\left[\begin{array}{l}{a}^{n}n{a}^{n-1}b\\ 0\text{}{a}^{n}\end{array}\right]\\ For\text{}\text{n}=\text{1,}\\ L.H.S.={\left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]}^{1}=\left[\begin{array}{l}a\text{}b\\ 0\text{}a\end{array}\right]\\ R.H.S=\left[\begin{array}{l}{a}^{1}1.{a}^{1-1}b\\ 0{a}^{1}\end{array}\right]=\left[\begin{array}{l}a\text{}ab\\ 0\text{}a\end{array}\right]\\ So,\text{\hspace{0.17em}}\\ \left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]=\left[\begin{array}{l}a\text{}ab\\ 0\text{}a\end{array}\right]\\ So,\text{\hspace{0.17em}}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}a\text{}b\\ 0\text{}a\end{array}\right]=\left[\begin{array}{l}aab\\ 0a\end{array}\right]\\ \therefore \text{P (n) is true for n}=1.\\ \text{Let P (n) be true for n}=\text{k}\text{.}\\ \therefore P\left(k\right):\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\left[\begin{array}{l}ab\\ 0a\end{array}\right]}^{k}=\left[\begin{array}{l}{a}^{k}k{a}^{k-1}b\\ 0\text{}{a}^{k}\end{array}\right]\text{\hspace{0.17em}}\text{\hspace{0.17em}}\dots \left(i\right)\\ \text{Multiple both sides by}\text{\hspace{0.17em}}\left[\begin{array}{l}a\text{}b\\ 0\text{}a\end{array}\right]\\ {\left[\begin{array}{l}a\text{}b\\ 0\text{}a\end{array}\right]}^{k}×\left[\begin{array}{l}a\text{}b\\ 0a\end{array}\right]=\left[\begin{array}{l}{a}^{k}k{a}^{k-1}b\\ 0\text{}{a}^{k}\end{array}\right]×\left[\begin{array}{l}ab\\ 0a\end{array}\right]\\ L.H.S.={\left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]}^{k}×\left[\begin{array}{l}a\text{}b\\ 0\text{}a\end{array}\right]\\ =\left[\begin{array}{l}{a}^{k}k{a}^{k-1}b\\ 0{a}^{k}\end{array}\right]×\left[\begin{array}{l}ab\\ 0\text{}a\end{array}\right]\\ \left[From\text{equation}\left(i\right),\text{\hspace{0.17em}}we\text{\hspace{0.17em}}get\right]\\ =\left[\begin{array}{l}{a}^{k+1}{a}^{k}b+k{a}^{k}b\\ 0{a}^{k+1}\end{array}\right]\\ =\left[\begin{array}{l}{a}^{k+1}\left(k+1\right){a}^{k}b\\ 0{a}^{k+1}\end{array}\right]\\ L.H.S.=\left[\begin{array}{l}{a}^{k}k{a}^{k-1}b\\ 0{a}^{k}\end{array}\right]×\left[\begin{array}{l}ab\\ 0a\end{array}\right]\\ =\left[\begin{array}{l}{a}^{k+1}b{a}^{k}+k{a}^{k}b\\ 0{a}^{k+1}\end{array}\right]\\ =\left[\begin{array}{l}{a}^{k+1}\left(k+1\right){a}^{k}b\\ 0{a}^{k+1}\end{array}\right]\\ \text{This shows P(n) is true for n = k + 1 then by principle of mathematical}\\ \text{induction, P(n)}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{is true for all positive integral values of n}\text{.}\end{array}$

Q.64

$\text{If \hspace{0.17em}\hspace{0.17em}A=}\left[\begin{array}{l}111\\ 111\\ 111\end{array}\right],\mathrm{prove}\mathrm{that}{\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{l}{3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}\\ {3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}\\ {3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}{3}^{\mathrm{n}–1}\end{array}\right],\mathrm{ }\mathrm{n}\in \mathrm{N}.$

Ans.

$\begin{array}{c}\mathrm{Let}\mathrm{P}\left(\mathrm{n}\right):{\mathrm{A}}^{\mathrm{n}}=\left[\begin{array}{l}{\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}\\ {\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}\\ {\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}{\mathrm{3}}^{\mathrm{n}-1}\end{array}\right]\mathrm{and}\mathrm{A}=\left[\begin{array}{l}11\mathrm{1}\\ 1\mathrm{1}\mathrm{1}\\ 1\mathrm{1}\mathrm{1}\end{array}\right]\\ \mathrm{For}\mathrm{n}=\mathrm{1}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=\mathrm{A}=\left[\begin{array}{l}11\mathrm{1}\\ 11\mathrm{1}\\ \mathrm{1}1\mathrm{1}\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left[\begin{array}{l}{\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}\\ {\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}\\ {\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}{\mathrm{3}}^{1-1}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{3}}^{0}{\mathrm{3}}^{0}{\mathrm{3}}^{\mathrm{0}}\\ {\mathrm{3}}^{0}{\mathrm{3}}^{0}{\mathrm{3}}^{\mathrm{0}}\\ {\mathrm{3}}^{0}{\mathrm{3}}^{0}{\mathrm{3}}^{\mathrm{0}}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}1\mathrm{1}\mathrm{1}\\ \mathrm{1}\mathrm{1}\mathrm{1}\\ \mathrm{1}1\mathrm{1}\end{array}\right]\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\mathrm{ }\mathrm{is}\mathrm{true}\mathrm{for}\mathrm{n}=1.\\ \mathrm{Let}\mathrm{P}\left(\mathrm{n}\right)\mathrm{be}\mathrm{true}\mathrm{for}\mathrm{n}=\mathrm{k},\mathrm{then}\\ \mathrm{P}\left(\mathrm{k}\right):{\mathrm{A}}^{\mathrm{k}}=\left[\begin{array}{l}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\end{array}\right] ...\left(\mathrm{i}\right)\\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}\mathrm{A},\mathrm{we}\mathrm{get}\\ {\mathrm{A}}^{\mathrm{k}}.\mathrm{A}=\left[\begin{array}{l}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\end{array}\right].\mathrm{A}\\ {\mathrm{A}}^{\mathrm{k}+1}=\left[\begin{array}{l}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\end{array}\right].\left[\begin{array}{l}111\\ 111\\ 111\end{array}\right]\\ \mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}=\left[\begin{array}{l}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\\ {\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}{\mathrm{3}}^{\mathrm{k}-1}\end{array}\right].\left[\begin{array}{l}111\\ 111\\ 111\end{array}\right]\\ =\left[\begin{array}{l}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}\\ {3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}\\ {3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}+{3}^{\mathrm{k}–1}\end{array}\right]\\ =\left[\begin{array}{l}3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}\\ 3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}\\ 3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}3.{3}^{\mathrm{k}–1}\end{array}\right]\\ =\left[\begin{array}{l}{3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}\\ {3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}\\ {3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}{3}^{\left(\mathrm{k}–1\right)+1}\end{array}\right]\\ \therefore \mathrm{P}\left(\mathrm{n}\right)\mathrm{ }\mathrm{is}\mathrm{true}\mathrm{forn}=\mathrm{k}+1\\ \mathrm{By}\mathrm{principle}\mathrm{of}\mathrm{ma}\mathrm{the}\mathrm{matical}\mathrm{induction}\mathrm{that}\mathrm{P}\left(\mathrm{n}\right) \mathrm{is}\mathrm{true}\mathrm{f}\mathrm{oral}\mathrm{ln}\in \mathrm{N}.\end{array}$

Q.65

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

Ans.

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

$\begin{array}{l}\mathrm{Find}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{if}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[\begin{array}{l}02\mathrm{y} \mathrm{z}\\ \mathrm{x} \mathrm{y}\mathrm{}-\mathrm{z}\\ \mathrm{x}-\mathrm{y} \mathrm{z}\end{array}\right]\\ \mathrm{satisfiy}\mathrm{the}\mathrm{equation}\mathrm{A}‘\mathrm{A}=\mathrm{I}.\end{array}$

Ans.

$\begin{array}{l} \mathrm{A}=\left[\begin{array}{l}02\mathrm{y} \mathrm{ }\mathrm{z}\\ \mathrm{x} \mathrm{y}-\mathrm{z}\\ \mathrm{x}-\mathrm{y} \mathrm{ }\mathrm{z}\end{array}\right]\\ \mathrm{A}‘=\left[\begin{array}{l}0\mathrm{ } \mathrm{x}\mathrm{ } \mathrm{x}\\ 2\mathrm{y} \mathrm{ }\mathrm{y}-\mathrm{y}\\ \mathrm{z}-\mathrm{z} \mathrm{z}\end{array}\right]\\ \mathrm{A}‘\mathrm{A}=\left[\begin{array}{l}0 \mathrm{x} \mathrm{x}\\ 2\mathrm{y} \mathrm{ }\mathrm{y}-\mathrm{y}\\ \mathrm{z}-\mathrm{z} \mathrm{ }\mathrm{z}\end{array}\right]\left[\begin{array}{l}02\mathrm{y} \mathrm{ }\mathrm{z}\\ \mathrm{x} \mathrm{y}-\mathrm{z}\\ \mathrm{x}-\mathrm{y} \mathrm{ }\mathrm{z}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}0+{\mathrm{x}}^{2}+{\mathrm{x}}^{2}\mathrm{ } 0+\mathrm{xy}-\mathrm{xy} 0-\mathrm{xz}+\mathrm{xz}\\ 0+\mathrm{xy}-\mathrm{xy}4{\mathrm{y}}^{2}+{\mathrm{y}}^{2}+{\mathrm{y}}^{2}2\mathrm{yz}-\mathrm{yz}-\mathrm{yz}\\ 0-\mathrm{xz}+\mathrm{xz}2\mathrm{yz}-\mathrm{zy}-\mathrm{yz}\mathrm{ } {\mathrm{z}}^{2}+{\mathrm{z}}^{2}+{\mathrm{z}}^{2}\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{ccc}2{\mathrm{x}}^{2}& 0& 0\\ 0& 6{\mathrm{y}}^{2}& 0\\ 0& 0& 3{\mathrm{z}}^{2}\end{array}\right]\\ \mathrm{Since},\mathrm{ }\mathrm{A}‘\mathrm{A}=\mathrm{I}\\ \mathrm{So},\\ \left[\begin{array}{l}2{\mathrm{x}}^{2} 0 0\\ 0 \mathrm{ }6{\mathrm{y}}^{2} 0\\ 0 0 \mathrm{ }3{\mathrm{z}}^{2}\end{array}\right]=\left[\begin{array}{l}100\\ 010\\ 001\end{array}\right]\\ ⇒2{\mathrm{x}}^{2}=1,\mathrm{ }6{\mathrm{y}}^{2}=1\mathrm{ }\mathrm{and}3{\mathrm{z}}^{2}=1\\ ⇒\mathrm{x}=±\frac{1}{\sqrt{2}},\mathrm{ }\mathrm{y}=±\frac{1}{\sqrt{6}} \mathrm{and} \mathrm{z}=±\frac{1}{\sqrt{3}}\end{array}$

Q.69

$Forwhatvalueofx:\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[1\text{}2\text{}1\right]\left[\begin{array}{l}1\text{}2\text{}0\\ 2\text{}0\text{}1\\ 1\text{}0\text{}2\end{array}\right]\left[\begin{array}{l}0\\ 2\\ x\end{array}\right]=0$

Ans.

$\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\text{1}\text{}\text{2}\text{}\text{1}\right]\left[\begin{array}{l}\text{1}\text{}\text{2}\text{}\text{0}\\ \text{2}\text{}\text{0}\text{}\text{1}\\ \text{1}\text{}\text{0}\text{}\text{2}\end{array}\right]\left[\begin{array}{l}\text{0}\\ \text{2}\\ \text{x}\end{array}\right]=\text{0}\\ \left[1+4+1\text{}2+0+0\text{}0+2+2\right]\left[\begin{array}{l}\text{0}\\ \text{2}\\ \text{x}\end{array}\right]=\text{0}\\ \text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[6\text{}2\text{}4\right]\left[\begin{array}{l}\text{0}\\ \text{2}\\ \text{x}\end{array}\right]=\text{0}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[0+4+4x\right]=0\\ \text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4+4x=0\\ ⇒\text{}\text{}\text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}x=-1\end{array}$

Q.70

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

Ans.

$\begin{array}{l}\mathrm{Since},\mathrm{A}=\left[\begin{array}{l} 31\\ -12\end{array}\right],\mathrm{ }\mathrm{then}\\ {\mathrm{A}}^{2}=\left[\begin{array}{l} 31\\ -12\end{array}\right]\left[\begin{array}{l} 31\\ -12\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }9-13+2\\ -3-2 -1+4\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }85\\ -5 3\end{array}\right]\\ \mathrm{L}.\mathrm{H}.\mathrm{S}.={\mathrm{A}}^{2}-5\mathrm{A}+7\mathrm{I}\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }85\\ -5 3\end{array}\right]-5\left[\begin{array}{l} 31\\ -12\end{array}\right]+7\left[\begin{array}{l}10\\ 01\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }85\\ -5 3\end{array}\right]-\left[\begin{array}{l} \mathrm{ }15\mathrm{ } 5\\ -510\end{array}\right]+\left[\begin{array}{l}70\\ 07\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l} \mathrm{ }8-15+75-5+0\\ -5+5+0 3-10+7\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}00\\ 00\end{array}\right]\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\end{array}$

Q.71

$\text{Find x, if [x-5-1]}\left[\begin{array}{l}102\\ 021\\ 203\end{array}\right]\left[\begin{array}{l}\mathrm{x}\\ 4\\ 1\end{array}\right]=0.$

Ans.

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

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}\left(\mathrm{a}\right) \mathrm{Matrix}\mathrm{corresponding}\mathrm{to}\mathrm{sale}\mathrm{price}\mathrm{of}\mathrm{each}\mathrm{product}=\left[\begin{array}{l}\mathrm{2}.50\\ \mathrm{1}.50\\ \mathrm{1}.00\end{array}\right]\\ \mathrm{The}\mathrm{total}\mathrm{revenue}\mathrm{collected}\mathrm{by}\mathrm{each}\mathrm{market}\mathrm{is}\mathrm{given}\mathrm{by}\\ \left[\begin{array}{l}10,0002,00018,000\\ 6,00020,0008,000\end{array}\right]\left[\begin{array}{l}\mathrm{2}.50\\ \mathrm{1}.50\\ \mathrm{1}.00\end{array}\right]\mathrm{=}\left[\begin{array}{l}25,000+3,000+18,000\\ 15,000+30,000+8,000\end{array}\right]\\ \mathrm{=}\left[\begin{array}{l}46,000\\ 53,000\end{array}\right]\\ \mathrm{Total}\mathrm{cost}\mathrm{prices}\mathrm{of}\mathrm{the}\mathrm{commodities}\mathrm{each}\mathrm{market}\mathrm{I}\mathrm{and}\mathrm{II}\mathrm{are}\\ ‘ 46,000, ‘ 53,000\mathrm{.}\\ \mathrm{}\mathrm{Total} \mathrm{revnue}\mathrm{}\mathrm{by}\mathrm{both}\mathrm{markets}=46,000+53,000\\ = \mathrm{}99,000\\ \left(\mathrm{b}\right)\mathrm{The}\mathrm{cost}\mathrm{price}\mathrm{of}\mathrm{commodities}\mathrm{x},\mathrm{y}\mathrm{and}\mathrm{z}\mathrm{are}\mathrm{respectively}‘ 2.00,\\ ‘1.00\mathrm{and}‘ 0.50\mathrm{cost}\mathrm{price}\mathrm{of}\mathrm{each}\mathrm{market}\mathrm{is}\mathrm{given}\mathrm{below}\\ \mathrm{ }\left[\begin{array}{l}10,0002,00018,000\\ 6,00020,0008,000\end{array}\right]\left[\begin{array}{l}\mathrm{2}.00\\ \mathrm{1}.00\\ \mathrm{0}.50\end{array}\right]\mathrm{=}\left[\begin{array}{l}20,000+2,000+9,000\\ 12,000+20,000+4,000\end{array}\right]\\ \mathrm{=}\left[\begin{array}{l}\mathrm{31000}\\ \mathrm{36000}\end{array}\right]\\ \mathrm{Total}\mathrm{cost}\mathrm{prices}\mathrm{of}\mathrm{the}\mathrm{commodities}\mathrm{each}\mathrm{market}\mathrm{I}\mathrm{and}\mathrm{II}\mathrm{are}31,000,\\ \mathrm{ZZ}36,000\mathrm{.}\\ \mathrm{Total} \mathrm{revnue}\mathrm{}\mathrm{by}\mathrm{both}\mathrm{markets}=31,000+ \mathrm{}36,000\\ = 67,000\\ \mathrm{Gross}\mathrm{profit}= \mathrm{}99,000-67,000\\ =\mathrm{}32,000\end{array}$

Q.73

$\text{Find the matrix X so that\hspace{0.17em}\hspace{0.17em}X}\left[\begin{array}{l}123\\ 456\end{array}\right]=\left[\begin{array}{l}-7-8-9\\ 2 4 6\end{array}\right]$

Ans.

$\begin{array}{l} \mathrm{X}\left[\begin{array}{l}123\\ 456\end{array}\right]=\left[\begin{array}{l}–7–8–9\\ 2 4 6\end{array}\right]\\ \mathrm{Let}\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{X}\mathrm{be}\mathrm{m}×\mathrm{n}.\\ \mathrm{Since},\mathrm{the}\mathrm{order}\mathrm{of}\left[\begin{array}{l}123\\ 456\end{array}\right]\mathrm{is}2×3\mathrm{and}\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{X}\\ \mathrm{is}\mathrm{m}×\mathrm{n}.\\ \mathrm{So},\mathrm{n}=2.\\ \mathrm{Now},\mathrm{ }\mathrm{the}\mathrm{}\mathrm{order}\mathrm{of}\left[\begin{array}{l}–7–8–9\\ 2 4 6\end{array}\right]\mathrm{is} 2×3\mathrm{and}\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{X}\mathrm{is}\mathrm{m}×\mathrm{n}.\\ \mathrm{So},\mathrm{m}=2.\\ \mathrm{Thus},\mathrm{order}\mathrm{of}\mathrm{matrix}\mathrm{X}\mathrm{is}2×2.\\ \mathrm{Let}\mathrm{X}=\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right], \mathrm{then}\\ \mathrm{ }\left[\begin{array}{l}\mathrm{a}\mathrm{b}\\ \mathrm{c}\mathrm{d}\end{array}\right]\mathrm{}\left[\begin{array}{l}123\\ 456\end{array}\right]\left[\begin{array}{l}–7–8–9\\ 2 4 6\end{array}\right]\mathrm{}=\\ \left[\begin{array}{l}\mathrm{a}+4\mathrm{b}2\mathrm{a}+5\mathrm{b}3\mathrm{a}+6\mathrm{b}\\ \mathrm{c}+4\mathrm{d}2\mathrm{c}+5\mathrm{d}3\mathrm{c}+6\mathrm{d}\end{array}\right]=\left[\begin{array}{l}–7–8–9\\ 2 4 6\end{array}\right]\\ ⇒ \mathrm{a}+4\mathrm{b}=-7 ...\left(\mathrm{i}\right)\\ \mathrm{ }2\mathrm{a}+5\mathrm{b}=-8 ...\left(\mathrm{ii}\right)\\ \mathrm{ }\mathrm{c}+4\mathrm{d}=2 \mathrm{ }...\left(\mathrm{iii}\right)\\ 2\mathrm{c}+5\mathrm{d}=4 . ..\left(\mathrm{iv}\right)\\ 3\mathrm{c}+6\mathrm{d}=6 ...\left(\mathrm{v}\right)\\ \mathrm{Solving}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}\\ \mathrm{a}=1,\mathrm{ }\mathrm{b}=-\mathrm{2}\\ \mathrm{value}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{satis}\mathrm{fies}3\mathrm{a}+6\mathrm{b}=-9.\\ \mathrm{Solving}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{and}\mathrm{equation}\left(\mathrm{iv}\right),\mathrm{we}\mathrm{get}\\ \mathrm{c}=2,\mathrm{ }\mathrm{d}=0\\ \mathrm{value}\mathrm{of}\mathrm{c}\mathrm{and}\mathrm{d}\mathrm{satisfies}3\mathrm{c}+6\mathrm{d}=6.\\ \mathrm{Hence}, \mathrm{X}=\left[\begin{array}{l}1-2\\ 2 0\end{array}\right]\end{array}$

Q.74

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

Ans.

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

Q.75

$\begin{array}{l}\mathbf{If}\mathbf{A}\mathrm{=}\left[\begin{array}{l}\mathrm{\alpha } \mathrm{\beta }\\ \mathrm{\gamma }-\mathrm{\alpha }\end{array}\right]\mathrm{}\mathrm{is}\mathrm{}\mathrm{such}\mathrm{that}{\mathrm{A}}^{2}=\mathrm{I},\mathrm{then}\\ \left(\mathrm{A}\right)1+{\mathrm{\alpha }}^{2}+\mathrm{\beta \gamma }=0\\ \left(\mathrm{B}\right)1-{\mathrm{\alpha }}^{2}+\mathrm{\beta \gamma }=0\\ \left(\mathrm{C}\right)1-{\mathrm{\alpha }}^{2}-\mathrm{\beta \gamma }=0\\ \left(\mathrm{D}\right)1+{\mathrm{\alpha }}^{2}-\mathrm{\beta \gamma }=0\end{array}$

Ans.

$\begin{array}{l} {\mathrm{A}}^{2}=\left[\begin{array}{l}\mathrm{\alpha } \mathrm{\beta }\\ \mathrm{\gamma }-\mathrm{\alpha }\end{array}\right]\left[\begin{array}{l}\mathrm{\alpha } \mathrm{\beta }\\ \mathrm{\gamma }-\mathrm{\alpha }\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}{\mathrm{\alpha }}^{2}+\mathrm{\beta \gamma } \mathrm{\alpha \beta }-\mathrm{\alpha \beta }\\ \mathrm{\alpha \gamma }-\mathrm{\alpha \gamma } \mathrm{ }\mathrm{\beta \gamma }+{\mathrm{\alpha }}^{2}\end{array}\right]\\ âˆµ \mathrm{ }{\mathrm{A}}^{2}=\left[\begin{array}{l}10\\ 01\end{array}\right]\\ \therefore \left[\begin{array}{l}10\\ 01\end{array}\right]=\left[\begin{array}{l}{\mathrm{\alpha }}^{2}+\mathrm{\beta \gamma }\mathrm{ } 0\\ 0 \mathrm{ }\beta \mathrm{\gamma }+{\mathrm{\alpha }}^{2}\end{array}\right]\\ ⇒{\mathrm{\alpha }}^{2}+\mathrm{\beta \gamma }=1\\ ⇒1-{\mathrm{\alpha }}^{2}-\mathrm{\beta \gamma }=0\\ \therefore \mathrm{Option}\mathrm{ }\mathrm{C}\mathrm{is}\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{.}\end{array}$

Q.76

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

Ans.

A is symmetric matrix then aij=aji

A is skew symmetric matrix then

aij =– aji

This implies that aij = aji = – aji

2aji = 0

Or aji = 0

Therefore, A is a zero matrix.

Option (B) is correct.

Q.77

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

Ans.

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