# NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.3

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 are an essential resource for students of Class 12 to help them get a step ahead in their preparation for the Class 12 board examinations in Mathematics. The Class 12 NCERT textbook on Mathematics is recommended by the CBSE. Several other boards also recommend the NCERT textbook in their syllabus. The 3rd chapter in the NCERT Mathematics book is on the topic of matrices. Matrices are extensively used in Mathematics and other fields like Optics, Quantum Mechanics, Economics, etc. In Mathematics, they help in solving linear equations, plotting graphs, etc. They are applied in the studies of reflection and refraction.

Class 12 Maths Chapter 3 Exercise 3.3 is based on the concepts of the transpose of Matrices and Symmetric and Skew-symmetric Matrices. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 along with the study material cover all the important concepts as per the syllabus prescribed by the CBSE for its board examinations. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 are designed in such a way by the expert teachers that they help get the students ready to easily solve the questions asked in the CBSE board examinations. Practising the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 also helps the students in preparing for various entrance examinations for college admissions like IIT-JEE, VITEEE, etc. Students should practice NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 to be thorough with the problems.

## NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 (Ex 3.3)

NCERT Solutions Class 12 Maths Chapter 3 Exercise 3.3 have been prepared by expert teachers in an easy and simplified manner that can help every student studying in Class 12 in solving the questions with ease. Before attempting to solve this exercise, one needs to be very clear about the previously discussed concepts. These include types of matrices like Square Matrices and Diagonal Matrices, Equality of Matrices, Operations on Matrices like Addition and their Properties, Scalar Multiplication of Matrices, and Multiplication of Two Matrices and their Properties. This exercise is based on the concepts of the transpose of a Matrix, properties of the transpose of a Matrix, symmetric Matrices and skew-symmetric Matrices.

This exercise contains 12 questions, out of which 2 are MCQs (Multiple Choice Questions). The rest of the questions are short-answer questions and long-answer questions. Six of these questions are based on the concepts of transpose and its properties, while the rest four are based on the concepts of symmetric and skew-symmetric matrices. All the questions in NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 should be solved and practised repetitively to get a hold of the concepts studied and to score better marks in examinations. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 also cover two important theorems on symmetric and skew-symmetric matrices. Questions are often asked in the CBSE board examinations based on these two theorems.

## Some important points to learn before solving Exercise 3.3

The important topics covered in Exercise 3.3 include:

• Transpose of a Matrix- It is an operator that helps in obtaining the inverse and adjoint of a matrix. It can be obtained by interchanging the row and column indices of the given matrix. The transpose of matrix A is denoted by Aᵀ.
• Symmetric Matrix- A square matrix is said to be symmetric if and only if it is equal to its transpose.
• Skew-symmetric Matrix- A square matrix is said to be skew-symmetric if it is equal to the negative of its transpose matrix.
• If A and B are two matrices, some of the properties of their transpose are:
1. (A’)’ = A.
2. (kA)’ = kA’
3. (A + B)’ = A’ + B’
4. (AB)’ = B’ A’

### Study NCERT Solutions Prepared By Highly Qualified Teachers

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 make it easier for students to accurately apply the concepts taught in the chapter. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 have been prepared by highly qualified teachers in a precise and error-free manner. They prepare these solutions taking into consideration the level of understanding that students in Class 12 have. The problems in this exercise have been solved step-wise to help in easily writing the answers in examinations. Examiners always emphasize writing step-wise answers in the examinations. The solutions in NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 help score full marks in the examinations. A lot of students find it difficult to understand and remember the vast number of formulas given in the NCERT textbook, and consequently are unable to apply them to the problems in the exercises. Hence, practising the stepwise solutions in NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 makes it easier to remember the formulas and apply them correctly. This gives students an advantage in their examinations as they have the right approach to solve the problems.

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Along with the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3, Extramarks also provides study material for Class 12 Mathematics. The study material helps in a thorough understanding of the concepts required for solving Exercise 3.3. This study material also helps students revise the chapter. It has been prepared by highly qualified teachers having years of experience in a brief and easy-to-understand manner. Extramarks provides revision notes, CBSE Sample Question Papers, CBSE past years’ question papers, solved exemplar problems, worksheets for practice and many more. So, students can find all their study needs in one place and do not have to look elsewhere for more. Extramarks also provides study material and practice papers for students preparing for various entrance examinations, including JEE Mains and JEE Advanced. Students can refer to the short notes and practice questions for a quick revision before they appear for their examinations.

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Extramarks has prepared the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 in a step-by-step manner that helps the Class 12 students thoroughly understand the problems even if they are attempting them for the first time. NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 by Extramarks also provides alternative solutions to some problems that are easier to understand. The chapter on matrices is an important part of competitive examinations like JEE Mains and JEE Advanced along with the CBSE board examinations. The performance in these examinations determines the career of the students, as college admissions are based on these examinations. This leaves them stressed by the immense pressure placed on them and sometimes makes them demotivated. But, thoroughly practising the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 along with the study material helps the students in understanding the concepts and retain them in their minds for a long time. This builds confidence in them as they approach multiple examinations.

### Study Material That Follows the NCERT Guidelines

NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 have been prepared by highly qualified teachers who have years of experience in CBSE board examinations. They prepare the NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 taking into account the latest guidelines provided by the NCERT. The NCERT Solutions for Class 12 Maths Chapter 3 Exercise 3.3 are modified and updated regularly as per the latest guidelines. So one does not have to worry about the changes made by the CBSE in its latest curriculum. Extramarks also encourages students to give their feedback and suggest any changes or improvements.

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### NCERT Solutions for Class 12 Maths Chapter 3 Other Exercises

 Chapter 3 Matrix Exercises Exercise 3.1 10 Questions & Solutions (5 Short Answers, 5 Long Answers) Exercise 3.2 22 Questions & Solutions (3 Short Answers, 19 Long Answers) Exercise 3.4 18 Questions & Solutions (18 Short Answers)

Q.1

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

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.2

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

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.3

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

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.4

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

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.5

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

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.6

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

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.7

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

Ans

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

Q.8

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

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.9

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

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.10

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

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.11 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.12

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

Ans

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