# NCERT Solutions Class 12 Mathematics Chapter 4 – Determinants

NCERT Solutions Class 12 Mathematics Chapter 4 Determinants is available online for the students to refer to and study on Extramarks. The chapter introduces determinants, their properties, and their applications. In addition, the chapter also includes various problems and a step-by-step solution to each problem that aids students in understanding the concept better. Furthermore, the solutions are built on the CBSE syllabus of the latest academic year and with reference to NCERT.

A determinant is a number that can only be calculated with the help of square matrices. If the students want to score more in linear equations, then the determinant is essential to learn. It helps to check whether the system of equations has a unique solution or not. Besides, students will get a glimpse of the previous chapter, matrices. This chapter also carries essential elements and helps to improve their calculations.

In the NCERT Solutions Class 12 Mathematics Chapter 4, students can understand the complex theories and formulas in an easy to understand language. These sub-topics include basic principles of determinant, cofactors, adjoint, and the inverse of a matrix. Besides, this solution has proper explanations and clear concepts to help students understand better.

Students looking for step-by-step solutions to the problems mentioned in ch 4 mathematics class 12 may register in Extramarks. Extramarks not only offers other NCERT Solutions for Class 12 but also from Class 1 to class 11. In addition, students may log in to the website for updated study material, revision notes, CBSE sample papers, important questions, and the latest notifications and news regarding CBSE exams.

### Key Topics Covered In NCERT Solutions Class 12 Mathematics Chapter 4

Extramarks NCERT Solutions for Class 12 Mathematics Chapter 4 covers all essential topics under determinants. Students will start by learning how to find the determinant of a matrix of order one and the properties of determinants. More precise calculations and solutions make them understand multiplications in positives instead of negatives. There are four exercises and three subtopics under NCERT Solutions for Class 12 Mathematics Chapter 4. In addition, students can refer to the study material offered by Extramarks as it offers better solutions.

Topics covered in Chapter 4 Determinant include

 Exercise Topics 4.1 Introduction 4.2 Determinant 4.3 Properties of Determinants 4.4 Area of Triangle 4.5 Minors and Cofactors 4.6 The Adjoint and Inverse of a Matrix 4.7 Applications of Determinants and Matrices

#### 4.1 Introduction

NCERT Solutions Class 12 Mathematics Chapter 4 introduces the students to Determinants and their basic concepts and theorems, wherein they will be introduced to the properties of determinants such as minors, cofactors, and applications of determinants. Further, the students will get a hold of topics like the inconsistency of the system of linear equations and solutions of linear equations.

#### 4.2 Determinant

This is one of the crucial sections. Students will understand the grassroots concepts; as learnt in the previous Chapter 3 Matrices, any matrix inside a mode is read as a determinant of a matrix. In addition, students can learn how to find the determinant of a matrix of order one, making it easy to comprehend. Students may refer to Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 as it provides straightforward and easy-to-understand calculations.

#### 4.3 Properties of Determinants

Under this topic, students will observe that prominent examples revolve around the properties of determinants. This section elaborates on how the statement properties are true for determinants of any order. The students will get to explore six different properties of determinants. Students may refer to the exercise problems to understand the properties better. Extramarks NCERT Solutions Class 12 Mathematics Chapter 4 offers step-by-step solutions for many exercises under this topic.

#### 4.4 Area of Triangle

This section guides students on calculating the area of triangles and the application of vertices. It also elaborates on particular topics such as positive quantity and absolute values of determinants. It also helps the students understand the positive and negative values of the determinant.

#### 4.5 Minors and Cofactors

In the NCERT Solutions Class 12 Mathematics Chapter 4, minors, cofactors and their main concepts are covered. In addition, this section will teach the students about five more expansion ways that help calculate the area of a triangle.

#### 4.6 The Adjoint and Inverse of a Matrix

Further in Chapter 4 Class 12, students will learn about the inverse and existence. It contains detailed explanations in terms of finding the adjoint of the matrix. There is a total of five theorems that helps to verify the concepts.

#### 4.7 Applications of Determinants and Matrices

NCERT Solutions Class 12 Mathematics Chapter 4 covers the essential concepts in applications of determinants and matrices. A student will learn how to check the consistency of the system of linear equations. Under this topic, students will learn and study the solution system of linear equations.

For detailed study material, students may refer to Extramarks, where step-by-step solutions on essential concepts, including consistent and inconsistent systems, are provided. With the help of Extramarks NCERT Solutions, the student can score more in determinants and matrices.

### NCERT Solutions Class 12 Mathematics Chapter 4 Exercise & Answer Solutions

NCERT Solutions Class 12 Mathematics Chapter 4 Determinant is available for students to refer to on the Extramarks website. It contains step-by-step solutions with detailed explanations. The students can solve complex problems with the easiest and most time-saving methods. Referring to our study material, students can make their basics strong and easily solve advanced theorems.

Click on the below links to view exercise-specific questions and solutions for NCERT Solutions Class 12 Mathematics Chapter 4 Determinant.

In addition to NCERT solutions class 12 mathematics chapter 4, students can also access other chapters of class 4. Furthermore, Extramarks offers NCERT solutions for other classes, which can be accessed by clicking on the respective link.

• 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

The NCERT Exemplar Class 12 Mathematics covers the essential subject knowledge. Once the students clear all the necessary topics under determinants, they can start preparing with an exemplar. There is a lot of importance to the topic as it appears in the competitive exams. Students can start preparing for their first-term CBSE exam with the help of this exemplar. It provides a complete explanatory solution for students with good knowledge of the subject.

At Extramarks, our subject matter experts offer their best solution with a proper step-by-step explanation. The exemplar consists of all the six exercises, i.e., introduction, determinant, properties of determinants, area of a triangle, minors and cofactors. In addition, the NCERT Exemplar contains various questions, multiple-choice type questions, and long answer type questions.

### Key Features of NCERT Solutions Class 12 Mathematics Chapter 4

Class 12 Mathematics Chapter 4 has some essential elements that are a part of competitive exams. The students will get a detailed explanation of the terms, definitions, rules, and properties of determinants. A basic understanding of these complex theories is essential for better understanding. Hence, NCERT Solutions Class 12 Mathematics Chapter 4 offers a precise solution for each example mentioned in the exercise.

Some of the key features of NCERT Solutions Class 12 Mathematics Chapter 4 include:

• The solutions cover the entire syllabus of CBSE Class 12 Mathematics Chapter 4 and help to provide a detailed explanation of all topics.
• The format of the NCERT Solutions is self-explanatory and clears all doubts a student may have while learning complex concepts.
• To score better in the examination and competitive entrance exams, students must grasp all the topics. Hence NCERT Solutions Class 12 Mathematics Chapter 4 consists of all important aspects of Determinants needed for the competitive entrance exam.
• The solution helps increase knowledge in main topics.

Q.1

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}1& \mathrm{sin\theta }& 1\\ -\mathrm{ }\mathrm{sin\theta }& 1& \mathrm{sin\theta }\\ -1& -\mathrm{sin\theta }& 1\end{array}\right],\mathrm{where}0\le \mathrm{\theta }\le 2\mathrm{\pi }.\mathrm{Then}\\ \left(\mathrm{A}\right)\mathrm{Det}\left(\mathrm{A}\right)=0\left(\mathrm{B}\right)\mathrm{Det}\left(\mathrm{A}\right)\in \left(2,\mathrm{\infty }\right)\\ \left(\mathrm{C}\right)\mathrm{Det}\left(\mathrm{A}\right)\in \left(2,4\right)\left(\mathrm{D}\right)\mathrm{Det}\left(\mathrm{A}\right)\in \left[2,4\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}\mathrm{1}& \mathrm{sin\theta }& \mathrm{1}\\ -\mathrm{ }\mathrm{sin\theta }& \mathrm{1}& \mathrm{sin\theta }\\ -\mathrm{1}& -\mathrm{sin\theta }& \mathrm{1}\end{array}\right],\mathrm{where}0\le \mathrm{\theta }\le 2\mathrm{\pi }.\mathrm{Then}\\ |\mathrm{A}|=|\begin{array}{ccc}\mathrm{1}& \mathrm{sin\theta }& \mathrm{1}\\ -\mathrm{ }\mathrm{sin\theta }& \mathrm{1}& \mathrm{sin\theta }\\ -\mathrm{1}& -\mathrm{sin\theta }& \mathrm{1}\end{array}|\\ =1\left(1+{\mathrm{sin}}^{2}\mathrm{\theta }\right)-\mathrm{sin\theta }\left(-\mathrm{sin\theta }+\mathrm{sin\theta }\right)+1\left({\mathrm{sin}}^{2}\mathrm{\theta }+1\right)\\ =2\left(1+{\mathrm{sin}}^{2}\mathrm{\theta }\right)\\ \mathrm{Now}, \mathrm{0}\le \mathrm{\theta }\le 2\mathrm{\pi }\\ ⇒\mathrm{0}\le \mathrm{sin\theta }\le 1\\ ⇒\mathrm{0}\le {\mathrm{sin}}^{2}\mathrm{\theta }\le 1\\ ⇒\mathrm{1}\le 1+{\mathrm{sin}}^{2}\mathrm{\theta }\le 2\\ ⇒\mathrm{2}\le 2\left(1+{\mathrm{sin}}^{2}\mathrm{\theta }\right)\le 4\\ \therefore \mathrm{ }\mathrm{Det}\left(\mathrm{A}\right)\in \left[2,4\right]\\ \mathrm{Thus},\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{D}\mathrm{.}\end{array}$

Q.2

$\begin{array}{l}\mathrm{If}\mathrm{x},\mathrm{y},\mathrm{z}\mathrm{are}\mathrm{non}\mathrm{zero}\mathrm{real}\mathrm{numbers},\mathrm{then}\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{matrix}\\ \mathrm{A}=\left[\begin{array}{ccc}\mathrm{x}& 0& 0\\ 0& \mathrm{y}& 0\\ 0& 0& \mathrm{z}\end{array}\right]\mathrm{ }\mathrm{is}\\ \left(\mathrm{A}\right)\left[\begin{array}{ccc}{\mathrm{x}}^{–1}& 0& 0\\ 0& {\mathrm{y}}^{–1}& 0\\ 0& 0& {\mathrm{z}}^{–1}\end{array}\right]\\ \left(\mathrm{B}\right)\mathrm{xyz}\left[\begin{array}{ccc}{\mathrm{x}}^{–1}& 0& 0\\ 0& {\mathrm{y}}^{–1}& 0\\ 0& 0& {\mathrm{z}}^{–1}\end{array}\right]\\ \left(\mathrm{C}\right)\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{x}& 0& 0\\ 0& \mathrm{y}& 0\\ 0& 0& \mathrm{z}\end{array}\right]\\ \left(\mathrm{D}\right)\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{matrix}\mathrm{ }\mathrm{is}\\ \mathrm{A}=\left[\begin{array}{ccc}\mathrm{x}& \mathrm{0}& \mathrm{0}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{z}\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{ccc}\mathrm{x}& \mathrm{0}& \mathrm{0}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{z}\end{array}|\\ =\mathrm{x}\left(\mathrm{yz}-0\right)-0+0\\ \mathrm{ }=\mathrm{xyz}\ne 0\\ \mathrm{Thus},\mathrm{inverse}\mathrm{of}\mathrm{matrix}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now}\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=\mathrm{yz}, {\mathrm{A}}_{\mathrm{12}}=0, \mathrm{ }{\mathrm{A}}_{\mathrm{13}}=0\\ {\mathrm{A}}_{\mathrm{21}}=0, {\mathrm{A}}_{\mathrm{22}}=\mathrm{xz},{\mathrm{A}}_{\mathrm{23}}=0\\ \mathrm{ }{\mathrm{A}}_{\mathrm{31}}=0, {\mathrm{A}}_{\mathrm{32}}=0, {\mathrm{A}}_{\mathrm{33}}=\mathrm{xy}\\ \mathrm{Adj} \mathrm{A}={\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]}^{‘}=\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]\\ \therefore {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj} \mathrm{A}\\ =\frac{1}{\mathrm{xyz}}\left[\begin{array}{ccc}\mathrm{yz}& 0& 0\\ 0& \mathrm{xz}& 0\\ 0& 0& \mathrm{xy}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{\mathrm{yz}}{\mathrm{xyz}}& 0& 0\\ 0& \frac{\mathrm{xz}}{\mathrm{xyz}}& 0\\ 0& 0& \frac{\mathrm{xy}}{\mathrm{xyz}}\end{array}\right]\\ =\left[\begin{array}{ccc}\frac{1}{\mathrm{x}}& 0& 0\\ 0& \frac{1}{\mathrm{y}}& 0\\ 0& 0& \frac{1}{\mathrm{z}}\end{array}\right]=\left[\begin{array}{ccc}{\mathrm{x}}^{-1}& 0& 0\\ 0& {\mathrm{y}}^{-1}& 0\\ 0& 0& {\mathrm{z}}^{-1}\end{array}\right]\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.3

$\begin{array}{l}If\text{​}a,b,careinA.P.,thenthedeterminant\\ |\begin{array}{ccc}x+2& x+3& x+2a\\ x+3& x+4& x+2b\\ x+4& x+5& x+2c\end{array}|is\\ \left(A\right)0\left(B\right)1\left(C\right)x\left(D\right)2x\end{array}$

Ans

$\begin{array}{l}|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}|\\ \mathrm{If}\mathrm{}\mathrm{a},\mathrm{b},\mathrm{c}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.,\mathrm{so}2\mathrm{b}=\mathrm{a}+\mathrm{c}\\ \mathrm{Then},\\ |\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}|=|\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+\left(\mathrm{a}+\mathrm{c}\right)\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}-{\mathrm{R}}_{\mathrm{2}}\mathrm{ }\mathrm{and}\mathrm{ }{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{ }=|\begin{array}{ccc}-1& -1& \mathrm{a}-\mathrm{c}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+\left(\mathrm{a}+\mathrm{c}\right)\\ 1& 1& \mathrm{c}-\mathrm{a}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{ }=|\begin{array}{ccc}0& 0& 0\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+\left(\mathrm{a}+\mathrm{c}\right)\\ 1& 1& \mathrm{c}-\mathrm{a}\end{array}|\\ \mathrm{ }=0 \left[\mathrm{Since},\mathrm{ }\mathrm{all}\mathrm{elements}\mathrm{of}{\mathrm{R}}_{\mathrm{1}}\mathrm{are}\mathrm{zero}\mathrm{.}\right]\\ \mathrm{Hence},\\ |\begin{array}{ccc}\mathrm{x}+2& \mathrm{x}+3& \mathrm{x}+2\mathrm{a}\\ \mathrm{x}+3& \mathrm{x}+4& \mathrm{x}+2\mathrm{b}\\ \mathrm{x}+4& \mathrm{x}+5& \mathrm{x}+2\mathrm{c}\end{array}|=0\\ \mathrm{Therefore},\mathrm{correct}\mathrm{option}\mathrm{is}\mathrm{A}\mathrm{.}\end{array}$

Q.4

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{the}\mathrm{following}\mathrm{equations}\\ \frac{2}{\mathrm{x}}+\frac{3}{\mathrm{y}}+\frac{10}{\mathrm{z}}=4\\ \frac{4}{\mathrm{x}}-\frac{6}{\mathrm{y}}+\frac{5}{\mathrm{z}}=1\\ \frac{6}{\mathrm{x}}+\frac{9}{\mathrm{y}}-\frac{20}{\mathrm{z}}=1\end{array}$

Ans

$\begin{array}{l}\frac{\mathrm{2}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{3}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{10}}{\mathrm{z}}=\mathrm{4}\\ \mathrm{ }\frac{\mathrm{4}}{\mathrm{x}}-\frac{\mathrm{6}}{\mathrm{y}}\mathrm{+}\frac{\mathrm{5}}{\mathrm{z}}=\mathrm{1}\\ \frac{\mathrm{6}}{\mathrm{x}}\mathrm{+}\frac{\mathrm{9}}{\mathrm{y}}-\frac{\mathrm{20}}{\mathrm{z}}=\mathrm{1}\\ \mathrm{Let}\mathrm{p}=\frac{1}{\mathrm{x}}\mathrm{q}=\frac{1}{\mathrm{y}}\mathrm{and}\mathrm{r}=\frac{1}{\mathrm{z}},\mathrm{then}\\ \mathrm{ }2\mathrm{p}+3\mathrm{q}+10\mathrm{r}=\mathrm{4}\\ 4\mathrm{p}-6\mathrm{q}+5\mathrm{r}=\mathrm{1}\\ 6\mathrm{p}+9\mathrm{q}-20\mathrm{r}=\mathrm{1}\\ \mathrm{The}\mathrm{system}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{as}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}\right],\mathrm{ }\mathrm{X}=\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 1\\ 1\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{ccc}2& 3& 10\\ 4& -6& 5\\ 6& 9& -20\end{array}|\\ =2\left(120-45\right)-3\left(-80-30\right)+10\left(36+36\right)\\ =150+330+720\\ =1200\\ \mathrm{Thus},\mathrm{}\mathrm{A}\mathrm{is}\mathrm{non}–\mathrm{singular}.\mathrm{Therefore},\mathrm{its}\mathrm{inverse}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}=75,\mathrm{ }{\mathrm{A}}_{\mathrm{12}}=110,\mathrm{ }{\mathrm{A}}_{\mathrm{13}}=72,\\ {\mathrm{A}}_{\mathrm{21}}=150,\mathrm{ }{\mathrm{A}}_{\mathrm{22}}=-100, {\mathrm{A}}_{\mathrm{23}}=0,\mathrm{ }\\ {\mathrm{A}}_{\mathrm{31}}=75,\mathrm{ }{\mathrm{A}}_{\mathrm{32}}=30, {\mathrm{A}}_{\mathrm{33}}=-24\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}75& 110& 72\\ 150& -100& 0\\ 75& 30& -24\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\\ \mathrm{Now}, \mathrm{ }\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ ⇒\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right] =\frac{1}{1200}\left[\begin{array}{ccc}75& 150& 75\\ 110& -100& 30\\ 72& 0& -24\end{array}\right]\left[\begin{array}{l}4\\ 1\\ 2\end{array}\right]\\ ⇒\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right] =\frac{1}{1200}\left[\begin{array}{l}300+150+150\\ 440-100+60\\ 288+0-48\end{array}\right]\\ ⇒\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right] =\frac{1}{1200}\left[\begin{array}{l}600\\ 400\\ 240\end{array}\right]\\ ⇒\left[\begin{array}{l}\mathrm{p}\\ \mathrm{q}\\ \mathrm{r}\end{array}\right] =\left[\begin{array}{l}\frac{1}{2}\\ \frac{1}{3}\\ \frac{1}{5}\end{array}\right]\\ ⇒\mathrm{p}=\frac{1}{2},\mathrm{ }\mathrm{q}=\frac{1}{3}\mathrm{ }\mathrm{and}\mathrm{ }\mathrm{r}=\frac{1}{5}\\ ⇒\mathrm{x}=\frac{1}{\mathrm{p}}=2,\mathrm{ }\mathrm{y}=\frac{1}{\mathrm{q}}=3 \mathrm{and} \mathrm{z}=\frac{1}{\mathrm{r}}=5\\ \mathrm{Hence},\mathrm{the}\mathrm{required}\mathrm{solution}\mathrm{is}\\ \mathrm{x}=2,\mathrm{y}=3\mathrm{and}\mathrm{z}=5\mathrm{.}\end{array}$

Q.5 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{sin\alpha }& \mathrm{cos\alpha }& \mathrm{cos}\left(\mathrm{\alpha }+\mathrm{\beta }\right)\\ \mathrm{sin\beta }& \mathrm{cos\beta }& \mathrm{cos}\left(\mathrm{\beta }+\mathrm{\delta }\right)\\ \mathrm{sin\gamma }& \mathrm{cos\gamma }& \mathrm{cos}\left(\mathrm{\gamma }+\mathrm{\delta }\right)\end{array}\right|=0$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{ccc}\mathrm{sin\alpha }& \mathrm{cos\alpha }& \mathrm{cos}\left(\mathrm{\alpha }+\mathrm{\delta }\right)\\ \mathrm{sin\beta }& \mathrm{cos\beta }& \mathrm{cos}\left(\mathrm{\beta }+\mathrm{\delta }\right)\\ \mathrm{sin\gamma }& \mathrm{cos\gamma }& \mathrm{cos}\left(\mathrm{\gamma }+\mathrm{\delta }\right)\end{array}|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{sin\delta C}}_{\mathrm{1}}\mathrm{and}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{cos\delta C}}_{\mathrm{2}},\mathrm{ }\mathrm{we}\mathrm{get}\\ =\frac{1}{\mathrm{sin\delta cos\delta }}|\begin{array}{ccc}\mathrm{sin\alpha sin\delta }& \mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }-\mathrm{sin\alpha sin\delta }\\ \mathrm{sin\beta sin\delta }& \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }-\mathrm{sin\beta sin\delta }\\ \mathrm{sin\gamma sin\delta }& \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }-\mathrm{sin\gamma sin\delta }\end{array}|\\ {\mathrm{C}}_{1}\to {\mathrm{C}}_{1}+{\mathrm{C}}_{3},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ =\frac{1}{\mathrm{sin\delta cos\delta }}|\begin{array}{ccc}\mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }& \mathrm{cos\alpha cos\delta }-\mathrm{sin\alpha sin\delta }\\ \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }& \mathrm{cos\beta cos\delta }-\mathrm{sin\beta sin\delta }\\ \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }& \mathrm{cos\gamma cos\delta }-\mathrm{sin\gamma sin\delta }\end{array}|\\ =\frac{1}{\mathrm{sin\delta cos\delta }}×0\left[\begin{array}{l}\because \mathrm{\Delta }=0,\mathrm{if}\mathrm{any}\mathrm{two}\mathrm{coloumns}\\ \mathrm{are}\mathrm{identical}\mathrm{.}\end{array}\right]\\ =0=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.6 Using properties of determinants prove that:

$\left|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 2& 3+2\mathrm{p}& 4+3\mathrm{p}+2\mathrm{q}\\ 3& 6+3\mathrm{p}& 10+6\mathrm{p}+3\mathrm{q}\end{array}\right|=1$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 2& 3+2\mathrm{p}& 4+3\mathrm{p}+2\mathrm{q}\\ 3& 6+3\mathrm{p}& 10+6\mathrm{p}+3\mathrm{q}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-2{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-3{\mathrm{R}}_{\mathrm{1}},\mathrm{ }\mathrm{we}\mathrm{get}\\ =|\begin{array}{ccc}1& 1+\mathrm{p}& 1+\mathrm{p}+\mathrm{q}\\ 0& 1& 2+\mathrm{p}\\ 0& 3& 7+3\mathrm{p}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =1|\begin{array}{cc}1& 2+\mathrm{p}\\ 3& 7+3\mathrm{p}\end{array}|-0+0\\ =7+3\mathrm{p}-6-3\mathrm{p}\\ =1=\mathrm{R}.\mathrm{H}.\mathrm{S}.\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.7 Using properties of determinants prove that:

$\left|\begin{array}{ccc}3\mathrm{a}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ -\mathrm{b}+\mathrm{a}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ -\mathrm{c}+\mathrm{a}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}\right|=3\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{ccc}3\mathrm{a}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ -\mathrm{b}+\mathrm{a}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ -\mathrm{c}+\mathrm{a}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{1}}\to {\mathrm{C}}_{\mathrm{1}}+{\mathrm{C}}_{\mathrm{2}}+{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ =|\begin{array}{ccc}\mathrm{a}+\mathrm{b}+\mathrm{c}& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}+\mathrm{c}& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}+\mathrm{c}& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}|\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{ccc}1& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ 1& 3\mathrm{b}& -\mathrm{b}+\mathrm{c}\\ 1& -\mathrm{c}+\mathrm{b}& 3\mathrm{c}\end{array}|\\ \mathrm{Apply}\mathrm{ }{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{1}\mathrm{and}{\mathrm{R}}_{3}\to {\mathrm{R}}_{3}-{\mathrm{R}}_{1},\mathrm{ }\mathrm{we}\mathrm{get}\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{ccc}1& -\mathrm{a}+\mathrm{b}& -\mathrm{a}+\mathrm{c}\\ 0& \mathrm{a}+2\mathrm{b}& \mathrm{a}-\mathrm{b}\\ 0& \mathrm{a}-\mathrm{c}& \mathrm{a}+2\mathrm{c}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left\{\left(\mathrm{a}+2\mathrm{b}\right)\left(\mathrm{a}+2\mathrm{c}\right)-\left(\mathrm{a}-\mathrm{b}\right)\left(\mathrm{a}-\mathrm{c}\right)\right\}\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left\{{\mathrm{a}}^{2}+2\mathrm{ac}+2\mathrm{ab}+4\mathrm{bc}-\left({\mathrm{a}}^{2}-\mathrm{ac}-\mathrm{ba}+\mathrm{bc}\right)\right\}\\ =\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left\{{\mathrm{a}}^{2}+2\mathrm{ac}+2\mathrm{ab}+4\mathrm{bc}-{\mathrm{a}}^{2}+\mathrm{ac}+\mathrm{ba}-\mathrm{bc}\right\}\\ =3\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{ab}+\mathrm{bc}+\mathrm{ca}\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Thus},\mathrm{the}\mathrm{given}\mathrm{result}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.8 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{x}& {\mathrm{x}}^{2}& 1+{\mathrm{px}}^{3}\\ \mathrm{y}& {\mathrm{y}}^{2}& 1+{\mathrm{py}}^{3}\\ \mathrm{z}& {\mathrm{z}}^{2}& 1+{\mathrm{pz}}^{3}\end{array}\right|=\left(1+\mathrm{pxyz}\right)\left(\mathrm{x}–\mathrm{y}\right)\left(\mathrm{y}–\mathrm{z}\right)\left(\mathrm{z}–\mathrm{x}\right)$

Ans

$\begin{array}{l}\text{L}\text{.H}\text{.S}\text{.}=|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ y& {y}^{2}& 1+p{y}^{3}\\ z& {z}^{2}& 1+p{z}^{3}\end{array}|\\ Applying{\text{R}}_{\text{3}}\to {\text{R}}_{\text{3}}-{\text{R}}_{\text{1}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}{\text{R}}_{\text{2}}\to {\text{R}}_{\text{2}}-{\text{R}}_{\text{1}},\text{\hspace{0.17em}}\text{\hspace{0.17em}}we\text{\hspace{0.17em}}\text{get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ y-x& {y}^{2}-{x}^{2}& \left(1+p{y}^{3}\right)-\left(1+p{x}^{3}\right)\\ z-x& {z}^{2}-{x}^{2}& \left(1+p{z}^{3}\right)-\left(1+p{x}^{3}\right)\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ 1& y+x& \left(p{y}^{3}-p{x}^{3}\right)\\ 1& z+x& \left(p{z}^{3}-p{x}^{3}\right)\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ 1& y+x& p\left({y}^{2}+yx+{x}^{2}\right)\\ 1& z+x& p\left({z}^{2}+zx+{x}^{2}\right)\end{array}|\\ Applying\text{\hspace{0.17em}}{\text{R}}_{\text{3}}\to {\text{R}}_{\text{3}}-{\text{R}}_{\text{2}},\text{we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ 1& y+x& p\left({y}^{2}+yx+{x}^{2}\right)\\ 0& z-y& p\left({z}^{2}-{y}^{2}+zx-yx\right)\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)\left(z-y\right)|\begin{array}{ccc}x& {x}^{2}& 1+p{x}^{3}\\ 1& y+x& p\left({y}^{2}+yx+{x}^{2}\right)\\ 0& 1& p\left(z+y+x\right)\end{array}|\\ Expanding{\text{along R}}_{\text{3}}\text{, we get}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)\left(z-y\right)\left\{\begin{array}{l}0-1|\begin{array}{cc}x& 1+p{x}^{3}\\ 1& p\left({y}^{2}+yx+{x}^{2}\right)\end{array}|\\ +p\left(z+y+x\right)|\begin{array}{cc}x& {x}^{2}\\ 1& y+x\end{array}|\end{array}\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)\left(z-y\right)\left[\begin{array}{l}-1\left\{xp\left({y}^{2}+yx+{x}^{2}\right)-\left(1+p{x}^{3}\right)\right\}\\ +p\left(z+y+x\right)\left\{x\left(y+x\right)-{x}^{2}\right\}\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)\left(z-y\right)\left\{\begin{array}{l}-px{y}^{2}-p{x}^{2}y-p{x}^{3}+1+p{x}^{3}\\ +p{x}^{2}y+px{y}^{2}+pxyz\end{array}\right\}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(y-x\right)\left(z-x\right)\left(z-y\right)\left(1+pxyz\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(1+pxyz\right)\left(x-y\right)\left(y-z\right)\left(z-x\right)=R.H.S.\\ Therefore,\text{the given result is proved}\text{.}\end{array}$

Q.9 Using properties of determinants prove that:

$\left|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^{2}& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }& {\mathrm{\beta }}^{2}& \mathrm{\gamma }+\mathrm{\alpha }\\ \mathrm{\gamma }& {\mathrm{\gamma }}^{2}& \mathrm{\alpha }+\mathrm{\beta }\end{array}\right|=\left(\mathrm{\beta }–\mathrm{\gamma }\right)\left(\mathrm{\gamma }–\mathrm{\alpha }\right)\left(\mathrm{\alpha }–\mathrm{\beta }\right)\left(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }\right)$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^{2}& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }& {\mathrm{\beta }}^{2}& \mathrm{\gamma }+\mathrm{\alpha }\\ \mathrm{\gamma }& {\mathrm{\gamma }}^{2}& \mathrm{\alpha }+\mathrm{\beta }\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{ }\mathrm{we}\mathrm{}\mathrm{get}\\ =|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^{2}& \mathrm{\beta }+\mathrm{\gamma }\\ \mathrm{\beta }-\mathrm{\alpha }& {\mathrm{\beta }}^{2}-{\mathrm{\alpha }}^{2}& \mathrm{\alpha }-\mathrm{\beta }\\ \mathrm{\gamma }-\mathrm{\alpha }& {\mathrm{\gamma }}^{2}-{\mathrm{\alpha }}^{2}& \mathrm{\alpha }-\mathrm{\gamma }\end{array}|\\ =\left(\mathrm{\beta }-\mathrm{\alpha }\right)\left(\mathrm{\gamma }-\mathrm{\alpha }\right)|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^{2}& \mathrm{\beta }+\mathrm{\gamma }\\ 1& \mathrm{\beta }+\mathrm{\alpha }& -1\\ 1& \mathrm{\gamma }+\mathrm{\alpha }& -1\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{2}},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ =\left(\mathrm{\beta }-\mathrm{\alpha }\right)\left(\mathrm{\gamma }-\mathrm{\alpha }\right)|\begin{array}{ccc}\mathrm{\alpha }& {\mathrm{\alpha }}^{2}& \mathrm{\beta }+\mathrm{\gamma }\\ 1& \mathrm{\beta }+\mathrm{\alpha }& -1\\ 0& \mathrm{\gamma }-\mathrm{\beta }& 0\end{array}|\\ \mathrm{Expanding}\mathrm{}\mathrm{along}{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ =\left(\mathrm{\beta }-\mathrm{\alpha }\right)\left(\mathrm{\gamma }-\mathrm{\alpha }\right)\left\{-\left(\mathrm{\gamma }-\mathrm{\beta }\right)\left(-\mathrm{\alpha }-\mathrm{\beta }-\mathrm{\gamma }\right)\right\}\\ =\left(\mathrm{\beta }-\mathrm{\gamma }\right)\left(\mathrm{\gamma }-\mathrm{\alpha }\right)\left(\mathrm{\alpha }-\mathrm{\beta }\right)\left(\mathrm{\alpha }+\mathrm{\beta }+\mathrm{\gamma }\right)=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.10

$\left|\mathbf{\text{Evaluate}}\begin{array}{ccc}1& \mathrm{x}& \mathrm{y}\\ 1& \mathrm{x}+\mathrm{y}& \mathrm{y}\\ 1& \mathrm{x}& \mathrm{x}+\mathrm{y}\end{array}\mathbf{\text{}}\right|.$

Ans

$\begin{array}{l}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{1}& \mathrm{x}& \mathrm{y}\\ \mathrm{1}& \mathrm{x}+\mathrm{y}& \mathrm{y}\\ \mathrm{1}& \mathrm{x}& \mathrm{x}+\mathrm{y}\end{array}\right|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\\ =\left|\begin{array}{ccc}\mathrm{1}& \mathrm{x}& \mathrm{y}\\ \mathrm{0}& \mathrm{y}& \mathrm{0}\\ \mathrm{0}& \mathrm{0}& \mathrm{x}\end{array}\right|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =1\left(\mathrm{xy}-0\right)-0+0\\ =\mathrm{xy}\end{array}$

Q.11

$\mathbf{\text{Evaluate}}|\begin{array}{ccc}x& y& x+y\\ y& x+y& x\\ x+y& x& y\end{array}|.$

Ans

$\begin{array}{l}\mathrm{\Delta }=|\begin{array}{ccc}\mathrm{x}& \mathrm{y}& \mathrm{x}+\mathrm{y}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{ }\mathrm{we}\mathrm{get}\\ =|\begin{array}{ccc}2\mathrm{x}+2\mathrm{y}& 2\mathrm{x}+2\mathrm{y}& 2\mathrm{x}+2\mathrm{y}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}|\\ =\mathrm{2}\left(\mathrm{x} +\mathrm{ }\mathrm{y}\right)|\begin{array}{ccc}\mathrm{1}& \mathrm{1}& \mathrm{1}\\ \mathrm{y}& \mathrm{x}+\mathrm{y}& \mathrm{x}\\ \mathrm{x}+\mathrm{y}& \mathrm{x}& \mathrm{y}\end{array}|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{1}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_{\mathrm{3}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\mathrm{2}\left(\mathrm{x} +\mathrm{ }\mathrm{y}\right)|\begin{array}{ccc}\mathrm{1}& \mathrm{0}& \mathrm{0}\\ \mathrm{y}& \mathrm{x}& \mathrm{x}-\mathrm{y}\\ \mathrm{x}+\mathrm{y}& -\mathrm{y}& -\mathrm{x}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\mathrm{2}\left(\mathrm{x} +\mathrm{ }\mathrm{y}\right)|\begin{array}{cc}\mathrm{x}& \mathrm{x}-\mathrm{y}\\ -\mathrm{y}& -\mathrm{x}\end{array}|\\ =\mathrm{2}\left(\mathrm{x} +\mathrm{ }\mathrm{y}\right)\left(-{\mathrm{x}}^{2}+\mathrm{xy}-{\mathrm{y}}^{2}\right)\\ =-\mathrm{2}\left(\mathrm{x} +\mathrm{ }\mathrm{y}\right)\left({\mathrm{x}}^{2}-\mathrm{xy}+{\mathrm{y}}^{2}\right)\\ =-\mathrm{2}\left({\mathrm{x}}^{3}+{\mathrm{y}}^{3}\right)\\ \end{array}$

Q.12

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right].\mathrm{Verify}\mathrm{that} \\ \left(\mathrm{i}\right)\mathrm{ }{\left[\mathrm{adj}\mathrm{ }\mathrm{A}\right]}^{–1}=\mathrm{adj}\mathrm{ }\left({\mathrm{A}}^{–1}\right)\left(\mathrm{ii}\right){\left({\mathrm{A}}^{–1}\right)}^{–1}=\mathrm{A}\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}|\\ =1\left(15-1\right)+2\left(-10-1\right)+1\left(-2-3\right)\\ =14-22-5=-13\ne 0\\ \mathrm{Now},\mathrm{ }\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=14,{\mathrm{A}}_{\mathrm{12}}=11,{\mathrm{A}}_{\mathrm{13}}=-5\\ {\mathrm{A}}_{\mathrm{21}}=11,{\mathrm{A}}_{\mathrm{22}}=4,{\mathrm{A}}_{\mathrm{23}}=-3\\ {\mathrm{A}}_{\mathrm{31}}=-5,{\mathrm{A}}_{\mathrm{32}}=-3,{\mathrm{A}}_{\mathrm{33}}=-1\\ \therefore \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]}^{‘}=\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]\\ \mathrm{So}, {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ \mathrm{ }=\frac{1}{-13}\left[\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}\right]\\ \mathrm{ }=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]\\ \left(\mathrm{i}\right)|\mathrm{Adj}\mathrm{ }\mathrm{A}|=|\begin{array}{ccc}14& 11& -5\\ 11& 4& -3\\ -5& -3& -1\end{array}|\\ =14\left(-4-9\right)-11\left(-11-15\right)-5\left(-33+20\right)\\ =-182+286+65\\ =169\\ \mathrm{Cofactors}\mathrm{of}\mathrm{Adj}\mathrm{A}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=-13,{\mathrm{A}}_{\mathrm{12}}=26,{\mathrm{A}}_{\mathrm{13}}=-13\\ {\mathrm{A}}_{\mathrm{21}}=26,{\mathrm{A}}_{\mathrm{22}}=-39,{\mathrm{A}}_{\mathrm{23}}=-13\\ {\mathrm{A}}_{\mathrm{31}}=-13,{\mathrm{A}}_{\mathrm{32}}=-13,{\mathrm{A}}_{\mathrm{33}}=-65\\ \mathrm{Adj}\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)={\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]}^{‘}\\ \mathrm{ }=\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ \therefore {\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)}^{-1}=\frac{1}{|\mathrm{Adj}\mathrm{ }\mathrm{A}|}\mathrm{Adj}\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)\\ \mathrm{ }=\frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ \mathrm{ }=\frac{1}{169}×13\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ {\mathrm{A}}^{-1}\mathrm{ }=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right]\end{array}$ $\begin{array}{l}Adj\text{\hspace{0.17em}}{A}^{-1}={\left[\begin{array}{ccc}\frac{-4}{169}-\frac{9}{169}& -\left(-\frac{11}{169}-\frac{15}{169}\right)& -\frac{33}{169}-\frac{20}{169}\\ -\left(\frac{-11}{169}-\frac{15}{169}\right)& -\frac{14}{169}-\frac{25}{169}& -\left(-\frac{42}{169}+\frac{55}{169}\right)\\ -\frac{33}{169}+\frac{20}{169}& -\left(-\frac{42}{169}+\frac{55}{169}\right)& \frac{56}{169}-\frac{121}{169}\end{array}\right]}^{‘}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{169}\left[\begin{array}{ccc}-13& 26& -13\\ 26& -39& -13\\ -13& -13& -65\end{array}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{13}{169}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ Adj\text{\hspace{0.17em}}{A}^{-1}={\left(Adj\text{\hspace{0.17em}}A\right)}^{-1}\\ Hence,\text{it is proved}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}Since,\\ {A}^{-1}=\frac{1}{13}\left[\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}\right]=\left[\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}\right]\\ and\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Adj\text{\hspace{0.17em}}A=\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ |{A}^{-1}|=|\begin{array}{ccc}-\frac{14}{13}& -\frac{11}{13}& \frac{5}{13}\\ -\frac{11}{13}& -\frac{4}{13}& \frac{3}{13}\\ \frac{5}{13}& \frac{3}{13}& \frac{1}{13}\end{array}|\\ \text{}={\left(\frac{1}{13}\right)}^{3}|\begin{array}{ccc}-14& -11& 5\\ -11& -4& 3\\ 5& 3& 1\end{array}|\\ \text{}={\left(\frac{1}{13}\right)}^{3}\left\{-14\left(-4-9\right)+11\left(-11-15\right)+5\left(-33+20\right)\right\}\\ \text{}={\left(\frac{1}{13}\right)}^{3}\left(182-286-65\right)\\ \text{}={\left(\frac{1}{13}\right)}^{3}\left(-169\right)=-\frac{1}{13}\\ \therefore {\left({A}^{-1}\right)}^{-1}=\frac{Adj\text{\hspace{0.17em}}{A}^{-1}}{|{A}^{-1}|}\\ \text{}=\frac{1}{\left(-\frac{1}{13}\right)}\frac{1}{13}\left[\begin{array}{ccc}-1& 2& -1\\ 2& -3& -1\\ -1& -1& -5\end{array}\right]\\ \text{}=\left[\begin{array}{ccc}1& -2& 1\\ -2& 3& 1\\ 1& 1& 5\end{array}\right]⇒{\left({A}^{-1}\right)}^{-1}=A\\ Hence,\text{it is proved}\text{.}\end{array}$

Q.13

${\text{If A}}^{–1}=|\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}|\mathrm{and}\mathrm{B}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right],\mathrm{find}{\left(\mathrm{AB}\right)}^{–1}.$

Ans

$\begin{array}{l}\mathrm{Since}, {\left(\mathrm{AB}\right)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\\ \mathrm{ }\mathrm{B}=\left[\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}\right]\\ \therefore |\mathrm{B}|=|\begin{array}{ccc}1& 2& -2\\ -1& 3& 0\\ 0& -2& 1\end{array}|\\ =1\left(3-0\right)-2\left(-1-0\right)-2\left(2-0\right)\\ =3+2-4=1\ne 0\\ \mathrm{Thus},{\mathrm{B}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{cofactors}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{11}}=3,{\mathrm{A}}_{\mathrm{12}}=1,{\mathrm{A}}_{\mathrm{13}}=2\\ {\mathrm{A}}_{\mathrm{21}}=2,{\mathrm{A}}_{\mathrm{22}}=1,{\mathrm{A}}_{\mathrm{23}}=2\\ {\mathrm{A}}_{\mathrm{31}}=6,{\mathrm{A}}_{\mathrm{32}}=2,{\mathrm{A}}_{\mathrm{33}}=5\\ \mathrm{Adj}\mathrm{ }\mathrm{B}={\left[\begin{array}{ccc}3& 1& 2\\ 2& 1& 2\\ 6& 2& 5\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\\ {\mathrm{B}}^{-1}=\frac{1}{|\mathrm{B}|}\mathrm{Adj}\mathrm{ }\mathrm{B}\\ =\frac{1}{1}\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]=\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\\ \mathrm{Now}, {\left(\mathrm{AB}\right)}^{-1}={\mathrm{B}}^{-1}{\mathrm{A}}^{-1}\\ =\left[\begin{array}{ccc}3& 2& 6\\ 1& 1& 2\\ 2& 2& 5\end{array}\right]\left[\begin{array}{ccc}3& -1& 1\\ -15& 6& -5\\ 5& -2& 2\end{array}\right]\\ =\left[\begin{array}{ccc}9-30+30& -3+12-12& 3-10+12\\ 3-15+10& -1+6-4& 1-5+4\\ 6-30+25& 2-12+10& 2-10+10\end{array}\right]\\ =\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right]\\ \mathrm{Thus},\mathrm{required}\mathrm{solution}\mathrm{is}\\ {\left(\mathrm{AB}\right)}^{-1}=\left[\begin{array}{ccc}9& -3& 5\\ -2& 1& 0\\ 1& 0& 2\end{array}\right]\end{array}$

Q.14

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\\ |\begin{array}{ccc}{\mathrm{a}}^{2}& \mathrm{bc}& \mathrm{ac}+{\mathrm{c}}^{2}\\ {\mathrm{a}}^{2}+\mathrm{ab}& {\mathrm{b}}^{2}& \mathrm{ac}\\ \mathrm{ab}& {\mathrm{b}}^{2}+\mathrm{bc}& {\mathrm{c}}^{2}\end{array}|=4{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}.=|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& \mathrm{bc}& \mathrm{ac}+{\mathrm{c}}^{\mathrm{2}}\\ {\mathrm{a}}^{\mathrm{2}}+\mathrm{ab}& {\mathrm{b}}^{\mathrm{2}}& \mathrm{ac}\\ \mathrm{ab}& {\mathrm{b}}^{\mathrm{2}}+\mathrm{bc}& {\mathrm{c}}^{\mathrm{2}}\end{array}|\\ \mathrm{Taking}\mathrm{out}\mathrm{common}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{from}{\mathrm{C}}_{\mathrm{1}},{\mathrm{C}}_{\mathrm{2}}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\mathrm{respectively}\mathrm{.}\\ \mathrm{ }=\mathrm{abc}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{a}+\mathrm{ }\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}-{\mathrm{R}}_{\mathrm{1}},\mathrm{ }{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}-{\mathrm{R}}_{\mathrm{1}}\mathrm{ }\mathrm{we}\mathrm{get}\\ \mathrm{ }=\mathrm{abc}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{ }\mathrm{b}& \mathrm{b}-\mathrm{c}& -\mathrm{c}\\ \mathrm{b}-\mathrm{a}& \mathrm{b}& -\mathrm{a}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=\mathrm{abc}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{ }\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{b}-\mathrm{a}& \mathrm{b}& -\mathrm{a}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{3}}\to {\mathrm{R}}_{\mathrm{3}}+{\mathrm{R}}_{\mathrm{2}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=\mathrm{abc}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{ }\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ 2\mathrm{b}& 2\mathrm{b}& 0\end{array}|\\ \mathrm{ }=2{\mathrm{ab}}^{2}\mathrm{c}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}& \mathrm{a}+\mathrm{c}\\ \mathrm{ }\mathrm{a}+\mathrm{b}& \mathrm{b}& \mathrm{a}\\ \mathrm{1}& \mathrm{1}& 0\end{array}|\\ \mathrm{Applying}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{\mathrm{2}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=2{\mathrm{ab}}^{2}\mathrm{c}|\begin{array}{ccc}\mathrm{a}& \mathrm{c}-\mathrm{a}& \mathrm{a}+\mathrm{c}\\ \mathrm{ }\mathrm{a}+\mathrm{b}& -\mathrm{a}& \mathrm{a}\\ \mathrm{1}& \mathrm{0}& 0\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \mathrm{ }=2{\mathrm{ab}}^{2}\mathrm{c}\left\{1\left(\mathrm{ac}-{\mathrm{a}}^{2}+{\mathrm{a}}^{2}+\mathrm{ac}\right)-0+0\right\}\\ \mathrm{ }=2{\mathrm{ab}}^{2}\mathrm{c}\left(2\mathrm{ac}\right)\\ \mathrm{ }=4{\mathrm{a}}^{2}{\mathrm{b}}^{2}{\mathrm{c}}^{2}=\mathrm{R}.\mathrm{H}.\mathrm{S}.\end{array}$

Q.15

$\begin{array}{l}\mathrm{Solve}\mathrm{thee}\mathrm{quation}\\ |\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}|=\mathbf{0},\mathrm{a}\ne \mathbf{0}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}:|\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}|=0,\mathrm{a}\ne 0.\\ |\begin{array}{ccc}\mathrm{x}+\mathrm{a}& \mathrm{x}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}|=\mathrm{0}\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{ }\mathrm{we}\mathrm{have}\\ ⇒ |\begin{array}{ccc}3\mathrm{x}+\mathrm{a}& 3\mathrm{x}+\mathrm{a}& 3\mathrm{x}+\mathrm{a}\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}|=\mathrm{0}\\ ⇒\left(3\mathrm{x}+\mathrm{a}\right)|\begin{array}{ccc}1& 1& 1\\ \mathrm{x}& \mathrm{x}+\mathrm{a}& \mathrm{x}\\ \mathrm{x}& \mathrm{x}& \mathrm{x}+\mathrm{a}\end{array}|=\mathrm{0}\\ \mathrm{Apply}{\mathrm{C}}_{\mathrm{2}}\to {\mathrm{C}}_{2}-{\mathrm{C}}_{1}\mathrm{and}{\mathrm{C}}_{\mathrm{3}}\to {\mathrm{C}}_{\mathrm{3}}-{\mathrm{C}}_{\mathrm{1}},\mathrm{}\mathrm{we}\mathrm{get}\\ ⇒\left(3\mathrm{x}+\mathrm{a}\right)|\begin{array}{ccc}1& 0& 0\\ \mathrm{x}& \mathrm{a}& 0\\ \mathrm{x}& 0& \mathrm{a}\end{array}|=\mathrm{0}\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ ⇒\left(3\mathrm{x}+\mathrm{a}\right)\left\{1\left({\mathrm{a}}^{2}-0\right)-0+0\right\}=0\\ ⇒ \mathrm{ }{\mathrm{a}}^{2}\left(3\mathrm{x}+\mathrm{a}\right)=0\\ ⇒ \left(3\mathrm{x}+\mathrm{a}\right)=0⇒ \mathrm{x}=-\frac{\mathrm{a}}{3}\\ \mathrm{Therefore},\mathrm{x}=-\frac{\mathrm{a}}{3}.\end{array}$

Q.16

$\begin{array}{l}\mathrm{If}\mathrm{a},\mathrm{b}\mathrm{and}\mathrm{c}\mathrm{are}\mathrm{real}\mathrm{numbers},\mathrm{and}\\ \mathrm{\Delta }=|\begin{array}{ccc}\mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}|=0,\\ \mathrm{Show}\mathrm{that}\mathrm{either}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\mathrm{or}\mathrm{a}=\mathrm{b}=\mathrm{c}.\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=|\begin{array}{ccc}\mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}|\\ \mathrm{Applying}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{R}}_{\mathrm{1}}+{\mathrm{R}}_{\mathrm{2}}+{\mathrm{R}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ =|\begin{array}{ccc}2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)& 2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)& 2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}|\\ =2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{ccc}1& 1& 1\\ \mathrm{c}+\mathrm{a}& \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}\\ \mathrm{a}+\mathrm{b}& \mathrm{b}+\mathrm{c}& \mathrm{c}+\mathrm{a}\end{array}|\\ \mathrm{Apply}\mathrm{ }{\mathrm{C}}_{2}\to {\mathrm{C}}_{2}-{\mathrm{C}}_{1},\mathrm{ }{\mathrm{C}}_{3}\to {\mathrm{C}}_{3}-{\mathrm{C}}_{1}\\ =2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)|\begin{array}{ccc}1& 0& 0\\ \mathrm{c}+\mathrm{a}& \mathrm{b}-\mathrm{c}& \mathrm{b}-\mathrm{a}\\ \mathrm{a}+\mathrm{b}& \mathrm{c}-\mathrm{a}& \mathrm{c}-\mathrm{b}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right).\left[1\left\{\left(\mathrm{b}-\mathrm{c}\right)\left(\mathrm{c}-\mathrm{b}\right)-\left(\mathrm{b}-\mathrm{a}\right)\left(\mathrm{c}-\mathrm{a}\right)\right\}-0+0\right]\\ =2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\overline{)\mathrm{bc}}-{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+\mathrm{cb}-\overline{)\mathrm{bc}}+\mathrm{ba}+\mathrm{ac}-{\mathrm{a}}^{2}\right)\\ =2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}\right)\\ \mathrm{But}\mathrm{ }\mathrm{\Delta }=0\\ \mathrm{So},\\ 2\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}\right)=0\\ \mathrm{Either} \mathrm{a}+\mathrm{b}+\mathrm{c}=0 \mathrm{or} -{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}=0\\ \mathrm{If}\mathrm{ }-{\mathrm{a}}^{2}-{\mathrm{b}}^{2}-{\mathrm{c}}^{2}+\mathrm{cb}+\mathrm{ba}+\mathrm{ac}=0\\ \mathrm{Then},\\ \mathrm{ }-2{\mathrm{a}}^{2}-2{\mathrm{b}}^{2}-2{\mathrm{c}}^{2}+2\mathrm{cb}+2\mathrm{ba}+2\mathrm{ac}=0\\ ⇒2{\mathrm{a}}^{2}+2{\mathrm{b}}^{2}+2{\mathrm{c}}^{2}-2\mathrm{cb}-2\mathrm{ba}-2\mathrm{ac}=0\\ ⇒{\left(\mathrm{a}-\mathrm{b}\right)}^{2}+{\left(\mathrm{b}-\mathrm{c}\right)}^{2}+{\left(\mathrm{c}-\mathrm{a}\right)}^{2}=0\\ ⇒{\left(\mathrm{a}-\mathrm{b}\right)}^{2}={\left(\mathrm{b}-\mathrm{c}\right)}^{2}={\left(\mathrm{c}-\mathrm{a}\right)}^{2}=0\left[\begin{array}{l}{\left(\mathrm{a}-\mathrm{b}\right)}^{2},{\left(\mathrm{b}-\mathrm{c}\right)}^{2},{\left(\mathrm{c}-\mathrm{a}\right)}^{2}\mathrm{}\\ \mathrm{are}\mathrm{non}–\mathrm{negative}\mathrm{.}\end{array}\right]\\ ⇒\mathrm{a}=\mathrm{b}=\mathrm{c}\\ \mathrm{Thus},\mathrm{either}\mathrm{a}+\mathrm{b}+\mathrm{c}=0\mathrm{or}\mathrm{a}=\mathrm{b}=\mathrm{c}.\end{array}$

Q.17

$\begin{array}{l}\mathrm{Evaluate}:\\ \mathrm{}\left|\begin{array}{ccc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }& -\mathrm{sin\alpha }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }& 0\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }& \mathrm{cos\alpha }\end{array}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=|\begin{array}{ccc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }& -\mathrm{sin\alpha }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }& \mathrm{0}\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }& \mathrm{cos\alpha }\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{ }\mathrm{we}\mathrm{ }\mathrm{get}\\ =-\mathrm{sin\alpha }|\begin{array}{l}-\mathrm{sin\beta }\mathrm{cos\beta }\\ \mathrm{sin\alpha cos\beta }\mathrm{sin\alpha sin\beta }\end{array}|-0|\begin{array}{cc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }\\ \mathrm{sin\alpha cos\beta }& \mathrm{sin\alpha sin\beta }\end{array}|\\ +\mathrm{cos\alpha }|\begin{array}{cc}\mathrm{cos\alpha cos\beta }& \mathrm{cos\alpha sin\beta }\\ -\mathrm{sin\beta }& \mathrm{cos\beta }\end{array}|\\ =-\mathrm{sin\alpha }\left(-{\mathrm{sin\alpha sin}}^{\mathrm{2}}\mathrm{\beta }-{\mathrm{sin\alpha cos}}^{2}\mathrm{\beta }\right)-0+\mathrm{cos\alpha }\left({\mathrm{cos\alpha cos}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{cos\alpha sin}}^{\mathrm{2}}\mathrm{\beta }\right)\\ ={\mathrm{sin}}^{\mathrm{2}}{\mathrm{\alpha sin}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{sin}}^{\mathrm{2}}{\mathrm{\alpha cos}}^{2}\mathrm{\beta }+{\mathrm{cos}}^{\mathrm{2}}{\mathrm{\alpha cos}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{cos}}^{\mathrm{2}}{\mathrm{\alpha sin}}^{\mathrm{2}}\mathrm{\beta }\\ ={\mathrm{sin}}^{\mathrm{2}}\mathrm{\alpha }\left({\mathrm{sin}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{cos}}^{2}\mathrm{\beta }\right)+{\mathrm{cos}}^{\mathrm{2}}\mathrm{\alpha }\left({\mathrm{cos}}^{\mathrm{2}}\mathrm{\beta }+{\mathrm{sin}}^{\mathrm{2}}\mathrm{\beta }\right)\left[\because {\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }=1\right]\\ ={\mathrm{sin}}^{\mathrm{2}}\mathrm{\alpha }\left(1\right)+{\mathrm{cos}}^{\mathrm{2}}\mathrm{\alpha }\left(1\right)\\ ={\mathrm{sin}}^{\mathrm{2}}\mathrm{\alpha }+{\mathrm{cos}}^{\mathrm{2}}\mathrm{\alpha }\\ =1=\mathrm{R}.\mathrm{H}.\mathrm{S}/\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.18

$\begin{array}{l}\mathrm{Without}\mathrm{expanding}\mathrm{the}\mathrm{determinant},\mathrm{prove}\mathrm{that}\\ \mathrm{}|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^{2}& \mathrm{bc}\\ \mathrm{b}& {\mathrm{b}}^{2}& \mathrm{ca}\\ \mathrm{c}& {\mathrm{c}}^{2}& \mathrm{ab}\end{array}|\mathrm{=}|\begin{array}{ccc}1& {\mathbf{a}}^{2}& {\mathbf{a}}^{3}\\ 1& {\mathbf{b}}^{2}& {\mathbf{b}}^{3}\\ 1& {\mathbf{c}}^{2}& {\mathbf{c}}^{3}\end{array}|\mathrm{.}\end{array}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^{\mathrm{2}}& \mathrm{bc}\\ \mathrm{b}& {\mathrm{b}}^{\mathrm{2}}& \mathrm{ca}\\ \mathrm{c}& {\mathrm{c}}^{\mathrm{2}}& \mathrm{ab}\end{array}|\\ \mathrm{Apply}{\mathrm{R}}_{\mathrm{1}}\to {\mathrm{aR}}_{\mathrm{1}},{\mathrm{R}}_{\mathrm{2}}\to {\mathrm{bR}}_{\mathrm{2}}\mathrm{ }\mathrm{and} {\mathrm{R}}_{\mathrm{3}}\to {\mathrm{cR}}_{\mathrm{3}},\mathrm{ }\mathrm{we}\mathrm{get}\\ =\frac{1}{\mathrm{abc}}|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}& \mathrm{abc}\\ {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}& \mathrm{abc}\\ {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}& \mathrm{abc}\end{array}|\\ =\frac{1}{\mathrm{abc}}.\mathrm{abc}|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}& \mathrm{1}\\ {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}& \mathrm{1}\\ {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}& \mathrm{1}\end{array}|\left[\mathrm{Taking}\mathrm{out}\mathrm{abc}\mathrm{from}{\mathrm{C}}_{\mathrm{3}}.\right]\\ =\left(-1\right)|\begin{array}{ccc}{\mathrm{a}}^{\mathrm{2}}& 1& {\mathrm{a}}^{\mathrm{3}}\\ {\mathrm{b}}^{\mathrm{2}}& 1& {\mathrm{b}}^{\mathrm{3}}\\ {\mathrm{c}}^{\mathrm{2}}& 1& {\mathrm{c}}^{\mathrm{3}}\end{array}|\left[{\mathrm{C}}_{2}↔{\mathrm{C}}_{3}\right]\\ ={\left(-1\right)}^{2}|\begin{array}{ccc}1& {\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}\\ 1& {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}\\ 1& {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}\end{array}|\left[{\mathrm{C}}_{1}↔{\mathrm{C}}_{2}\right]\\ =|\begin{array}{ccc}1& {\mathrm{a}}^{\mathrm{2}}& {\mathrm{a}}^{\mathrm{3}}\\ 1& {\mathrm{b}}^{\mathrm{2}}& {\mathrm{b}}^{\mathrm{3}}\\ 1& {\mathrm{c}}^{\mathrm{2}}& {\mathrm{c}}^{\mathrm{3}}\end{array}|=\mathrm{R}.\mathrm{H}.\mathrm{S}\mathrm{.}\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{proved}\mathrm{.}\end{array}$

Q.19

$\begin{array}{l}\mathrm{Prove}\mathrm{that}\mathrm{the}\mathrm{determinant}\mathrm{}|\begin{array}{ccc}\mathrm{x}& \mathrm{sin\theta }& \mathrm{cos\theta }\\ -\mathrm{sin\theta }& -\mathrm{x}& 1\\ \mathrm{cos\theta }& 1& \mathrm{x}\end{array}|\mathrm{}\mathrm{is}\\ \mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\end{array}$

Ans

$\begin{array}{l}|\begin{array}{ccc}\mathrm{x}& \mathrm{sin\theta }& \mathrm{cos\theta }\\ -\mathrm{sin\theta }& -\mathrm{x}& \mathrm{1}\\ \mathrm{cos\theta }& \mathrm{1}& \mathrm{x}\end{array}|\\ \mathrm{Expanding}\mathrm{along}{\mathrm{R}}_{\mathrm{1}},\mathrm{we}\mathrm{get}\\ =\mathrm{x}\left(-{\mathrm{x}}^{2}-1\right)-\mathrm{sin\theta }\left(-\mathrm{xsin\theta }-\mathrm{cos\theta }\right)+\mathrm{cos\theta }\left(-\mathrm{sin\theta }+\mathrm{xcos\theta }\right)\\ =-{\mathrm{x}}^{3}-\mathrm{x}+{\mathrm{xsin}}^{2}\mathrm{\theta }+\mathrm{sin\theta cos\theta }-\mathrm{cos\theta sin\theta }+{\mathrm{xcos}}^{2}\mathrm{\theta }\\ =-{\mathrm{x}}^{3}-\mathrm{x}+\mathrm{x}\left({\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }\right)\\ =-{\mathrm{x}}^{3}-\mathrm{x}+\mathrm{x}\left(1\right)\left[{\mathrm{sin}}^{2}\mathrm{\theta }+{\mathrm{cos}}^{2}\mathrm{\theta }=1\right]\\ =-{\mathrm{x}}^{3},\mathrm{which}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\\ \mathrm{Therefore},\mathrm{determinant}\mathrm{is}\mathrm{independent}\mathrm{of}\mathrm{\theta }\mathrm{.}\end{array}$

Q.20 The cost of 4 kg onion, 3 kg wheat and 2 kg rice is ₹60. The cost of 2 kg onion, 4 kg wheat and 6 kg rice is ₹90. The cost of 6 kg onion 2 kg wheat and 3 kg rice is ₹70. Find cost of each item per kg by matrix method.

Ans

$\begin{array}{l}\text{Let cost of 1 kg onions be ₹ x, 1 kg wheat be ₹ y and}\\ \text{1 kg rice be ₹z}\text{. Then, the given situation can be represented}\\ \text{by a system of equations as:}\\ \text{4x + 3y + 2z = 60}\\ \text{2x + 4y + 6z = 90}\\ \text{6x + 2y + 3z = 70}\\ \text{This system of equations can be written in the form of}\\ \text{AX = B as given below:}\\ \text{A}=\left[\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]\text{ }\text{and B=}\left[\begin{array}{l}60\\ 90\\ 70\end{array}\right]\\ \left|A\right|=\left|\begin{array}{ccc}4& 3& 2\\ 2& 4& 6\\ 6& 2& 3\end{array}\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}}=4\left(12-12\right)-3\left(6-36\right)+2\left(4-24\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}}=0+90-40=50\ne 0\\ Now,\text{\hspace{0.17em}}\\ {A}_{11}=0,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{12}=30,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{13}=-20\\ {A}_{21}=-5,\text{\hspace{0.17em}}{A}_{22}=0,\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}_{23}=10\\ {A}_{31}=10,\text{\hspace{0.17em}}{A}_{32}=-20,\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{33}=10\\ \therefore Adj\text{\hspace{0.17em}}A={\left[\begin{array}{ccc}0& 30& -20\\ -5& 0& 10\\ 10& -20& 10\end{array}\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}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ \therefore {A}^{-1}=\frac{1}{\left|A\right|}Adj\text{\hspace{0.17em}}A\\ =\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\\ Now,\\ \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={A}^{-1}B\\ =\frac{1}{50}\left[\begin{array}{ccc}0& -5& 10\\ 30& 0& -20\\ -20& 10& 10\end{array}\right]\left[\begin{array}{l}60\\ 90\\ 70\end{array}\right]\\ =\frac{1}{50}\left[\begin{array}{l}0-450+700\\ 1800+0-1400\\ -1200+900+700\end{array}\right]=\frac{1}{50}\left[\begin{array}{l}250\\ 400\\ 400\end{array}\right]\\ ⇒\left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\left[\begin{array}{l}5\\ 8\\ 8\end{array}\right]\\ ⇒x=5,\text{\hspace{0.17em}}\text{\hspace{0.17em}}y=8\text{\hspace{0.17em}}and\text{\hspace{0.17em}}z=8.\\ \text{Thus, the cost of 1 kg onions is ₹5, 1 kg wheat is ₹ 8}\\ \text{and that of 1 kg rice is ₹8}\text{.}\end{array}$

Q.21

$\begin{array}{l}\mathbf{If} \mathbf{A}=\left|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right|,\mathrm{ }\mathrm{find}{{\mathrm{A}}^{–}}^{1}.\mathrm{ }{\mathrm{UsingA}}^{–1}\mathrm{solve}\mathrm{the}\mathrm{system}\mathrm{of}\\ \mathrm{equations}:\\ 2\mathrm{x}-3\mathrm{y}+5\mathrm{z}=11\\ 3\mathrm{x}+2\mathrm{y}-4\mathrm{z}=-5\\ \mathrm{x}+\mathrm{y}-2\mathrm{z}=-3\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}|\\ =2\left(-4+4\right)+3\left(-6+4\right)+5\left(3-2\right)\\ =0-6+5\\ =-1\ne 0\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=0, {\mathrm{A}}_{12}=2,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=-1, \mathrm{ }{\mathrm{A}}_{22}=-9, {\mathrm{A}}_{23}=-5\\ {\mathrm{A}}_{31}=2, {\mathrm{A}}_{32}=23, {\mathrm{A}}_{33}=13\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}0& 2& 1\\ -1& -9& -5\\ 2& 23& 13\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{-1}\left[\begin{array}{ccc}0& -1& 2\\ 2& -9& 23\\ 1& -5& 13\end{array}\right]\\ \mathrm{ }\therefore {\mathrm{A}}^{-1}=\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]...\left(\mathrm{i}\right)\\ \mathrm{Now},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\\ \mathrm{of}\mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right],\mathrm{ }\mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}11\\ -5\\ -3\end{array}\right]\\ \mathrm{The}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{given}\mathrm{by}\\ \therefore \mathrm{X}=\mathrm{ }{\mathrm{A}}^{-1}\mathrm{B}\\ =\left[\begin{array}{ccc}0& 1& -2\\ -2& 9& -23\\ -1& 5& -13\end{array}\right]\left[\begin{array}{l}11\\ -5\\ -3\end{array}\right]\left[\mathrm{Using}\mathrm{equation}\left(\mathrm{i}\right)\right]\\ =\left[\begin{array}{l}0-5+6\\ -22-45+69\\ -11-25+39\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\left[\begin{array}{l}1\\ 2\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=1,\mathrm{ }\mathrm{y}=2\mathrm{and}\mathrm{z}=3.\end{array}$

Q.22 Solve system of linear equations, using matrix method,

x – y + 2z = 7
3x + 4y – 5z = – 5
2x – y + 3z = 12

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{ }\mathrm{x}-\mathrm{y}+2\mathrm{z}=7\\ 3\mathrm{x}+4\mathrm{y}-5\mathrm{z}=-5\\ \mathrm{ }2\mathrm{x}-\mathrm{y}+3\mathrm{z}=12\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& -1& \mathrm{ }2\\ 3& 4& -5\\ 2& -1& 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l} \mathrm{ }7\\ -5\\ \mathrm{ }12\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}1& -1& 2\\ 3& 4& -5\\ 2& -1& 3\end{array}|\\ =1\left(12-5\right)+1\left(9+10\right)+2\left(-3-8\right)\\ =7+19-22=4\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=7, {\mathrm{A}}_{12}=-19,{\mathrm{A}}_{13}=-11\\ {\mathrm{A}}_{21}=1, \mathrm{ }{\mathrm{A}}_{22}=-1, {\mathrm{A}}_{23}=-1\\ {\mathrm{A}}_{31}=-3, {\mathrm{A}}_{32}=11, {\mathrm{A}}_{33}=7\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}7& -19& -11\\ 1& -1& -1\\ -3& 11& 7\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{4}\left[\begin{array}{ccc}7& 1& -3\\ -19& -1& 11\\ -11& -1& 7\end{array}\right]\left[\begin{array}{l} \mathrm{ }7\\ -5\\ \mathrm{ }12\end{array}\right]\\ =\frac{1}{4}\left[\begin{array}{l}49-5-36\\ -133+5+132\\ -77+5+84\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}8\\ 4\\ 12\end{array}\right]\mathrm{ }=\left[\begin{array}{l}2\\ 1\\ 3\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=1\mathrm{and}\mathrm{z}=3.\end{array}$

Q.23 Solve system of linear equations, using matrix method,

2x + 3y + 3z = 5
x – 2y + z = – 4
3x – y – 2z = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{ }2\mathrm{x}+3\mathrm{y}+3\mathrm{z}=5\\ \mathrm{ }\mathrm{x}-2\mathrm{y}+\mathrm{z}=-4\\ \mathrm{ }3\mathrm{x}-\mathrm{y}-2\mathrm{z}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}2& 3& \mathrm{ }3\\ 1& -2& 1\\ 3& -1& -2\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l} \mathrm{ }5\\ -4\\ \mathrm{ }3\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}2& 3& \mathrm{ }3\\ 1& -2& 1\\ 3& -1& -2\end{array}|=2\left(4+1\right)-3\left(-2-3\right)+3\left(-1+6\right)\\ =10+15+15=40\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=5, {\mathrm{A}}_{12}=5,{\mathrm{A}}_{13}=5\\ {\mathrm{A}}_{21}=3, \mathrm{ }{\mathrm{A}}_{22}=-13, {\mathrm{A}}_{23}=11\\ {\mathrm{A}}_{31}=9, {\mathrm{A}}_{32}=1, {\mathrm{A}}_{33}=-7\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}5& 5& 5\\ 3& -13& 11\\ 9& 1& -7\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{40}\left[\begin{array}{ccc}5& 3& 9\\ 5& -13& 1\\ 5& 11& -7\end{array}\right]\left[\begin{array}{l} \mathrm{ }5\\ -4\\ \mathrm{ }3\end{array}\right]\\ =\frac{1}{40}\left[\begin{array}{l}25-12+27\\ 25+52+3\\ 25-44-21\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{40}\left[\begin{array}{l} \mathrm{ }40\\ \mathrm{ }80\\ -40\end{array}\right]\\ =\left[\begin{array}{l}1\\ 2\\ -1\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=1,\mathrm{ }\mathrm{y}=2\mathrm{and}\mathrm{z}=-1.\end{array}$

Q.24 Solve system of linear equations, using matrix method,

x – y + z = 4
2x + y – 3z = 0
x + y + z = 2

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{x}-\mathrm{y}+\mathrm{z}=4\\ \mathrm{ }2\mathrm{x}+\mathrm{y}-3\mathrm{z}=0\\ \mathrm{ }\mathrm{x}+\mathrm{y}+\mathrm{z}=2\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& -1& \mathrm{ }1\\ 2& 1& -3\\ 1& 1& 1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 0\\ 2\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}1& -1& \mathrm{ }1\\ 2& 1& -3\\ 1& 1& 1\end{array}|\\ =1\left(1+3\right)+1\left(2+3\right)+1\left(2-1\right)\\ =4+5+1=10\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}_{11}=4,{\mathrm{A}}_{12}=-5,{\mathrm{A}}_{13}=1\\ {\mathrm{A}}_{21}=2, \mathrm{ }{\mathrm{A}}_{22}=0,{\mathrm{A}}_{23}=-2\\ {\mathrm{A}}_{31}=2, {\mathrm{A}}_{32}=5,{\mathrm{A}}_{33}=3\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{ccc}4& -5& 1\\ 2& 0& -2\\ 2& 5& 3\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ =\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\\ \therefore \mathrm{X}=\mathrm{ }{\mathrm{A}}^{-1}\mathrm{B}\\ =\frac{1}{10}\left[\begin{array}{ccc}4& 2& 2\\ -5& 0& 5\\ 1& -2& 3\end{array}\right]\left[\begin{array}{l}4\\ 0\\ 2\end{array}\right]\\ =\frac{1}{10}\left[\begin{array}{l}16+0+4\\ -20+0+10\\ 4+0+6\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]=\frac{1}{10}\left[\begin{array}{l} \mathrm{ }20\\ -10\\ \mathrm{ }10\end{array}\right]\\ =\left[\begin{array}{l}2\\ -1\\ 1\end{array}\right]\\ \mathrm{Hence},\mathrm{ }\mathrm{x}=2,\mathrm{ }\mathrm{y}=-1\mathrm{and}\mathrm{z}=1.\end{array}$

Q.25 Solve system of linear equations, using matrix method,

2x + y + z = 1
x – 2y – z = 3/2
3y – 5z = 9

Ans

$\begin{array}{l}The\text{given system of equations is:}\\ \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}}2x+y+z=1\\ \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}}x-2y-z=\frac{3}{2}\\ \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}}3y-5z=9\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{ccc}2& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ 1& -2& -1\\ 0& 3& -5\end{array}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\\ z\end{array}\right]\text{and B=}\left[\begin{array}{l}1\\ \frac{3}{2}\\ 9\end{array}\right]\\ Now,\\ |A|=|\begin{array}{ccc}2& 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\\ 1& -2& -1\\ 0& 3& -5\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\left(10+3\right)-1\left(-5-0\right)+1\left(3-0\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}}=26+5+3=34\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ Now,\text{\hspace{0.17em}}\\ {A}_{11}=13,{A}_{12}=5,{A}_{13}=3\\ {A}_{21}=8,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{22}=-10,{A}_{23}=-6\\ {A}_{31}=1,\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}_{32}=3,{A}_{33}=-5\\ Adj\text{\hspace{0.17em}}A={\left[\begin{array}{ccc}13& 5& 3\\ 8& -10& -6\\ 1& 3& -5\end{array}\right]}^{‘}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\frac{1}{|A|}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\\ \therefore X=\text{\hspace{0.17em}}{A}^{-1}B\\ \text{}=\frac{1}{34}\left[\begin{array}{ccc}13& 8& 1\\ 5& -10& 3\\ 3& -6& -5\end{array}\right]\left[\begin{array}{l}1\\ \frac{3}{\begin{array}{l}2\\ 9\end{array}}\end{array}\right]\\ \text{}=\frac{1}{34}\left[\begin{array}{l}13+12+9\\ 5-15+27\\ 3-9-45\end{array}\right]\\ \left[\begin{array}{l}x\\ y\\ z\end{array}\right]=\frac{1}{34}\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}34\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}17\\ -51\end{array}\right]\text{\hspace{0.17em}}=\left[\begin{array}{l}1\\ \frac{1}{2}\\ -\frac{3}{2}\end{array}\right]\\ Hence,\text{\hspace{0.17em}}x=1,\text{\hspace{0.17em}}y=\frac{1}{2}\text{and z}=-\frac{3}{2}.\end{array}$

Q.26 Solve system of linear equations, using matrix method,

5x + 2y = 3
3x + 2y = 5

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{ }5\mathrm{x}+2\mathrm{y}=3\\ \mathrm{ }3\mathrm{x}+2\mathrm{y}=5\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}5& 2\\ 3& 2\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}3\\ 5\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}5& 2\\ 3& 2\end{array}|\\ =10-6\\ =4\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{cc} 2& -3\\ -2& 5\end{array}\right]}^{‘}\\ \mathrm{ }=\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{–1}=\frac{1}{|\mathrm{A}|}\left(\mathrm{Adj}\mathrm{A}\right)\\ \mathrm{ }=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\\ \therefore \mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ \mathrm{ }=\frac{1}{4}\left[\begin{array}{cc}2& -2\\ -3& 5\end{array}\right]\left[\begin{array}{l}3\\ 5\end{array}\right]\\ \mathrm{ }=\frac{1}{4}\left[\begin{array}{l}6-10\\ -9+25\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\frac{1}{4}\left[\begin{array}{l}-4\\ 16\end{array}\right]⇒\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{l}-1\\ 4\end{array}\right]\\ \therefore \mathrm{x}=-1 \mathrm{and}\mathrm{ }\mathrm{y}=4.\end{array}$

Q.27 Solve system of linear equations, using matrix method,

4x – 3y = 3
3x – 5y = 7

Ans

$\begin{array}{l}The\text{given system of equations is:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4x-3y=3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3x-5y=7\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{cc}4& -3\\ 3& -5\end{array}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}3\\ 7\end{array}\right]\\ Now,\\ |A|=|\begin{array}{cc}4& -3\\ 3& -5\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-20+9\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-11\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}-5& -3\\ 3& \text{\hspace{0.17em}}\text{\hspace{0.17em}}4\end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{-1}}=\frac{1}{|A|}\left(\text{Adj A}\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-11}\left[\begin{array}{cc}-5& 3\\ -3& 4\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\\ \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}}X={A}^{-1}B\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}5& -3\\ 3& -4\end{array}\right]\left[\begin{array}{l}3\\ 7\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{l}15-21\\ 9-28\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-6\\ -19\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{-6}{11}\\ \frac{-19}{11}\end{array}\right]\\ \therefore x=\frac{-6}{11}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}y=\frac{-19}{11}.\end{array}$

Q.29 Solve system of linear equations, using matrix method,

2x – y = –2
3x + 4y = 3

Ans

$\begin{array}{l}The\text{given system of equations is:}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2x-y=-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}3x+4y=3\\ \text{The given system of equations can be written in the form of}\\ \text{AX}=\text{B, where}\\ \text{A}=\left[\begin{array}{cc}2& -1\\ 3& 4\end{array}\right],\text{\hspace{0.17em}}\text{\hspace{0.17em}}X=\left[\begin{array}{l}x\\ y\end{array}\right]\text{and B}=\left[\begin{array}{l}-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ Now,\\ |A|=|\begin{array}{cc}2& -1\\ 3& 4\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=8+3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11\ne 0\\ \therefore \text{A is a non-sigular}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{matrix}\text{.}\\ {\text{Therefore, A}}^{\text{-1}}\text{exists}\text{.}\\ \text{Now,}\\ \text{Adj A}={\left[\begin{array}{cc}4& -3\\ 1& \text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}\right]}^{‘}=\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}^{\text{-1}}=\frac{1}{|A|}\left(\text{Adj A}\right)\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\\ \therefore \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}}X={A}^{-1}B\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{cc}4& 1\\ -3& 2\end{array}\right]\left[\begin{array}{l}-2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{11}\left[\begin{array}{l}-8+3\\ 6+6\end{array}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\frac{1}{11}\left[\begin{array}{l}-5\\ 12\end{array}\right]\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left[\begin{array}{l}x\\ y\end{array}\right]=\left[\begin{array}{l}\frac{-5}{11}\\ \frac{12}{11}\end{array}\right]\\ \therefore x=\frac{-5}{11}\text{\hspace{0.17em}}\text{\hspace{0.17em}}and\text{\hspace{0.17em}}y=\frac{12}{11}.\end{array}$

Q.30 Solve system of linear equations, using matrix method,

5x + 2y = 4
7x + 3y = 5

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ 5\mathrm{x}+2\mathrm{y}=4\\ 7\mathrm{x}+3\mathrm{y}=5\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}5& 2\\ 7& 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}4\\ 5\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}5& 2\\ 7& 3\end{array}|\\ =15-14\\ =1\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{cc}3& -7\\ -2& 5\end{array}\right]}^{‘}=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \mathrm{ }{\mathrm{A}}^{–1}=\frac{1}{|\mathrm{A}|}\left(\mathrm{Adj}\mathrm{A}\right)\\ \mathrm{ }=\frac{1}{1}\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\\ \therefore \mathrm{ }\mathrm{X}={\mathrm{A}}^{-1}\mathrm{B}\\ =\left[\begin{array}{cc}3& -2\\ -7& 5\end{array}\right]\left[\begin{array}{l}4\\ 5\end{array}\right]\\ =\left[\begin{array}{l}12-10\\ -28+25\end{array}\right]\\ \left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]=\left[\begin{array}{l}2\\ -3\end{array}\right]\\ \therefore \mathrm{x}=2 \mathrm{and}\mathrm{ }\mathrm{y}=-3.\end{array}$

Q.31 Examine the consistency of the system of equations.

5x – y + 4z = 5
2x + 3y + 5z = 2
5x –2y + 6z = –1

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ 5\mathrm{x}-\mathrm{y}+4\mathrm{z}=5\\ 2\mathrm{x}+3\mathrm{y}+5\mathrm{z}=2\\ \mathrm{ }5\mathrm{x}-2\mathrm{y}+6\mathrm{z}=-1\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}5& -1& 4\\ 2& 3& 5\\ 5& -2& 6\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 2\\ -1\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}5& -1& 4\\ 2& 3& 5\\ 5& -2& 6\end{array}|\\ =5\left(18+10\right)+1\left(12-25\right)+4\left(-4-15\right)\\ =140-13-76=51\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.32 Examine the consistency of the system of equations.

3x – y – 2z = 2
2y – z = – 1
3x –5y = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{ }3\mathrm{x}-\mathrm{y}-2\mathrm{z}=2\\ \mathrm{ }2\mathrm{y}-\mathrm{z}=-1\\ 3\mathrm{x}-5\mathrm{y}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}2\\ -1\\ 3\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}3& -1& -2\\ 0& 2& -1\\ 3& -5& 0\end{array}|\\ =3\left(0-5\right)+1\left(0+3\right)-2\left(0-6\right)\\ =-15+3+12=0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{sigular}\mathrm{matrix}\mathrm{.}\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{cc}2& -1\\ -5& 0\end{array}|=-5\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{cc}0& -1\\ 3& 0\end{array}|=-3\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{cc}0& 2\\ 3& -5\end{array}|=-6\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{cc}-1& -2\\ -5& 0\end{array}|=10\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{cc}3& -2\\ 3& 0\end{array}|=6\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{cc}3& -1\\ 3& -5\end{array}|=12\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{cc}-1& -2\\ 2& -1\end{array}|=5\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{cc}3& -2\\ 0& -1\end{array}|=3\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{cc}3& -1\\ 0& 2\end{array}|=6\\ \left(\mathrm{Adj}\mathrm{A}\right)={\left[\begin{array}{ccc}-5& -3& -6\\ 10& 6& 12\\ 5& 3& 6\end{array}\right]}^{‘}=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\\ \left(\mathrm{Adj}\mathrm{A}\right)\mathrm{B}=\left[\begin{array}{ccc}-5& 10& 5\\ -3& 6& 3\\ -6& 12& 6\end{array}\right]\left[\begin{array}{l}2\\ -1\\ 3\end{array}\right]\\ =\left[\begin{array}{l}-10-10+15\\ -6-6+9\\ -12-12+18\end{array}\right]\\ =\left[\begin{array}{l}-5\\ -3\\ -6\end{array}\right]\ne \mathrm{O}\\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{does}\\ \mathrm{not}\mathrm{exist}.\mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{inconsistent}\mathrm{.}\end{array}$

Q.33 Examine the consistency of the system of equations.

x + y + z = 1
2x + 3y + 2z = 2
ax + ay + 2az = 4

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{x}+\mathrm{y}+\mathrm{z}=1\\ \mathrm{ }2\mathrm{x}+3\mathrm{y}+2\mathrm{z}=2\\ \mathrm{ax}+\mathrm{ay}+2\mathrm{az}=4\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ \mathrm{a}& \mathrm{a}& 2\mathrm{a}\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\\ \mathrm{z}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}1\\ 2\\ 4\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{ccc}1& 1& 1\\ 2& 3& 2\\ \mathrm{a}& \mathrm{a}& 2\mathrm{a}\end{array}|\\ =1\left(6\mathrm{a}-2\mathrm{a}\right)-1\left(4\mathrm{a}-2\mathrm{a}\right)+1\left(2\mathrm{a}-3\mathrm{a}\right)\\ =4\mathrm{a}-2\mathrm{a}-\mathrm{a}=\mathrm{a}\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.34 Examine the consistency of the system of equations.

x + 3y = 5
2x + 6y = 8

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{ }\mathrm{x}+3\mathrm{y}=5\\ 2\mathrm{x}+6\mathrm{y}=8\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}1& 3\\ 2& 6\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 8\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}1& 3\\ 2& 6\end{array}|\mathrm{ }=6-6\\ =0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{sigular}\mathrm{matrix}\mathrm{.}\\ \left(\mathrm{Adj}\mathrm{A}\right)={\left[\begin{array}{cc}6& -2\\ -3& 1\end{array}\right]}^{‘}=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]\\ \left(\mathrm{Adj}\mathrm{A}\right)\mathrm{B}=\left[\begin{array}{cc}6& -3\\ -2& 1\end{array}\right]\left[\begin{array}{l}5\\ 8\end{array}\right]\\ =\left[\begin{array}{l}30-24\\ -10+8\end{array}\right]\\ =\left[\begin{array}{l}6\\ -2\end{array}\right]\ne \mathrm{O}\\ \mathrm{Thus},\mathrm{the}\mathrm{solution}\mathrm{of}\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{does}\\ \mathrm{not}\mathrm{exist}.\mathrm{Hence},\mathrm{the}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{inconsistent}\mathrm{.}\end{array}$

Q.35 Examine the consistency of the system of equations

2x – y = 5
x + y = 4

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ 2\mathrm{x}-\mathrm{y}=5\\ \mathrm{x}+\mathrm{y}=4\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}2& -1\\ 1& 1\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}5\\ 4\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}2& -1\\ 1& 1\end{array}|\\ =2+1\\ =3\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular} \mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \mathrm{Hence},\mathrm{the}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}\mathrm{consistent}\mathrm{.}\end{array}$

Q.36 Examine the consistency of the system of equations.
x + 2y = 2
2x + 3y = 3

Ans

$\begin{array}{l}\mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{is}:\\ \mathrm{x}+2\mathrm{y}=2\\ 2\mathrm{x}+3\mathrm{y}=3\\ \mathrm{The}\mathrm{given}\mathrm{system}\mathrm{of}\mathrm{equations}\mathrm{can}\mathrm{be}\mathrm{written}\mathrm{in}\mathrm{the}\mathrm{form}\mathrm{of}\\ \mathrm{AX}=\mathrm{B},\mathrm{where}\\ \mathrm{A}=\left[\begin{array}{cc}1& 2\\ 2& 3\end{array}\right], \mathrm{X}=\left[\begin{array}{l}\mathrm{x}\\ \mathrm{y}\end{array}\right]\mathrm{and}\mathrm{B}=\left[\begin{array}{l}2\\ 3\end{array}\right]\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}1& 2\\ 2& 3\end{array}|\\ =3-4\\ =-1\ne 0\\ \therefore \mathrm{A}\mathrm{is}\mathrm{a}\mathrm{non}–\mathrm{sigular}\mathrm{ }\mathrm{matrix}\mathrm{.}\\ \mathrm{Therefore},{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{.}\\ \end{array}$

Hence, the given system of equations is consistent.

Q.37

$\begin{array}{l}\mathrm{If}\mathrm{A}\mathrm{is}\mathrm{aninvertible}\mathrm{matrix}\mathrm{of}\mathrm{order}2,\mathrm{then} \mathrm{det}\left({\mathrm{A}}^{–1}\right) \mathrm{is}\\ \mathrm{equalto} \phantom{\rule{0ex}{0ex}}\mathrm{ }\left(\mathrm{A}\right)\mathrm{Det}\left(\mathrm{A}\right)\phantom{\rule{0ex}{0ex}}\mathrm{ }\left(\mathrm{B}\right)\frac{1}{\mathrm{Det}\left(\mathrm{A}\right)}\phantom{\rule{0ex}{0ex}}\left(\mathrm{C}\right)1\phantom{\rule{0ex}{0ex}}\left(\mathrm{D}\right)0\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{A}\mathrm{is}\mathrm{an}\mathrm{invertible}\mathrm{matrix},\mathrm{then}{\mathrm{A}}^{–1}\mathrm{exists}\mathrm{and}{\mathrm{A}}^{–1}=\frac{1}{|\mathrm{A}|}\mathrm{adjA}.\\ \mathrm{Let}\mathrm{}\mathrm{a}\mathrm{matrix}\mathrm{of}\mathrm{order}2\mathrm{be}\mathrm{as}\mathrm{A}=\left[\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right],\\ \mathrm{then}\mathrm{}|\mathrm{A}|=\mathrm{ad}-\mathrm{bc}\mathrm{and}\mathrm{Adj}\mathrm{A}=\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]\\ \mathrm{Now},\\ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{A}\\ =\frac{1}{|\mathrm{A}|}\left[\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}\right]\\ =\left[\begin{array}{cc}\frac{\mathrm{d}}{|\mathrm{A}|}& -\frac{\mathrm{b}}{|\mathrm{A}|}\\ -\frac{\mathrm{c}}{|\mathrm{A}|}& \frac{\mathrm{a}}{|\mathrm{A}|}\end{array}\right]\\ \therefore \mathrm{ }|{\mathrm{A}}^{-1}|=|\begin{array}{cc}\frac{\mathrm{d}}{|\mathrm{A}|}& -\frac{\mathrm{b}}{|\mathrm{A}|}\\ -\frac{\mathrm{c}}{|\mathrm{A}|}& \frac{\mathrm{a}}{|\mathrm{A}|}\end{array}|\\ =\frac{1}{{|\mathrm{A}|}^{2}}|\begin{array}{cc}\mathrm{d}& -\mathrm{b}\\ -\mathrm{c}& \mathrm{a}\end{array}|\\ =\frac{1}{{|\mathrm{A}|}^{2}}\left(\mathrm{ad}-\mathrm{bc}\right)\\ =\frac{1}{{|\mathrm{A}|}^{2}}|\mathrm{A}|\\ =\frac{1}{|\mathrm{A}|}\\ \therefore \mathrm{det}\left({\mathrm{A}}^{-1}\right)=\frac{1}{\mathrm{det}\left(\mathrm{A}\right)}\\ \mathrm{Thus},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{B}\right).\end{array}$

Q.38 Let A be a nonsingular square matrix of order 3 x 3. Then |adj A| is equal to

(A) |A|
(B) |A|2
(C) |A|3
(D) 3|A|

Ans

$\begin{array}{l}\mathrm{Since},\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)\mathrm{A}=|\mathrm{A}|\mathrm{I}\\ =\left[\begin{array}{ccc}|\mathrm{A}|& 0& 0\\ 0& |\mathrm{A}|& 0\\ 0& 0& |\mathrm{A}|\end{array}\right]\\ ⇒ \mathrm{ }|\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)\mathrm{A}|=|\begin{array}{ccc}|\mathrm{A}|& 0& 0\\ 0& |\mathrm{A}|& 0\\ 0& 0& |\mathrm{A}|\end{array}|\\ ⇒ \mathrm{ }|\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)||\mathrm{A}|={|\mathrm{A}|}^{3}|\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}|\\ ⇒ \mathrm{ }|\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)||\mathrm{A}|\mathrm{ }={|\mathrm{A}|}^{3}\left(\mathrm{I}\right)\\ \therefore \mathrm{ }|\left(\mathrm{Adj}\mathrm{ }\mathrm{A}\right)|={|\mathrm{A}|}^{2}\\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{option}\mathrm{is}\left(\mathrm{B}\right).\end{array}$

Q.39

$\begin{array}{l}\mathrm{If}\mathrm{A}= \left|\begin{array}{ccc} \mathrm{ }\mathbf{2}& -\mathbf{1}& 1\\ -\mathbf{1}& \mathrm{ }\mathbf{2}& -\mathbf{1}\mathbf{}\mathbf{}\\ \mathrm{ }\mathbf{1}& -\mathbf{1}& \mathbf{2}\end{array}\right|,\mathrm{verify}\mathrm{that}\\ {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+9\mathrm{A}-4\mathrm{I}=\mathrm{O}\mathrm{and}\mathrm{hence}\mathrm{find}{\mathrm{A}}^{–1}.\end{array}$

Ans

$\begin{array}{l} \mathrm{A}=\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]\\ {\mathrm{A}}^{\mathrm{2}}=\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]\\ =\left[ \begin{array}{ccc}4+1+1& \mathrm{ }-2-2-1& 2+1+2\\ -2-2-1& \mathrm{ }1+4+1& -1-2-2\\ 2+1+2& -1-2-2& 1+1+4\end{array}\right]\\ =\left[\begin{array}{ccc} \mathrm{ }6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]\\ {\mathrm{A}}^{\mathrm{3}}={\mathrm{A}}^{\mathrm{2}}\mathrm{A}\\ =\left[\begin{array}{ccc} \mathrm{ }6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]\\ =\left[ \begin{array}{ccc}12+5+5& \mathrm{ }-6-10-5& \mathrm{6}+\mathrm{5}+\mathrm{10}\\ -\mathrm{10}-\mathrm{6}-\mathrm{5}& \mathrm{ }5+12+5& -\mathrm{5}-6-10\\ \mathrm{10}+5+6& -5-\mathrm{10}-6& \mathrm{5}+5+12\end{array}\right]=\left[ \begin{array}{ccc}\mathrm{22}& \mathrm{ }-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& 22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \mathrm{22}\end{array}\right]\\ \therefore {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{2}+9\mathrm{A}-4\mathrm{I}\\ =\mathrm{}\left[ \begin{array}{ccc}\mathrm{22}& \mathrm{ }-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& 22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \mathrm{22}\end{array}\right]-6\left[\begin{array}{ccc} \mathrm{ }6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]+9\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]\\ -4\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \mathrm{ }1\end{array}\right]\\ =\mathrm{}\left[ \begin{array}{ccc}\mathrm{22}& \mathrm{ }-\mathrm{21}& \mathrm{21}\\ -\mathrm{21}& 22& -\mathrm{21}\\ \mathrm{21}& -\mathrm{21}& \mathrm{22}\end{array}\right]-\left[\begin{array}{ccc}36& -30& 30\\ -30& 36& -30\\ 30& -30& \mathrm{ }36\end{array}\right]+\left[\begin{array}{ccc}18& -9& 9\\ -9& 18& -9\\ 9& -9& \mathrm{ }18\end{array}\right]\\ -\left[\begin{array}{ccc}4& 0& 0\\ 0& 4& 0\\ 0& 0& \mathrm{ }4\end{array}\right]\\ =\left[ \begin{array}{ccc}\mathrm{40}& \mathrm{ }-\mathrm{30}& \mathrm{30}\\ -\mathrm{30}& 40& -\mathrm{30}\\ \mathrm{30}& -\mathrm{30}& \mathrm{40}\end{array}\right]-\left[\begin{array}{ccc}40& -30& 30\\ -30& 40& -30\\ 30& -30& \mathrm{ }40\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\mathrm{O}\\ \mathrm{Thus},\mathrm{ }{\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{2}+5\mathrm{A}+11\mathrm{I}=0.\\ \mathrm{Hence},\mathrm{it}\mathrm{is}\mathrm{verified}\mathrm{.}\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+9\mathrm{A}-4\mathrm{I}=\mathrm{O}\\ ⇒\left(\mathrm{AAA}\right){\mathrm{A}}^{-1}-6\left(\mathrm{AA}\right){\mathrm{A}}^{-1}+9{\mathrm{AA}}^{-1}-4\mathrm{ }{\mathrm{IA}}^{-1}=\mathrm{O}\\ \left[\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}|\mathrm{A}|\ne 0\right]\\ ⇒\mathrm{AA}\left({\mathrm{AA}}^{-1}\right)-6\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+9\mathrm{I}=4{\mathrm{A}}^{-1}\\ ⇒ {\mathrm{A}}^{2}\mathrm{I}-6\mathrm{AI}+9\mathrm{I}=4{\mathrm{A}}^{-1}\\ ⇒ \mathrm{ }\frac{1}{4}\mathrm{ }\left({\mathrm{A}}^{2}-6\mathrm{A}+9\mathrm{I}\right)={\mathrm{A}}^{-1}\\ ⇒ {\mathrm{A}}^{-1}=\frac{1}{4}\mathrm{ }\left({\mathrm{A}}^{2}-6\mathrm{A}+9\mathrm{I}\right)...\left(\mathrm{i}\right)\\ \mathrm{Now},\\ {\mathrm{A}}^{2}-6\mathrm{A}+9\mathrm{I}=\left[\begin{array}{ccc} \mathrm{ }6& -5& 5\\ -5& 6& -5\\ 5& -5& 6\end{array}\right]-6\left[ \begin{array}{ccc} 2& \mathrm{ }-\mathrm{1}& \mathrm{1}\\ -\mathrm{1}& 2& -\mathrm{1}\\ \mathrm{1}& -\mathrm{1}& \mathrm{2}\end{array}\right]+9\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\end{array}$ $\begin{array}{l}\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}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}-5& 5\\ -5& \text{\hspace{0.17em}}\text{\hspace{0.17em}}6& -5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}5& -5& 6\end{array}\right]-\left[\begin{array}{ccc}12& -6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}6\\ -6& 12& -6\\ 6& -6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}12\end{array}\right]+\left[\begin{array}{ccc}9& 0& 0\\ 0& 9& 0\\ 0& 0& 9\end{array}\right]\\ \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}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}15& \text{\hspace{0.17em}}\text{\hspace{0.17em}}-5& 5\\ -5& \text{\hspace{0.17em}}\text{\hspace{0.17em}}15& -5\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}5& -5& 15\end{array}\right]-\left[\begin{array}{ccc}12& -6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}6\\ -6& 12& -6\\ 6& -6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}12\end{array}\right]\text{\hspace{0.17em}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3& 1& -1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 3& 1\\ -1& 1& 3\end{array}\right]\\ From\text{equation}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\left(i\right),\text{we have}\\ \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{A}}^{\text{-1}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3& 1& -1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 3& 1\\ -1& 1& 3\end{array}\right]\end{array}$

Q.40

$\begin{array}{l}\mathrm{For}\mathrm{the}\mathrm{matrix}\mathrm{A}=\left[ \begin{array}{ccc}1& 1& 1\\ 1& 2& -\mathbf{3}\\ 2& -\mathbf{1}& \mathbf{3}\end{array}\right],\mathrm{show}\mathrm{that}\\ {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+5\mathrm{A}+11\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}.\end{array}$

Ans

$\begin{array}{l}\mathrm{A}=\left[ \begin{array}{ccc}\mathrm{1}& 1& \mathrm{1}\\ \mathrm{1}& 2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \mathrm{3}\end{array}\right]\\ {\mathrm{A}}^{\mathrm{2}}=\left[ \begin{array}{ccc}\mathrm{1}& 1& \mathrm{1}\\ \mathrm{1}& 2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \mathrm{3}\end{array}\right]\left[ \begin{array}{ccc}\mathrm{1}& 1& \mathrm{1}\\ \mathrm{1}& 2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \mathrm{3}\end{array}\right]\\ =\left[ \begin{array}{ccc}1+1+2& \mathrm{ }1+2-\mathrm{1}& \mathrm{1}-3+3\\ 1+2-\mathrm{6}& \mathrm{ }1+4+3& 1-\mathrm{6}-\mathrm{9}\\ \mathrm{2}-1+6& 2-2-3& 2+3+9\end{array}\right]\\ =\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]\\ {\mathrm{A}}^{\mathrm{3}}={\mathrm{A}}^{\mathrm{2}}\mathrm{A}\\ =\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]\left[ \begin{array}{ccc}\mathrm{1}& 1& \mathrm{1}\\ \mathrm{1}& 2& -\mathrm{3}\\ \mathrm{2}& -\mathrm{1}& \mathrm{3}\end{array}\right]\\ =\left[ \begin{array}{ccc}4+2+2& \mathrm{ }4+4-1& \mathrm{4}-\mathrm{6}+\mathrm{3}\\ -\mathrm{3}+\mathrm{8}-\mathrm{28}& \mathrm{ }-3+16+14& -\mathrm{3}-24-42\\ \mathrm{7}-3+28& 7-\mathrm{6}-14& \mathrm{7}+9+42\end{array}\right]\\ =\left[ \begin{array}{ccc}\mathrm{8}& 7& \mathrm{1}\\ -\mathrm{23}& 27& -\mathrm{69}\\ \mathrm{32}& -\mathrm{13}& \mathrm{58}\end{array}\right]\\ \therefore {\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{2}+5\mathrm{A}+11\mathrm{I}\\ =\mathrm{}\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]-6\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& \mathrm{ }14\end{array}\right]+5\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& \mathrm{ }3\end{array}\right]\\ +11\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& \mathrm{ }1\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}8& 7& 1\\ -23& 27& -69\\ 32& -13& 58\end{array}\right]-\left[\begin{array}{ccc}24& 12& 6\\ -18& 48& -84\\ 42& -18& \mathrm{ }84\end{array}\right]+\left[\begin{array}{ccc}5& 5& 5\\ 5& 10& -15\\ 10& -5& \mathrm{ }15\end{array}\right]\\ +\left[\begin{array}{ccc}11& 0& 0\\ 0& 11& 0\\ 0& 0& \mathrm{ }11\end{array}\right]\\ =\left[\begin{array}{ccc}8-24+5+11& 7-12+5+0& 1-6+5+0\\ -23+18+5+0& 27-48+10+11& -69+84-15+0\\ 32-42+10+0& -13+18-5+0& 58-84+15+11\end{array}\right]\\ =\mathrm{}\left[\begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\right]=\mathrm{O}\\ \mathrm{Thus},\mathrm{ }{\mathrm{A}}^{\mathrm{3}}-6{\mathrm{A}}^{2}+5\mathrm{A}+11\mathrm{I}=0.\\ \mathrm{Now},\mathrm{ }\\ {\mathrm{A}}^{3}-6{\mathrm{A}}^{2}+5\mathrm{A}+11\mathrm{I}=\mathrm{O}\\ ⇒\left(\mathrm{AAA}\right){\mathrm{A}}^{-1}-6\left(\mathrm{AA}\right){\mathrm{A}}^{-1}+5{\mathrm{AA}}^{-1}+11\mathrm{ }{\mathrm{IA}}^{-1}=\mathrm{O}\\ \left[\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}|\mathrm{A}|\ne 0\right]\\ ⇒\mathrm{AA}\left({\mathrm{AA}}^{-1}\right)-6\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+5\mathrm{I}=-11{\mathrm{A}}^{-1}\\ ⇒ {\mathrm{A}}^{2}\mathrm{I}-6\mathrm{AI}+5\mathrm{I}=-11{\mathrm{A}}^{-1}\\ ⇒ \frac{1}{-11}\mathrm{ }\left({\mathrm{A}}^{2}-6\mathrm{A}+5\mathrm{I}\right)={\mathrm{A}}^{-1}\\ ⇒ {\mathrm{A}}^{-1}=\frac{1}{-11}\mathrm{ }\left({\mathrm{A}}^{2}-6\mathrm{A}+5\mathrm{I}\right)...\left(\mathrm{i}\right)\\ \mathrm{Now},\\ {\mathrm{A}}^{2}-6\mathrm{A}+5\mathrm{I}=\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]-6\left[\begin{array}{ccc}1& 1& 1\\ 1& 2& -3\\ 2& -1& 3\end{array}\right]+5\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 1\end{array}\right]\\ =\left[\begin{array}{ccc}4& 2& 1\\ -3& 8& -14\\ 7& -3& 14\end{array}\right]-\left[\begin{array}{ccc}6& 6& 6\\ 6& 12& -18\\ 12& -6& 18\end{array}\right]+\left[\begin{array}{ccc}5& 0& 0\\ 0& 5& 0\\ 0& 0& 5\end{array}\right]\\ \end{array}$ $\begin{array}{l}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}9& 2& 1\\ -3& 13& -14\\ 7& -3& 19\end{array}\right]-\left[\begin{array}{ccc}6& 6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}6\\ 6& 12& -18\\ 12& -6& \text{\hspace{0.17em}}\text{\hspace{0.17em}}18\end{array}\right]\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}3& -4& -5\\ -9& 1& 4\\ -5& 3& 1\end{array}\right]\\ From\text{equation}\left(i\right),\text{we have:}\\ \text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\frac{1}{-11}\text{\hspace{0.17em}}\left({A}^{2}-6A+5I\right)\\ \text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\frac{1}{-11}\left[\begin{array}{ccc}3& -4& -5\\ -9& 1& 4\\ -5& 3& 1\end{array}\right]=\frac{1}{11}\left[\begin{array}{ccc}-3& 4& 5\\ 9& -1& -4\\ 5& -3& -1\end{array}\right]\end{array}$

Q.41

$\begin{array}{l}\mathrm{For}\mathrm{matrix} \mathrm{A}=\left[\begin{array}{l}32\\ 11\end{array}\right],\mathrm{find}\mathrm{the}\mathrm{number}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{such}\mathrm{that}\\ {\mathrm{A}}^{2}+\mathrm{aA}+\mathrm{bI}=\mathrm{O}.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{matrix} \mathrm{A}=\left[\begin{array}{l}\mathrm{3}\mathrm{2}\\ \mathrm{1}\mathrm{1}\end{array}\right],\\ {\mathrm{A}}^{2}=\left[\begin{array}{l}3\mathrm{2}\\ \mathrm{1}\mathrm{1}\end{array}\right]\left[\begin{array}{l}\mathrm{3}\mathrm{2}\\ 1\mathrm{1}\end{array}\right]\\ =\left[\begin{array}{l}9+26+2\\ 3+12+1\end{array}\right]\\ =\left[\begin{array}{cc}11& 8\\ 4& 3\end{array}\right]\\ \mathrm{Now}, {\mathrm{A}}^{\mathrm{2}}+\mathrm{aA}+\mathrm{bI}=\mathrm{O}\\ \mathrm{AA}\left({\mathrm{A}}^{-1}\right)+{\mathrm{aAA}}^{-1}=-{\mathrm{bIA}}^{-1}\left[\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}|\mathrm{A}|\ne 0\right]\\ ⇒ \mathrm{ }\mathrm{A}\left({\mathrm{AA}}^{-1}\right)+\mathrm{aI}=-{\mathrm{bIA}}^{-1}\\ ⇒ \mathrm{ }\mathrm{A}\left(\mathrm{I}\right)+\mathrm{aI}=-{\mathrm{bIA}}^{-1}\\ ⇒ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{-\mathrm{b}}\left(\mathrm{A}+\mathrm{aI}\right)\\ \mathrm{Now},\\ |\mathrm{A}|=|\begin{array}{cc}3& 2\\ 1& 1\end{array}|\\ =3-2=1\\ ⇒ \mathrm{ }{\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{AdjA}\\ =\frac{1}{1}\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have} {\mathrm{A}}^{-1}=\frac{1}{-\mathrm{b}}\left(\left[\begin{array}{l}3\mathrm{2}\\ 1\mathrm{1}\end{array}\right]+\mathrm{a}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\right)\\ ⇒ \mathrm{ }\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]=\frac{1}{-\mathrm{b}}\left[\begin{array}{cc}3+\mathrm{a}& 2\\ 1& 1+\mathrm{a}\end{array}\right]\\ ⇒ \mathrm{ }\left[\begin{array}{cc}1& -2\\ -1& 3\end{array}\right]=\left[\begin{array}{cc}\frac{3+\mathrm{a}}{-\mathrm{b}}& \frac{2}{-\mathrm{b}}\\ \frac{1}{-\mathrm{b}}& \frac{1+\mathrm{a}}{-\mathrm{b}}\end{array}\right]\\ \mathrm{Comparing}\mathrm{the}\mathrm{corresponding}\mathrm{elements}\mathrm{of}\mathrm{two}\mathrm{matrices},\mathrm{we}\mathrm{have}\\ -1=\frac{1}{-\mathrm{b}}⇒\mathrm{b}=1\\ \mathrm{and} \mathrm{ }1=\frac{3+\mathrm{a}}{-\mathrm{b}}⇒-\mathrm{b}=3+\mathrm{a}\\ ⇒ -1=3+\mathrm{a}\\ ⇒ \mathrm{a}=-4\\ \mathrm{Thus},\mathrm{the}\mathrm{required}\mathrm{value}\mathrm{of}\mathrm{a}\mathrm{and}\mathrm{b}\mathrm{are}-4\mathrm{and}1\mathrm{.}\end{array}$

Q.42

$\text{If\hspace{0.17em}\hspace{0.17em}A =}\left|\begin{array}{l} 3\mathrm{ }1\\ -12\end{array}\right|\text{},\mathrm{show}\mathrm{that}{\mathrm{A}}^{2}-5\mathrm{A}+7\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}.$

Ans

$\begin{array}{l}\mathrm{Given}, \mathrm{A}=\left[\begin{array}{l} \mathrm{3} 1\\ -\mathrm{1}\mathrm{2}\end{array}\right],\mathrm{show}\mathrm{that}{\mathrm{A}}^{\mathrm{2}}–5\mathrm{A}+7\mathrm{I}=\mathrm{O}.\mathrm{Hence}\mathrm{find}{\mathrm{A}}^{–1}\mathrm{.}\\ {\mathrm{A}}^{\mathrm{2}}=\left[\begin{array}{l} \mathrm{3} \mathrm{1}\\ -\mathrm{1}\mathrm{2}\end{array}\right]\left[\begin{array}{l} \mathrm{3} \mathrm{1}\\ -\mathrm{1}\mathrm{2}\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{9}-1 3+\mathrm{2}\\ -\mathrm{3}-2-\mathrm{1}+\mathrm{4}\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }85\\ -53\end{array}\right]\\ \therefore {\mathrm{A}}^{\mathrm{2}}-5\mathrm{A}+7\mathrm{I}=\left[\begin{array}{l} \mathrm{ }85\\ -53\end{array}\right]-5\left[\begin{array}{l} \mathrm{3} \mathrm{1}\\ -1\mathrm{2}\end{array}\right]+7\left[\begin{array}{l}10\\ 01\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }85\\ -53\end{array}\right]-\left[\begin{array}{l} 15 5\\ -\mathrm{5}\mathrm{10}\end{array}\right]+\left[\begin{array}{l}70\\ 07\end{array}\right]\\ =\left[\begin{array}{l} \mathrm{ }8-15+75-5+0\\ -5+5+03-10+7\end{array}\right]\\ =\left[\begin{array}{l}\mathrm{ }00\\ 00\end{array}\right]\\ \mathrm{Hence},\\ \mathrm{ }{\mathrm{A}}^{\mathrm{2}}-5\mathrm{A}+7\mathrm{I}=\mathrm{O}\\ \therefore \mathrm{A}.\mathrm{A}-5\mathrm{A}=-7\mathrm{I}\\ ⇒\mathrm{A}.\mathrm{A}\left({\mathrm{A}}^{-1}\right)-5{\mathrm{AA}}^{-1}=-7\mathrm{I}\left({\mathrm{A}}^{-1}\right)\\ \left[\mathrm{Post}–\mathrm{multiplying}\mathrm{by}{\mathrm{A}}^{–1}\mathrm{as}|\mathrm{A}|\ne 0\right]\\ ⇒\mathrm{A}\left({\mathrm{AA}}^{-1}\right)-5\mathrm{I}=-7{\mathrm{A}}^{-1}\\ ⇒\mathrm{AI}-5\mathrm{I}=-7{\mathrm{A}}^{-1}\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{-7}\left(\mathrm{A}-5\mathrm{I}\right)\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{7}\left(5\mathrm{I}-\mathrm{A}\right)\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{7}\left(5\left[\begin{array}{l}10\\ 01\end{array}\right]-\left[\begin{array}{l} 3 1\\ -1\mathrm{2}\end{array}\right]\right)\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{7}\left(\left[\begin{array}{l}50\\ 05\end{array}\right]-\left[\begin{array}{l} 3 1\\ -1\mathrm{2}\end{array}\right]\right)\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{7}\left[\begin{array}{cc}5-3& 0-1\\ 0+1& 5-2\end{array}\right]\\ ⇒{\mathrm{A}}^{-1}=\frac{1}{7}\left[\begin{array}{cc}2& -1\\ 1& 3\end{array}\right]\end{array}$

Q.43

$\text{Let\hspace{0.17em}\hspace{0.17em}A=}\left|\begin{array}{l} 3\mathrm{ }1\\ -12\end{array}\right|\mathrm{and} \mathrm{B}=\left|\begin{array}{l}68\\ 79\end{array}\right|.\mathrm{ }\mathrm{Verify}\mathrm{that}{\left(\mathrm{AB}\right)}^{–1}={\mathrm{B}}^{–1}{\mathrm{A}}^{–1}.$

Ans

$\begin{array}{l}Let\text{A}=\left[\begin{array}{cc}3& 7\\ 2& 5\end{array}\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}}|A|=|\begin{array}{cc}3& 7\\ 2& 5\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=15-14=1\\ Now,\text{\hspace{0.17em}}\\ {A}_{11}={\left(-1\right)}^{1+1}5=5\\ {A}_{12}={\left(-1\right)}^{1+2}2=-2\\ {A}_{21}={\left(-1\right)}^{2+1}7=-7\\ {A}_{22}={\left(-1\right)}^{2+2}3=3\\ Adj\text{\hspace{0.17em}}A={\left[\begin{array}{cc}5& -2\\ -7& 3\end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{A}^{-1}=\frac{1}{|A|}.Adj\text{\hspace{0.17em}}A\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}5& -7\\ -2& 3\end{array}\right]\\ Let\text{B}=\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\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}}|B|=|\begin{array}{cc}6& 8\\ 7& 9\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=54-56=-2\\ Now,\text{\hspace{0.17em}}\\ {B}_{11}={\left(-1\right)}^{1+1}9=9\\ {B}_{12}={\left(-1\right)}^{1+2}7=-7\\ {B}_{21}={\left(-1\right)}^{2+1}8=-8\\ {B}_{22}={\left(-1\right)}^{2+2}6=6\\ Adj\text{\hspace{0.17em}}B={\left[\begin{array}{cc}9& -7\\ -8& 6\end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}{B}^{-1}=\frac{1}{|B|}.Adj\text{\hspace{0.17em}}B\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{2}\left[\begin{array}{cc}9& -8\\ -7& 6\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-\frac{9}{2}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{7}{2}& -3\end{array}\right]\\ {B}^{-1}{A}^{-1}=\left[\begin{array}{cc}-\frac{9}{2}& \text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\frac{7}{2}& -3\end{array}\right]\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}5-7\\ -2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}-\frac{45}{2}-8\frac{63}{2}+12\\ \frac{35}{2}+6-\frac{49}{2}-9\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right]\dots \left(1\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}}AB=\left[\begin{array}{cc}3& 7\\ 2& 5\end{array}\right]\left[\begin{array}{cc}6& 8\\ 7& 9\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}18+49& 24+63\\ 12+35& 16+45\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}67& 87\\ 47& 61\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|AB|=|\begin{array}{cc}67& 87\\ 47& 61\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4087-4089\text{\hspace{0.17em}}=-2\\ Adj\left(AB\right)={\left[\begin{array}{cc}61& -47\\ -87& 67\end{array}\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\right]\\ {\left(AB\right)}^{-1}=-\frac{1}{2}\left[\begin{array}{cc}61& -87\\ -47& 67\end{array}\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{cc}-\frac{61}{2}& \frac{87}{2}\\ \frac{47}{2}& -\frac{67}{2}\end{array}\right]\dots \left(2\right)\\ Thus,\text{from equation}\left(1\right)\text{and equation}\left(2\right),\text{we have}\\ {\left(AB\right)}^{-1}={B}^{-1}{A}^{-1}\\ Hence\text{proved}\text{.}\end{array}$

Q.44 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1\mathrm{ } 0 2\\ 0\mathrm{cos\alpha } \mathrm{ }s\mathrm{in\alpha }\\ 3\mathrm{sin\alpha }-\mathrm{cos\alpha }\end{array}\right|$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{cos}\alpha & \mathrm{sin}\alpha \\ 0& \mathrm{sin}\alpha & -\mathrm{cos}\alpha \end{array}\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}}\text{\hspace{0.17em}}|A|=|\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{ccc}1& 0& 0\\ 0& \mathrm{cos}\alpha & \mathrm{sin}\alpha \\ 0& \mathrm{sin}\alpha & -\mathrm{cos}\alpha \end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\left(-{\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha \right)+0\left(0-0\right)+0\left(0-0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-1\ne 0\\ So,\text{the inverse of A exists}\text{.}\\ \text{Now,}\\ {\text{A}}_{\text{11}}={\left(-1\right)}^{1+1}|\begin{array}{l}\text{\hspace{0.17em}}\mathrm{cos}\alpha \mathrm{sin}\alpha \\ \mathrm{sin}\alpha \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\mathrm{cos}\alpha \end{array}|\\ \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}}=-{\mathrm{cos}}^{2}\alpha -{\mathrm{sin}}^{2}\alpha \\ \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}}=-1\\ {\text{A}}_{\text{12}}={\left(-1\right)}^{1+2}|\begin{array}{l}0\text{}\mathrm{sin}\alpha \\ 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}-\mathrm{cos}\alpha \end{array}|\\ \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}}=0\\ {\text{A}}_{\text{13}}={\left(-1\right)}^{1+3}|\begin{array}{l}0\text{}\mathrm{cos}\alpha \\ 0sin\alpha \end{array}|\text{\hspace{0.17em}}=0\\ {\text{A}}_{\text{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}00\\ sin\alpha -\mathrm{cos}\alpha \end{array}|\text{\hspace{0.17em}}=0\\ {A}_{\text{22}}={\left(-1\right)}^{2+2}|\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 0\text{}-\mathrm{cos}\alpha \end{array}|\text{\hspace{0.17em}}=-\mathrm{cos}\alpha \\ {\text{A}}_{\text{23}}={\left(-1\right)}^{2+3}|\begin{array}{l}10\\ 0\mathrm{sin}\alpha \end{array}|\\ \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}}=-\mathrm{sin}\alpha \\ {\text{A}}_{\text{31}}={\left(-1\right)}^{3+1}|\begin{array}{l}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{}0\\ \mathrm{cos}\alpha \text{}\mathrm{sin}\alpha \end{array}|\\ \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}}=0\\ {\text{A}}_{\text{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 0\mathrm{sin}\alpha \end{array}|\\ \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}}=-sin\alpha \\ {\text{A}}_{\text{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}10\\ 0\text{}\mathrm{cos}\alpha \end{array}|\\ \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}}=\mathrm{cos}\alpha \\ Adj\text{\hspace{0.17em}}A=\text{\hspace{0.17em}}{\left[\begin{array}{ccc}-1& 0& 0\\ 0& -\mathrm{cos}\alpha & -sin\alpha \\ 0& \text{\hspace{0.17em}}-sin\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{cos}\alpha \end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1& 0& 0\\ 0& -\mathrm{cos}\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}-sin\alpha \\ 0& -sin\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{cos}\alpha \end{array}\right]\\ {A}^{-1}=\frac{1}{|A|}Adj\text{\hspace{0.17em}}A\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=-1\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}-1& 0& 0\\ 0& -\mathrm{cos}\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}-sin\alpha \\ 0& -sin\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}\mathrm{cos}\alpha \end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1& 0& 0\\ 0& \mathrm{cos}\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}sin\alpha \\ 0& sin\alpha & \text{\hspace{0.17em}}\text{\hspace{0.17em}}-\mathrm{cos}\alpha \end{array}\right]\end{array}$

Q.45 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1-1\mathrm{ }2\\ 0\mathrm{ } \mathrm{ }2-3\\ 3-2 4\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}\right]\\ \mathrm{ }|\mathrm{A}|=| \mathrm{ }\begin{array}{ccc}1& -1& 2\\ 0& 2& -3\\ 3& -2& 4\end{array}|\\ \mathrm{ }=1\left(8-6\right)+1\left(0+9\right)+2\left(0-6\right)\\ \mathrm{ }=2+9\mathrm{ }-\mathrm{ }12\\ \mathrm{ }=-1\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{l} 2 -\mathrm{3}\\ - 2 4\end{array}|\\ \mathrm{ }=2\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{l}0-3\\ 3 4\end{array}|\\ \mathrm{ }=-\mathrm{ }9\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{l}0\mathrm{ } 2\\ 3-\mathrm{ }2\end{array}|\\ \mathrm{ }=0-6\\ \mathrm{ }=-6\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}-12\\ -24\end{array}|\\ \mathrm{ }=-\left(-4+4\right)\mathrm{ }=0\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{l}12\\ 34\end{array}|\mathrm{ }=-2\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{l}1-1\\ 3-2\end{array}|\\ \mathrm{ }=-\left(-2+3\right) =-1\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{l}-1\mathrm{ } \mathrm{ }2\\ \mathrm{ }2-3\end{array}| =-1\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}1 2\\ 0-3\end{array}|\\ \mathrm{ }=3\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}1-1\\ 0 2\end{array}|\\ \mathrm{ }=2\\ \mathrm{Adj}\mathrm{ }\mathrm{A}=\mathrm{ }{\left[\begin{array}{ccc}2& -9& -6\\ 0& -2& -1\\ -1& \mathrm{ }3& 2\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc} \mathrm{ }2& 0& -1\\ -9& -2& 3\\ -6& -1& 2\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =-1\left[\begin{array}{ccc} \mathrm{ }2& 0& -1\\ -9& -2& 3\\ -6& -1& 2\end{array}\right]\\ =\left[\begin{array}{ccc}-2& 0& 1\\ 9& 2& -3\\ 6& 1& -2\end{array}\right]\end{array}$

Q.46 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l} \mathrm{ }2 \mathrm{ }13\\ \mathrm{ }4-10\\ -7\mathrm{ } 21\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{l} 2 13\\ 4-1\mathrm{0}\\ -7 \mathrm{2}\mathrm{1}\end{array}\right]\\ \mathrm{ }|\mathrm{A}|=|\begin{array}{l} 2 \mathrm{ }13\\ 4-\mathrm{1}\mathrm{0}\\ -7 2\mathrm{1}\end{array}|\\ \mathrm{ }=2\left(-1-0\right)-1\left(4-0\right)+3\left(8-7\right)\\ \mathrm{ }=-2-4 + 3\\ \mathrm{ }=-3\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{l}-1 0\\ 2\mathrm{1}\end{array}|\\ \mathrm{ }=-1\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{l} \mathrm{ }40\\ -71\end{array}|\\ \mathrm{ }=-\mathrm{ }4\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{l} 4-1\\ -7\mathrm{ } \mathrm{ }2\end{array}|\\ \mathrm{ }=8-7\\ \mathrm{ }=1\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}13\\ 21\end{array}|\\ \mathrm{ }=-\left(1-6\right)\\ \mathrm{ }=5\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{l} \mathrm{ }23\\ -71\end{array}|\\ \mathrm{ }=23\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{l} \mathrm{ }2-1\\ -7\mathrm{ } \mathrm{ }2\end{array}|\\ \mathrm{ }=-\left(4+7\right)\\ \mathrm{ }=-11\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{l} \mathrm{ }13\\ -10\end{array}|\\ \mathrm{ }=3\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}23\\ 40\end{array}|\\ \mathrm{ }=12\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}2 1\\ 4-1\end{array}|\\ \mathrm{ }=-6\\ \mathrm{Adj}\mathrm{ }\mathrm{A}=\mathrm{ }{\left[\begin{array}{ccc}-1& -4& 1\\ 5& 23& -11\\ 3& 12& -6\end{array}\right]}^{‘}\\ =\left[\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\\ =-\frac{1}{3}\left[\begin{array}{ccc}-1& 5& 3\\ -4& 23& 12\\ 1& -11& -6\end{array}\right]\end{array}$

Q.47 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}10 0\\ 3\mathrm{}3\mathrm{ } 0\\ 52-1\end{array}\right|$

Ans

$\begin{array}{l}Let\text{\hspace{0.17em}}\text{\hspace{0.17em}}A=\left[\begin{array}{l}\text{1}\text{}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \text{3}\text{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\\ \text{5}\text{2}-\text{1}\end{array}\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}}\text{\hspace{0.17em}}|A|=|\begin{array}{l}\text{1}\text{}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ \text{3}\text{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\\ \text{5}\text{}\text{2}-\text{1}\end{array}|\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\left(-3-0\right)-0\left(-3-0\right)+0\left(6-15\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-3\ne 0\\ So,\text{the inverse of A exists}\text{.}\\ \text{Now,}\\ {\text{A}}_{\text{11}}={\left(-1\right)}^{1+1}|\begin{array}{l}\text{3}\text{ }\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\\ \text{2}-\text{1}\end{array}|=-3\\ {\text{A}}_{\text{12}}={\left(-1\right)}^{1+2}|\begin{array}{l}3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 5-1\end{array}|=3\\ {\text{A}}_{\text{13}}={\left(-1\right)}^{1+3}|\begin{array}{l}33\\ 52\end{array}|\\ \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}}=6-15\text{\hspace{0.17em}}=-9\\ {\text{A}}_{\text{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 2-1\end{array}|\text{\hspace{0.17em}}=0\\ {\text{A}}_{\text{22}}={\left(-1\right)}^{2+2}|\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 5-1\end{array}|\text{\hspace{0.17em}}=-1\\ {\text{A}}_{\text{23}}={\left(-1\right)}^{2+3}|\begin{array}{l}1\text{}0\\ 52\end{array}|\\ \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(2-0\right)\text{\hspace{0.17em}}=-2\\ {\text{A}}_{\text{31}}={\left(-1\right)}^{3+1}|\begin{array}{l}0\text{}0\\ 30\end{array}|\text{\hspace{0.17em}}=0\\ {\text{A}}_{\text{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}10\\ 30\end{array}|\text{\hspace{0.17em}}=0\\ {\text{A}}_{\text{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}10\\ 3\text{}3\end{array}|\text{\hspace{0.17em}}=3\\ Adj\text{\hspace{0.17em}}A={\left[\begin{array}{ccc}-3& 3& -9\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& -1& -2\\ 0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& \text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]}^{‘}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{ccc}-3& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ 3& -1& 0\\ -9& -2& \text{\hspace{0.17em}}3\end{array}\right]\\ {A}^{-1}=\frac{1}{|A|}Adj\text{\hspace{0.17em}}A\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=-\frac{1}{3}\left[\begin{array}{ccc}-3& \text{\hspace{0.17em}}\text{\hspace{0.17em}}0& 0\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}3& -1& 0\\ -9& -2& \text{\hspace{0.17em}}3\end{array}\right]\end{array}$

Q.48 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}1\mathrm{}23\\ 024\\ 005\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{l}123\\ \mathrm{0}\mathrm{2}\mathrm{4}\\ \mathrm{0}\mathrm{0}\mathrm{5}\end{array}\right]\\ \mathrm{ }|\mathrm{A}|=|\begin{array}{l}\mathrm{1}23\\ \mathrm{0}2\mathrm{4}\\ 00\mathrm{5}\end{array}|\\ \mathrm{ }=1\left(10-0\right)-2\left(0-0\right)+3\left(0-0\right)\\ \mathrm{ }=10\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{l}\mathrm{2}\mathrm{4}\\ 0\mathrm{5}\end{array}|\mathrm{ }=10\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{l}04\\ 05\end{array}|\mathrm{ }=0\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{l}02\\ 00\end{array}|\mathrm{ }=0\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}23\\ 05\end{array}|\mathrm{ }=-10\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{l}13\\ 05\end{array}|\mathrm{ }=5\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{l}12\\ 00\end{array}| =0\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{l}12\\ 02\end{array}|\mathrm{ }=2\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}13\\ 04\end{array}|\mathrm{ }=-4\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}12\\ 02\end{array}|\mathrm{ }=2\end{array}$ $Adj A= 10 0 0 −10 5 0 2 −4 2 ‘ = 10 −10 2 0 5 −4 0 0 2 A −1 = 1 A Adj A = 1 10 10 −10 2 0 5 −4 0 0 2 MathType@MTEF@5@5@+= feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbwvMCKf MBHbqedmvETj2BSbqefm0B1jxALjhiov2DaerbuLwBLnhiov2DGi1B TfMBaebbnrfifHhDYfgasaacH8MrFz0xbbf9q8WrFfeuY=Hhbbf9v8 qqaqFr0xc9pk0xbba9q8WqFfea0=yr0RYxir=Jbba9q8aq0=yq=He9 q8qqQ8frFve9Fve9Ff0dmeaabaqaaiaacaGaaeqabaWaaqaafaaakq aabeqaaiaadgeacaWGKbGaamOAaiaaykW7caWGbbGaeyypa0ZaamWa aeaafaqabeWadaaabaGaaGymaiaaicdaaeaacaaIWaaabaGaaGimaa qaaiaaykW7caaMc8UaeyOeI0IaaGymaiaaicdaaeaacaaI1aaabaGa aGimaaqaaiaaikdaaeaacaaMc8UaaGPaVlabgkHiTiaaisdaaeaaca aMc8UaaGPaVlaaikdaaaaacaGLBbGaayzxaaWaaWbaaSqabeaacaGG NaaaaaGcbaGaaCzcaiaaykW7caaMc8Uaeyypa0ZaamWaaeaafaqabe WadaaabaGaaGymaiaaicdaaeaacaaMc8UaaGPaVlabgkHiTiaaigda caaIWaaabaGaaGOmaaqaaiaaicdaaeaacaaI1aaabaGaeyOeI0IaaG inaaqaaiaaicdaaeaacaaIWaaabaGaaGPaVlaaikdaaaaacaGLBbGa ayzxaaaabaGaamyqamaaCaaaleqabaGamGjGgkHiTiacyciIXaaaaO Gaeyypa0ZaaSaaaeaacaaIXaaabaWaaqWaaeaacaWGbbaacaGLhWUa ayjcSdaaaiaadgeacaWGKbGaamOAaiaaykW7caWGbbaabaGaaCzcai aaykW7caaMc8Uaeyypa0ZaaSaaaeaacaaIXaaabaGaaGymaiaaicda aaWaamWaaeaafaqabeWadaaabaGaaGymaiaaicdaaeaacaaMc8UaaG PaVlabgkHiTiaaigdacaaIWaaabaGaaGOmaaqaaiaaicdaaeaacaaI 1aaabaGaeyOeI0IaaGinaaqaaiaaicdaaeaacaaIWaaabaGaaGPaVl aaikdaaaaacaGLBbGaayzxaaaaaaa@918A@$

Q.49 Find the inverse of each of the matrices (if it exists)

$\left|\begin{array}{l}-15\\ -32\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{l}-\mathrm{1}5\\ -\mathrm{3}\mathrm{2}\end{array}\right]\\ \mathrm{ }|\mathrm{A}|=|\begin{array}{l}-\mathrm{1}5\\ -3\mathrm{2}\end{array}|\\ \mathrm{ }=-2+15\\ \mathrm{ }=13\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}2 =2\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}\left(-3\right) =3\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}\left(5\right)\mathrm{ }=-5\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}\left(-1\right)\mathrm{ }=-1\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{l} \mathrm{ }2 \mathrm{ }3\\ -5-1\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}2-5\\ 3-1\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\mathrm{ }=\frac{1}{13}\left[\begin{array}{l}2-5\\ 3-1\end{array}\right]\\ \end{array}$

Q.50 Find the inverse of each of the matrices (if it exists)

$\text{[}\begin{array}{l}2-2\\ 4 \mathrm{ }3\end{array}\right]$

Ans

$\begin{array}{l}\mathrm{Let} \mathrm{A}=\left[\begin{array}{l}2-\mathrm{2}\\ 4 3\end{array}\right]\\ \mathrm{ }|\mathrm{A}|=|\begin{array}{l}2-2\\ 4 \mathrm{ }3\end{array}|\\ \mathrm{ }=6+8\\ \mathrm{ }=14\ne 0\\ \mathrm{So},\mathrm{the}\mathrm{inverse}\mathrm{of}\mathrm{A}\mathrm{exists}\mathrm{.}\\ \mathrm{Now},\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}3=3\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}4=-4\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}\left(-2\right)=2\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}2=2\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{l}3-4\\ 2 \mathrm{ }2\end{array}\right]}^{‘}\\ =\left[\begin{array}{l} \mathrm{ }32\\ -42\end{array}\right]\\ {\mathrm{A}}^{-1}=\frac{1}{|\mathrm{A}|}\mathrm{Adj}\mathrm{ }\mathrm{A}\mathrm{ }=\frac{1}{14}\left[\begin{array}{l} \mathrm{ }32\\ -42\end{array}\right]\end{array}$

$\left|\begin{array}{l}1-1 2\\ 3 \mathrm{ }0-2\\ 1 \mathrm{ }0 \mathrm{ }3\end{array}\right|$

Ans

$\begin{array}{l}Let\text{​}\text{A}=\left[\begin{array}{l}\text{1}-\text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\\ \text{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}-\text{2}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}\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}}|A|=|\begin{array}{l}\text{1}-\text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\\ \text{3}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}-\text{2}\\ \text{1}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{0}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{3}\end{array}|\\ \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}}=1\left(0-0\right)+1\left(9+2\right)+2\left(0-0\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=11\\ {A}_{11}={\left(-1\right)}^{1+1}|\begin{array}{l}0-2\\ 0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}|=0,\\ {A}_{12}={\left(-1\right)}^{1+2}|\begin{array}{l}3-2\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}|=-11,\\ {A}_{13}={\left(-1\right)}^{1+3}|\begin{array}{l}30\\ 10\end{array}|=0,\\ {A}_{21}={\left(-1\right)}^{2+1}|\begin{array}{l}-12\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}03\end{array}|=3\\ {A}_{22}={\left(-1\right)}^{2+2}|\begin{array}{l}12\\ 13\end{array}|=1\\ {A}_{23}={\left(-1\right)}^{2+3}|\begin{array}{l}1-1\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\end{array}|=-1\\ {A}_{31}={\left(-1\right)}^{3+1}|\begin{array}{l}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0-2\end{array}|=2\\ {A}_{32}={\left(-1\right)}^{3+2}|\begin{array}{l}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 3-2\end{array}|=8\\ {A}_{33}={\left(-1\right)}^{3+3}|\begin{array}{l}1-1\\ 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\end{array}|=3\\ \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}}Adj\text{\hspace{0.17em}}A={\left[\begin{array}{l}0-11\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\\ 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1-1\\ 2\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\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}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ -11\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ A\left(Adj\text{\hspace{0.17em}}A\right)=\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ -11\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\left[\begin{array}{l}1-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0-2\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}0+9+20+0+00-6+6\\ -11+3+8\text{}11+0+0-2 2-2+24\\ 0-3+30+0+00+2+9\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}110\text{}0\\ 0\text{}110\\ 0011\end{array}\right]\\ Also,\\ \left(Adj\text{\hspace{0.17em}}A\right)A=\left[\begin{array}{l}1-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ 3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0-2\\ 1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\left[\begin{array}{l}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\\ -11\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}1\text{\hspace{0.17em}}\text{\hspace{0.17em}}8\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}0\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}3\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}0+11+03-1-2\text{}2-8+6\\ 0+0+09+0+26+0-6\\ 0+0+0\text{}3+0-3\text{}2+0+9\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}1100\\ 0\text{}11\text{}0\\ 0011\end{array}\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}}|A|I=11\left[\begin{array}{l}1\text{}0\text{}0\\ 0\text{}1\text{}0\\ 001\end{array}\right]\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left[\begin{array}{l}11\text{}00\\ 011\text{}0\\ 0011\end{array}\right]⇒A\left(Adj\text{\hspace{0.17em}}A\right)=\left(Adj\text{\hspace{0.17em}}A\right)A=|A|I\end{array}$

$\left|\begin{array}{l} 2 \mathrm{ }3\\ -4-6\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{}\mathrm{A}=\left[\begin{array}{l} \mathrm{ }2 \mathrm{ }3\\ -4-6\end{array}\right]\\ |\mathrm{A}|=|\begin{array}{l} 2 3\\ -4-6\end{array}|\\ =-12+12=0\\ {\mathrm{A}}_{11}={\left(-1\right)}^{1+1}\left(-6\right)=-6,\\ {\mathrm{A}}_{12}={\left(-1\right)}^{1+2}\left(-4\right)=4\\ {\mathrm{A}}_{21}={\left(-1\right)}^{2+1}3=-3\\ {\mathrm{A}}_{22}={\left(-1\right)}^{2+2}\left(2\right)=2\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{l}-64\\ -32\end{array}\right]}^{‘}\\ =\left[\begin{array}{l}-6-3\\ \mathrm{ }4 2\end{array}\right]\\ \mathrm{A}.\left(\mathrm{AdjA}\right)=\left[\begin{array}{l} \mathrm{ }2 \mathrm{ }3\\ -4\mathrm{}-6\end{array}\right]\left[\begin{array}{l}-6 -3\\ \mathrm{ }4 \mathrm{ }2\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-12+12-6+6\\ 24-24 12-12\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}00\\ 00\end{array}\right]=0\\ \left(\mathrm{AdjA}\right).\mathrm{A}=\left[\begin{array}{l}-6\mathrm{ }-3\\ \mathrm{ }4 2\end{array}\right]\left[\begin{array}{l} \mathrm{ }2 3\\ -4-6\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}-12+12-18+18\\ 12-1212-12\end{array}\right]\\ \mathrm{ }=\left[\begin{array}{l}00\\ 00\end{array}\right]=0\\ \mathrm{ }|\mathrm{A}|\mathrm{I}=0\left[\begin{array}{l}10\\ 01\end{array}\right]=0\\ \mathrm{Thus},\\ \mathrm{A}.\left(\mathrm{AdjA}\right)=\left(\mathrm{AdjA}\right).\mathrm{A}=|\mathrm{A}|\mathrm{I}\end{array}$

Q.53 Find adjoint of each of the matrices

$\left|\begin{array}{l} 1-12\\ \mathrm{ }2 35\\ -2 01\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l} \mathrm{1}-\mathrm{1}\mathrm{2}\\ \mathrm{2} \mathrm{ }3\mathrm{5}\\ -2 0\mathrm{1}\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have}\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}|\begin{array}{l}35\\ 01\end{array}|\\ =3\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}|\begin{array}{l} \mathrm{ }25\\ -2\mathrm{}1\end{array}|\\ \mathrm{ }=-\left(2+10\right)\\ \mathrm{ }=-12\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}|\begin{array}{l} \mathrm{ }23\\ -20\end{array}|\\ \mathrm{ }=\left(0+6\right)\\ \mathrm{ }=6\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}|\begin{array}{l}-12\\ \mathrm{ }01\end{array}|\\ =1\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}|\begin{array}{l} \mathrm{ }12\\ -21\end{array}|\\ \mathrm{ }=5\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}|\begin{array}{l} 1-1\\ -2 \mathrm{ }0\end{array}|\\ \mathrm{ }=-1\left(0-2\right)\\ \mathrm{ }=2\\ {\mathrm{A}}_{\mathrm{31}}={\left(-1\right)}^{3+1}|\begin{array}{l}-12\\ \mathrm{ }35\end{array}|\\ =-5-6\\ =-11\\ {\mathrm{A}}_{\mathrm{32}}={\left(-1\right)}^{3+2}|\begin{array}{l}12\\ 25\end{array}|\\ =-\left(5-4\right)\\ =-1\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}|\begin{array}{l}1-1\\ 2 \mathrm{ }3\end{array}|\\ =3+2\\ =5\\ \mathrm{Adj}\mathrm{A}={\left[\begin{array}{l} \mathrm{3}-12\mathrm{6}\\ 1 5\mathrm{2}\\ -11-1\mathrm{5}\end{array}\right]}^{‘}=\left[\begin{array}{l} \mathrm{ }3 1-11\\ -12 5-1\\ \mathrm{ }6\mathrm{ } 2 \mathrm{ }5\end{array}\right]\end{array}$

Q.54 Find adjoint of each of the matrices

$\left|\begin{array}{l}12\\ 34\end{array}\right|$

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{A}=\left[\begin{array}{l}1\mathrm{2}\\ \mathrm{3}\mathrm{4}\end{array}\right]\\ \mathrm{We}\mathrm{}\mathrm{have}\\ {\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{1+1}4\\ =4\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{1+2}3\\ \mathrm{ }=-3\\ {\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{2+1}2\\ =-2\\ {\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{2+2}1\\ \mathrm{ }=1\\ \mathrm{Adj}\mathrm{ }\mathrm{A}={\left[\begin{array}{l} \mathrm{ }4-3\\ -2 \mathrm{ }1\end{array}\right]}^{‘}\\ =\left[\begin{array}{l} \mathrm{ }4-2\\ -3\mathrm{ } 1\end{array}\right]\end{array}$

Q.55

$\begin{array}{l}\mathrm{If}\mathrm{\Delta }=|\begin{array}{l}{\mathrm{a}}_{11}{\mathrm{a}}_{12}{\mathrm{a}}_{13}\\ {\mathrm{a}}_{21}{\mathrm{a}}_{22}{\mathrm{a}}_{23}\\ {\mathrm{a}}_{31}{\mathrm{a}}_{32}{\mathrm{a}}_{33}\end{array}|\mathrm{}\mathrm{and}{\mathrm{A}}_{\mathrm{ij}}\mathrm{is}\mathrm{Co}\mathrm{factor}\mathrm{so}{\mathrm{fa}}_{\mathrm{ij}},\mathrm{then}\mathrm{value}\\ \mathrm{of}\mathrm{\Delta }\mathrm{is}\mathrm{given}\mathrm{by}\\ \left(\mathrm{A}\right){\mathrm{a}}_{11}{\mathrm{A}}_{31}+{\mathrm{a}}_{12}{\mathrm{A}}_{32}+{\mathrm{a}}_{13}{\mathrm{A}}_{33}\left(\mathrm{B}\right) \mathrm{ }{\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{12}{\mathrm{A}}_{21}+{\mathrm{a}}_{13}{\mathrm{A}}_{31}\\ \left(\mathrm{C}\right) {\mathrm{a}}_{21}{\mathrm{A}}_{11}+{\mathrm{a}}_{22}{\mathrm{A}}_{12}+{\mathrm{a}}_{23}{\mathrm{A}}_{13}\left(\mathrm{D}\right) \mathrm{ }{\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{31}{\mathrm{A}}_{31}\end{array}$

Ans

$\begin{array}{l}\mathrm{Since},\mathrm{we}\mathrm{know}\mathrm{that}\\ \mathrm{\Delta }=\mathrm{Sum}\mathrm{of}\mathrm{product}\mathrm{of}\mathrm{elements}\mathrm{of}{\mathrm{C}}_{\mathrm{1}}\mathrm{with}\mathrm{their}\mathrm{corresponding}\mathrm{cofactors}\\ \mathrm{ }={\mathrm{a}}_{11}{\mathrm{A}}_{11}+{\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{31}{\mathrm{A}}_{31}\\ \mathrm{Thus},\mathrm{the}\mathrm{option}\mathrm{D}\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.56

$\begin{array}{l}\mathrm{Using}\mathrm{Co}\mathrm{factors}\mathrm{of}\mathrm{element}\mathrm{sof}\mathrm{third}\mathrm{column},\mathrm{evaluate}\\ \mathrm{\Delta }=\left|\begin{array}{l}1\mathrm{x}\mathrm{yz}\\ 1\mathrm{y}\mathrm{zx}\\ 1\mathrm{z}\mathrm{xy}\end{array}\right|.\end{array}$

Ans

$\begin{array}{l}\mathrm{Given}\mathrm{determinant}\mathrm{is}\\ \mathrm{\Delta }=|\begin{array}{l}\mathrm{1}\mathrm{x}\mathrm{yz}\\ 1\mathrm{y}\mathrm{zx}\\ 1\mathrm{z}\mathrm{xy}\end{array}|\\ \mathrm{Minors}\mathrm{of}\mathrm{elements}\mathrm{of}\mathrm{third}\mathrm{column}\mathrm{are}\\ {\mathrm{M}}_{\mathrm{13}}=|\begin{array}{l}1\mathrm{y}\\ 1\mathrm{z}\end{array}|=\mathrm{z}-\mathrm{y}\\ {\mathrm{M}}_{\mathrm{23}}=|\begin{array}{l}1\mathrm{x}\\ 1\mathrm{z}\end{array}|=\mathrm{z}-\mathrm{x}\\ {\mathrm{M}}_{\mathrm{33}}=|\begin{array}{l}1\mathrm{x}\\ 1\mathrm{y}\end{array}|=\mathrm{y}-\mathrm{x}\\ \mathrm{Cofactors}\mathrm{of}\mathrm{elements}\mathrm{of}\mathrm{third}\mathrm{column}\mathrm{are}\\ {\mathrm{A}}_{\mathrm{13}}={\left(-1\right)}^{1+3}{\mathrm{M}}_{13}\\ =\mathrm{z}-\mathrm{y}\\ {\mathrm{A}}_{\mathrm{23}}={\left(-1\right)}^{2+3}{\mathrm{M}}_{23}\\ =-\left(\mathrm{z}-\mathrm{x}\right)\\ =\mathrm{x}-\mathrm{z}\\ {\mathrm{A}}_{\mathrm{33}}={\left(-1\right)}^{3+3}{\mathrm{M}}_{33}\\ =\mathrm{y}-\mathrm{x}\\ \mathrm{Expanding}\mathrm{the}\mathrm{determinant}\mathrm{along}{\mathrm{C}}_{\mathrm{3}},\mathrm{we}\mathrm{get}\\ \mathrm{\Delta }={\mathrm{a}}_{13}{\mathrm{A}}_{13}+{\mathrm{a}}_{23}{\mathrm{A}}_{23}+{\mathrm{a}}_{33}{\mathrm{A}}_{33}\\ =\mathrm{yz}\left(\mathrm{z}-\mathrm{y}\right)+\mathrm{zx}\left(\mathrm{x}-\mathrm{z}\right)+\mathrm{xy}\left(\mathrm{y}-\mathrm{x}\right)\\ ={\mathrm{yz}}^{2}-{\mathrm{y}}^{2}\mathrm{z}+{\mathrm{zx}}^{2}-{\mathrm{z}}^{2}\mathrm{x}+{\mathrm{xy}}^{2}-{\mathrm{x}}^{2}\mathrm{y}\\ ={\mathrm{zx}}^{2}-{\mathrm{y}}^{2}\mathrm{z}-{\mathrm{x}}^{2}\mathrm{y}+{\mathrm{xy}}^{2}+{\mathrm{yz}}^{2}-{\mathrm{z}}^{2}\mathrm{x}\\ =\mathrm{z}\left({\mathrm{x}}^{2}-{\mathrm{y}}^{2}\right)+\mathrm{xy}\left(\mathrm{y}-\mathrm{x}\right)+{\mathrm{z}}^{2}\left(\mathrm{y}-\mathrm{x}\right)\\ =\left(\mathrm{x}-\mathrm{y}\right)\left\{\mathrm{z}\left(\mathrm{x}+\mathrm{y}\right)-\mathrm{xy}-{\mathrm{z}}^{2}\right\}\\ =\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{zx}+\mathrm{zy}-\mathrm{xy}-{\mathrm{z}}^{2}\right)\\ =\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{zx}-\mathrm{xy}+\mathrm{zy}-{\mathrm{z}}^{2}\right)\\ =\left(\mathrm{x}-\mathrm{y}\right)\left\{-\mathrm{x}\left(\mathrm{y}-\mathrm{z}\right)+\mathrm{z}\left(\mathrm{y}-\mathrm{z}\right)\right\}\\ =\left(\mathrm{x}-\mathrm{y}\right)\left(\mathrm{y}-\mathrm{z}\right)\left(\mathrm{z}-\mathrm{x}\right)\end{array}$

Q.57

$\begin{array}{l}\mathrm{Using}\mathrm{Co}\mathrm{factors}\mathrm{of}\mathrm{element}\mathrm{so}\mathrm{f}\mathrm{second}\mathrm{row},\mathrm{evaluate}\\ \mathrm{\Delta }=|\begin{array}{l}538\\ 201\\ 123\end{array}|\end{array}$

Ans

$\begin{array}{l}\mathrm{\Delta }=|\begin{array}{l}\mathrm{5}\mathrm{3}\mathrm{8}\\ 2\mathrm{0}\mathrm{1}\\ \mathrm{1}2\mathrm{3}\end{array}|\\ \mathrm{Minors}\mathrm{corresponding}\mathrm{to}\mathrm{second}\mathrm{row}:\\ {\mathrm{M}}_{\mathrm{21}}=|\begin{array}{l}38\\ 23\end{array}|\\ \mathrm{ }=9-16\\ \mathrm{ }=-7\\ \because \mathrm{Cofactor}\mathrm{ }{\mathrm{A}}_{\mathrm{ij}}={\left(-1\right)}^{\mathrm{i}+\mathrm{j}}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore {\mathrm{A}}_{21}={\left(-1\right)}^{2+1}{\mathrm{M}}_{21}\\ =\left(-1\right)\left(-7\right)\\ =7\\ {\mathrm{M}}_{\mathrm{22}}=|\begin{array}{l}58\\ 13\end{array}|\\ \mathrm{ }=15-8\\ \mathrm{ }=7\\ {\mathrm{A}}_{22}={\left(-1\right)}^{2+2}{\mathrm{M}}_{22}\\ =7\\ {\mathrm{M}}_{\mathrm{23}}=|\begin{array}{l}53\\ 12\end{array}|\\ \mathrm{ }=10-3\\ \mathrm{ }=7\\ \mathrm{ }{\mathrm{A}}_{23}={\left(-1\right)}^{2+3}{\mathrm{M}}_{23}\\ \mathrm{ }=\left(-1\right)\left(7\right)\\ \mathrm{ }=-7\\ \mathrm{Expanding}\mathrm{the}\mathrm{determinant}\mathrm{along}{\mathrm{R}}_{\mathrm{2}},\mathrm{we}\mathrm{get}\\ \mathrm{\Delta }={\mathrm{a}}_{21}{\mathrm{A}}_{21}+{\mathrm{a}}_{22}{\mathrm{A}}_{22}+{\mathrm{a}}_{23}{\mathrm{A}}_{23}\\ \mathrm{ }=2×7+0×7+1×-7\\ \mathrm{ }=14+0-7\\ \mathrm{ }=7\\ \end{array}$

Q.58 Write Minors and Cofactors of the elements of following determinants:

$\text{(i)}\text{}\left|\begin{array}{l}100\\ 010\\ 001\end{array}\right|$ $\text{(ii)}\text{}\left|\begin{array}{l}10\mathrm{ }\mathrm{ }4\\ 35-1\\ 01\mathrm{ }\mathrm{ }2\end{array}\right|$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) |\begin{array}{l}1\mathrm{0}\mathrm{0}\\ 01\mathrm{0}\\ 00\mathrm{1}\end{array}|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}\\ =|\begin{array}{l}10\\ 01\end{array}|\\ =1\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}\\ =|\begin{array}{l}00\\ 01\end{array}|\\ =0\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{13}}={\mathrm{M}}_{13}\\ =|\begin{array}{l}01\\ 00\end{array}|\\ =0\\ \mathrm{ }\mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}\\ =|\begin{array}{l}00\\ 01\end{array}|\\ =0\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}\\ =|\begin{array}{l}10\\ 01\end{array}|\\ =1\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{23}}={\mathrm{M}}_{23}\\ =|\begin{array}{l}10\\ 00\end{array}|\\ =0\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{31}}={\mathrm{M}}_{31}\\ =|\begin{array}{l}00\\ 10\end{array}|\\ =0\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{32}}={\mathrm{M}}_{32}\\ =|\begin{array}{l}10\\ 00\end{array}|\\ =0\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{33}}={\mathrm{M}}_{33}\\ =|\begin{array}{l}10\\ 01\end{array}|\\ =1\\ \mathrm{Now},\mathrm{cofactor}\mathrm{of}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}},\mathrm{so}\\ {\mathrm{A}}_{\mathrm{ij}}={\left(-1\right)}^{\mathrm{i}+\mathrm{j}}{\mathrm{M}}_{\mathrm{ij}}\\ {\mathrm{A}}_{11}={\left(-1\right)}^{1+1}{\mathrm{M}}_{11}\\ =1\\ {\mathrm{A}}_{12}={\left(-1\right)}^{1+2}{\mathrm{M}}_{12}\\ \mathrm{ }=0\\ {\mathrm{A}}_{13}={\left(-1\right)}^{1+3}{\mathrm{M}}_{13}\\ \mathrm{ }=0\\ {\mathrm{A}}_{21}={\left(-1\right)}^{2+1}{\mathrm{M}}_{21}\\ \mathrm{ }=0\end{array}$ $\begin{array}{l}{A}_{22}={\left(-1\right)}^{2+2}{M}_{22}\\ \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}}=1\\ {A}_{23}={\left(-1\right)}^{2+3}{M}_{23}\\ \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}}=0\\ {A}_{31}={\left(-1\right)}^{3+1}{M}_{31}\\ \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}}=0\\ {A}_{32}={\left(-1\right)}^{3+2}{M}_{32}\\ \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}}=0\\ {A}_{33}={\left(-1\right)}^{3+3}{M}_{33}\\ \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}}=1\\ \left(\text{ii}\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}|\begin{array}{l}\text{1}\text{}\text{0}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{4}\\ \text{3}\text{}\text{5}\text{}-\text{1}\\ \text{0}\text{}\text{1}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{2}\end{array}|\\ Minor{\text{of the element a}}_{\text{ij}}{\text{is M}}_{\text{ij}}\\ \therefore Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{11}}={M}_{11}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}5\text{}-1\\ 1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=10+1\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=11\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{12}}={M}_{12}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}3\text{}-1\\ 0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}2\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=6\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{13}}={M}_{13}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}3\text{}5\\ 0\text{}1\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=3\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{21}}={M}_{21}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}0\text{}4\\ 1\text{}2\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-4\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{22}}={M}_{22}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}1\text{}4\\ 0\text{}2\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=2\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{23}}={M}_{23}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}1\text{}0\\ 0\text{}1\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{31}}={M}_{31}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}0\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ 5\text{}-1\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-20\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{32}}={M}_{32}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}1\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}4\\ 3\text{}-1\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-13\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}Minor\text{\hspace{0.17em}}{\text{of element a}}_{\text{33}}={M}_{33}\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=|\begin{array}{l}1\text{}0\\ 3\text{}5\end{array}|\\ \text{}\text{}\text{}\text{}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\\ Now,{\text{cofactor of a}}_{\text{ij}}{\text{is A}}_{\text{ij}}\text{, so}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}{\text{A}}_{\text{ij}}={\left(-1\right)}^{i+j}{M}_{ij}\\ {A}_{11}={\left(-1\right)}^{1+1}{M}_{11}\\ \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}}=11\\ {A}_{12}={\left(-1\right)}^{1+2}{M}_{12}\\ \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}}=-6\\ {A}_{13}={\left(-1\right)}^{1+3}{M}_{13}\\ \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}}=3\\ {A}_{21}={\left(-1\right)}^{2+1}{M}_{21}\\ \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(-1\right)\left(-4\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=4\\ {A}_{22}={\left(-1\right)}^{2+2}{M}_{22}\\ \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}}=2\\ \end{array}$ $\begin{array}{l}{A}_{23}={\left(-1\right)}^{2+3}{M}_{23}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-1\\ {A}_{31}={\left(-1\right)}^{3+1}{M}_{31}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=-20\\ {A}_{32}={\left(-1\right)}^{3+2}{M}_{32}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=\left(-1\right)\left(-1\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=13\\ {A}_{33}={\left(-1\right)}^{3+3}{M}_{33}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5\end{array}$

Q.59 Write Minors and Cofactors of the elements of following determinants:

$\left(\mathrm{i}\right)\left|\begin{array}{l}2-4\\ 0 3\end{array}\right|\left(\mathrm{ii}\right)\left|\begin{array}{l}\mathrm{a}\mathrm{c}\\ \mathrm{b}\mathrm{d}\end{array}\right|$

Ans

$\begin{array}{l}\left(\mathrm{i}\right) |\begin{array}{l}2-4\\ 0 \mathrm{ }3\end{array}|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}=3,\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}=0,\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}=-4,\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}=2.\\ \mathrm{Cofactor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}}={\left(-1\right)}^{\left(\mathrm{i}+\mathrm{j}\right)}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{ }{\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{\left(1+1\right)}{\mathrm{M}}_{11}=3\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{\left(1+2\right)}{\mathrm{M}}_{12}=0\\ \mathrm{ }{\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{\left(2+1\right)}{\mathrm{M}}_{21}=4\\ \mathrm{ }{\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{\left(2+2\right)}{\mathrm{M}}_{22}=2\\ \left(\mathrm{ii}\right) |\begin{array}{l}\mathrm{a}\mathrm{c}\\ \mathrm{b}\mathrm{d}\end{array}|\\ \mathrm{Minor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{11}}={\mathrm{M}}_{11}=\mathrm{d},\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{12}}={\mathrm{M}}_{12}=\mathrm{b},\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{21}}={\mathrm{M}}_{21}=\mathrm{c},\\ \mathrm{Minor}\mathrm{ }\mathrm{of}\mathrm{element}{\mathrm{a}}_{\mathrm{22}}={\mathrm{M}}_{22}=\mathrm{a}.\\ \mathrm{Cofactor}\mathrm{of}\mathrm{the}\mathrm{element}{\mathrm{a}}_{\mathrm{ij}}\mathrm{is}{\mathrm{A}}_{\mathrm{ij}}={\left(-1\right)}^{\left(\mathrm{i}+\mathrm{j}\right)}{\mathrm{M}}_{\mathrm{ij}}\\ \therefore \mathrm{ }{\mathrm{A}}_{\mathrm{11}}={\left(-1\right)}^{\left(1+1\right)}{\mathrm{M}}_{11}=\mathrm{d}\\ {\mathrm{A}}_{\mathrm{12}}={\left(-1\right)}^{\left(1+2\right)}{\mathrm{M}}_{12}=-\mathrm{b}\\ \mathrm{ }{\mathrm{A}}_{\mathrm{21}}={\left(-1\right)}^{\left(2+1\right)}{\mathrm{M}}_{21}=-\mathrm{c}\\ \mathrm{ }{\mathrm{A}}_{\mathrm{22}}={\left(-1\right)}^{\left(2+2\right)}{\mathrm{M}}_{22}=\mathrm{a}\end{array}$

Q.60 If area of triangle is 35 sq. units with vertices (2, -6), (5, 4) and (k, 4). Then k is
(A) 12 (B) –2 (C) –12, –2 (D) 12, –2

Ans

$\begin{array}{l}\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=± 35 \left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(2,-6\right),\\ \left(5,4\right),\left(\mathrm{k},4\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}2-61\\ 5 \mathrm{ }41\\ \mathrm{k} 41\end{array}|\\ ±35=\frac{1}{2}\left\{2\left(4-4\right)-\left(-6\right)\left(5-\mathrm{k}\right)+1\left(20-4\mathrm{k}\right)\right\}\\ ±35=\frac{1}{2}\left(0+30-6\mathrm{k}+20-4\mathrm{k}\right)\\ ±70=-10\mathrm{k}+50\\ ⇒\mathrm{ }-10\mathrm{k}+50=70 \mathrm{or} -10\mathrm{k}+50=-70\\ ⇒\mathrm{k}=\frac{20}{-10}=-2\mathrm{or}\mathrm{k}=\frac{-120}{-10}=12\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}12\mathrm{or}-2.\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{D}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.61 (i) Find equation of line joining (1, 2) and (3, 6) using determinants.
(ii) Find equation of line joining (3, 1) and (9, 3) using determinants.

Ans

$\begin{array}{l}\left(i\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Point A}\left(1,2\right)\text{and B}\left(3,6\right)\text{are joining each other and a point P}\left(x,y\right)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{triangle ABP}=\text{0}\\ \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}}|\begin{array}{l}121\\ 36\text{}1\\ xy\text{}1\end{array}|=0\\ ⇒\frac{1}{2}\left\{1\left(6-y\right)-2\left(3-x\right)+1\left(3y-6x\right)\right\}=0\\ ⇒\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}6-y-6+2x+3y-6x=0\\ ⇒\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}}\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}}\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}}2y-4x=0\\ ⇒\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}}y=2x\\ \text{Thus, the equation of line joining two points (1, 2) and}\left(3,6\right)\\ is\text{​}\text{y}=\text{2x}\text{.}\\ \left(ii\right)\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{Point A}\left(3,\text{\hspace{0.17em}}1\right)\text{and B}\left(9,3\right)\text{are joining each other and a point P}\left(x,y\right)\text{lies}\\ \text{on it, so A, P and B are collinear and area of triangle formed by these}\\ \text{points is zero}\text{. Then}\\ Area\text{\hspace{0.17em}}of\text{\hspace{0.17em}}\text{triangle ABP}=\text{0}\\ \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}}|\begin{array}{l}31\text{}1\\ 9\text{}31\\ x\text{}y1\end{array}|=0\\ ⇒\frac{1}{2}\left\{3\left(3-y\right)-1\left(9-x\right)+1\left(9y-3x\right)\right\}=0\\ ⇒\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}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}9-3y-9+x+9y-3x=0\\ ⇒\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}}\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}}\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}}6y-2x=0\\ ⇒\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}}3\text{\hspace{0.17em}}y=x\\ \text{Thus, the equation of line joining two points (1, 2) and}\left(3,6\right)\\ is\text{​}\text{x}-\text{3y}=\text{0}\text{.}\end{array}$

b

(i) (k, 0), (4, 0), (0, 2) (ii) (–2, 0), (0, 4), (0, k)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{ }\mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=±4 \mathrm{ }\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(\mathrm{k},0\right),\\ \left(4,0\right),\left(0,2\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}\mathrm{k}01\\ 401\\ 021\end{array}|\\ ±4=\frac{1}{2}\left\{\mathrm{k}\left(0-2\right)-0\left(4-0\right)+1\left(8-0\right)\right\}\\ ±4=\frac{1}{2}\left(-2\mathrm{k}-0+8\right)\\ ±8=-2\mathrm{k}+8\\ ⇒\mathrm{ }-2\mathrm{k}+8=8 \mathrm{or} -2\mathrm{k}+8=-8\\ ⇒\mathrm{k}=0\mathrm{}\mathrm{or}\mathrm{k}=\frac{-16}{-2}=8\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}0\mathrm{or}8\mathrm{.}\\ \left(\mathrm{ii}\right) \mathrm{Given}:\mathrm{area}\mathrm{of}\mathrm{triangle}=±4 \mathrm{ }\left(\begin{array}{l}\because \mathrm{Since}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{is}\mathrm{absolute}\\ \mathrm{value}\mathrm{of}\mathrm{\Delta }\end{array}\right)\\ \mathrm{Then}\mathrm{the}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(-2,0\right),\\ \left(0,4\right),\left(0,\mathrm{k}\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}-201\\ \mathrm{ }041\\ \mathrm{ }0\mathrm{k}1\end{array}|\\ ±4=\frac{1}{2}\left\{-2\left(4-\mathrm{k}\right)-0\left(0-0\right)+1\left(0-0\right)\right\}\\ ±4=\frac{1}{2}\left(-8+2\mathrm{k}-0+0\right)\\ ±8=2\mathrm{k}-8\\ ⇒\mathrm{ }2\mathrm{k}-8=8 \mathrm{or} 2\mathrm{k}-8=-8\\ ⇒\mathrm{k}=\frac{16}{2}=\mathrm{ }8\mathrm{}\mathrm{or}\mathrm{k}=\frac{0}{2}=0\\ \mathrm{Thus},\mathrm{value}\mathrm{of}\mathrm{k}\mathrm{is}\mathrm{either}8\mathrm{or}0\mathrm{.}\end{array}$

Q.63 Show that points A(a, b + c), B(b, c + a), C(c, a + b) are collinear.

Ans

$\begin{array}{l}\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\mathrm{A}\left(\mathrm{a},\mathrm{b}+\mathrm{c}\right),\\ \mathrm{B}\left(\mathrm{b},\mathrm{c}+\mathrm{a}\right),\mathrm{C}\left(\mathrm{c},\mathrm{a}+\mathrm{b}\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{ccc}\mathrm{a}& \mathrm{b}+\mathrm{c}& 1\\ \mathrm{b}& \mathrm{c}+\mathrm{a}& 1\\ \mathrm{c}& \mathrm{a}+\mathrm{b}& 1\end{array}|\\ =\frac{1}{2}\left\{\mathrm{a}\left(\mathrm{c}+\mathrm{a}-\mathrm{a}-\mathrm{b}\right)-\left(\mathrm{b}+\mathrm{c}\right)\left(\mathrm{b}-\mathrm{c}\right)+1\left(\mathrm{ba}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}-\mathrm{ca}\right)\right\}\\ =\frac{1}{2}\left\{\mathrm{ac}-\mathrm{ab}-\left({\mathrm{b}}^{2}-{\mathrm{c}}^{2}\right)+\left(\mathrm{ba}+{\mathrm{b}}^{2}-{\mathrm{c}}^{2}-\mathrm{ca}\right)\right\}\\ =\frac{1}{2}\left(\overline{)\mathrm{ac}}-\overline{)\mathrm{ab}}-\overline{){\mathrm{b}}^{2}}+\overline{){\mathrm{c}}^{2}}+\overline{)\mathrm{ba}}+\overline{){\mathrm{b}}^{2}}-\overline{){\mathrm{c}}^{2}}-\overline{)\mathrm{ca}}\right)\\ =0\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{A},\mathrm{B}\mathrm{and}\mathrm{C}\mathrm{is}\mathrm{zero},\mathrm{so}\mathrm{these}\\ \mathrm{points}\mathrm{are}\mathrm{collinear}\mathrm{.}\end{array}$

Q.64 Find area of the triangle with vertices at the point given in each of the following:
(i) (1, 0), (6, 0), (4, 3)
(ii) (2, 7), (1, 1), (10, 8)
(iii) (–2, –3), (3, 2), (–1, –8)

Ans

$\begin{array}{l}\left(\mathrm{i}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(1,0\right),\left(6,0\right),\left(4,3\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}101\\ 601\\ 431\end{array}|\\ =\frac{1}{2}\left\{1\left(0-3\right)-0\left(6-4\right)+1\left(18-0\right)\right\}\\ =\frac{1}{2}\left(-3-0+18\right)\\ =\frac{1}{2}\left(15\right)\\ =\frac{15}{2}\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \left(\mathrm{ii}\right)\mathrm{Theare}\mathrm{a}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(2,7\right),\left(1,1\right),\left(10,8\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}271\\ 111\\ 1081\end{array}|\\ =\frac{1}{2}\left\{2\left(1-8\right)-7\left(1-10\right)+1\left(8-10\right)\right\}\\ =\frac{1}{2}\left(-14+63-2\right)\\ =\frac{1}{2}\left(47\right)\\ =\frac{47}{2}\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \left(\mathrm{iii}\right)\mathrm{The}\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{formed}\mathrm{by}\mathrm{given}\mathrm{vertices}\left(-2,-3\right),\\ \left(3,2\right),\left(-1,-8\right)\\ \mathrm{\Delta }=\frac{1}{2}|\begin{array}{l}-2-31\\ 3 \mathrm{ }21\\ -1-81\end{array}|\\ =\frac{1}{2}\left\{-2\left(2+8\right)-\left(-3\right)\left(3+1\right)+1\left(-24+2\right)\right\}\\ =\frac{1}{2}\left(-20+12-22\right)\\ =\frac{1}{2}\left(-30\right)\\ =-15\mathrm{ }\mathrm{sq}.\mathrm{ }\mathrm{units}\\ \mathrm{Since},\mathrm{area}\mathrm{of}\mathrm{triangle}\mathrm{can}\mathrm{never}\mathrm{be}\mathrm{negative}.\\ \mathrm{So},\mathrm{area}\mathrm{of}\mathrm{triangle}= 15\mathrm{square}\mathrm{units}\mathrm{.}\end{array}$

Q.65 Which of the following is correct
(A) Determinant is a square matrix.
(B) Determinant is a number associated to a matrix.
(C) Determinant is a number associated to a square matrix.
(D) None of these.

Ans

To every square matrix A = [aij] of order n, we can associate a number (real or complex) called determinant of the square matrix A, where aij = (i,j)th element of A.
Therefore, option (C) is correct

Q.66 Let A b a square matrix of order 3 x 3, then |kA| is equal to

(A) k|A| (B) k2|A| (C) k3|A| (D) 3k|A|

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{ }\mathrm{A}\mathrm{be}\mathrm{a}\mathrm{square}\mathrm{matrix}\mathrm{of}\mathrm{order}3×3,\\ \mathrm{then} \mathrm{A}=\left[\begin{array}{l}{\mathrm{a}}_{1}{\mathrm{a}}_{2}{\mathrm{a}}_{3}\\ {\mathrm{b}}_{1}{\mathrm{b}}_{2}{\mathrm{b}}_{3}\\ {\mathrm{c}}_{1}{\mathrm{c}}_{2}{\mathrm{c}}_{3}\end{array}\right]\\ \mathrm{and}\\ \mathrm{kA}=\left[\begin{array}{l}{\mathrm{ka}}_{1}{\mathrm{ka}}_{2}{\mathrm{ka}}_{3}\\ {\mathrm{kb}}_{1}{\mathrm{kb}}_{2}{\mathrm{kb}}_{3}\\ {\mathrm{kc}}_{1}{\mathrm{kc}}_{2}{\mathrm{kc}}_{3}\end{array}\right]\\ |\mathrm{kA}|=|\begin{array}{l}{\mathrm{ka}}_{1}{\mathrm{ka}}_{2}{\mathrm{ka}}_{3}\\ {\mathrm{kb}}_{1}{\mathrm{kb}}_{2}{\mathrm{kb}}_{3}\\ {\mathrm{kc}}_{1}{\mathrm{kc}}_{2}{\mathrm{kc}}_{3}\end{array}|\\ ={\mathrm{k}}^{3}|\begin{array}{l}{\mathrm{a}}_{1}{\mathrm{a}}_{2}{\mathrm{a}}_{3}\\ {\mathrm{b}}_{1}{\mathrm{b}}_{2}{\mathrm{b}}_{3}\\ {\mathrm{c}}_{1}{\mathrm{c}}_{2}{\mathrm{c}}_{3}\end{array}|\left[\mathrm{Taking}\mathrm{k}\mathrm{common}\mathrm{from}{\mathrm{R}}_{\mathrm{1}}, {\mathrm{R}}_{\mathrm{2}}\mathrm{and}{\mathrm{R}}_{\mathrm{3}}.\right]\\ ={\mathrm{k}}^{3}|\mathrm{A}|\\ \mathrm{Thus},\mathrm{option}\left(\mathrm{C}\right)\mathrm{is}\mathrm{correct}\mathrm{.}\end{array}$

Q.67 By using properties of determinants, show that:

$\left|\begin{array}{l}1+{\mathrm{a}}^{2}\mathrm{ab}\mathrm{ab}\\ \mathrm{ab}{\mathrm{b}}^{2}+1\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^{2}+1\end{array}\right|=1+{\mathrm{a}}^{2}+{\mathrm{b}}^{2}+{\mathrm{c}}^{2}$

Ans

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}\mathrm{.}=|\begin{array}{l}\mathrm{1}+{\mathrm{a}}^{2}\mathrm{ab}\mathrm{ac}\\ \mathrm{ab}{\mathrm{b}}^{\mathrm{2}}+\mathrm{1}\mathrm{bc}\\ \mathrm{ca}\mathrm{cb}{\mathrm{c}}^{\mathrm{2}}+\mathrm{1}\end{array}|\\ \mathrm{Applying} {\mathrm{R}}_{1}\to \frac{1}{\mathrm{a}}{\mathrm{R}}_{1},{\mathrm{R}}_{2}\to \frac{1}{\mathrm{b}}{\mathrm{R}}_{2},{\mathrm{R}}_{3}\to \frac{1}{\mathrm{c}}{\mathrm{R}}_{3}\\ =\mathrm{abc}|\begin{array}{l}\frac{\mathrm{1}}{\mathrm{a}}+\mathrm{a}\mathrm{b}\mathrm{c}\\ \mathrm{a}\mathrm{b}+\frac{\mathrm{1}}{\mathrm{b}}\mathrm{c}\\ \mathrm{a}\mathrm{b}\mathrm{c}+\frac{\mathrm{1}}{\mathrm{c}}\end{array}|\\ \mathrm{Applying} {\mathrm{R}}_{1}\to {\mathrm{R}}_{1}-{\mathrm{R}}_{2},{\mathrm{R}}_{2}\to {\mathrm{R}}_{2}-{\mathrm{R}}_{3}\\ =\mathrm{abc}|\begin{array}{l}\frac{1}{}\end{array}\end{array}$