# NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5

These NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 about Determinants are specifically curated by the Extramarks Maths faculty expert which is detailed and easily comprehensible by the Class 12 students. This will help students practice and solve problems associated with the NCERT solutions Class 12 Maths chapter 4 exercise 4.5. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 provide students with simple and effective examples along with solutions that make it easy for the students to grasp the concept efficiently. The students can analyse the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 for better preparation for their board examination which can give the best results.

## NCERT Solutions for Class 12 Maths Chapter 4- Determinants Exercise 4.5

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, and NCERT solutions for other classes and subjects like NCERT solutions Class 11, NCERT solutions Class 10, NCERT Solutions Class 9, NCERT Solutions Class 8, NCERT Solutions Class 7, NCERT Solutions Class 6. It also has NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 till NCERT Solutions Class 1.  These NCERT solutions are for different subjects which are meant for all the NCERT students of different standards from 1 to 12. Extramarks also provides a convenient way for students to access these materials without them having to do any extra hard work.

NCERT Solutions for Class 12 Maths Chapter 4 consists of solutions and elucidation of what Determinants are and what is their significance. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 of Chapter 4 deals with all the conditions and problems related to the Adjoint and Inverse of a Matrix. Extramarks also has other NCERT Solutions for Class 12 Maths Chapter 4 Exercises apart from Exercise 4.5 for Class 12 Maths NCERT, which explains how to evaluate Determinants, to use the properties of Determinants for solving problems, to use Determinants for finding out Areas of Triangles with Vertices, Values, Equations of Lines etc. These solutions will stand out to the Class 12 students because they will give students a proper understanding of Chapter 4: Determinants as a whole. Hence, Extramarks gives importance to each and every topic just like NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, providing study materials for other Exercises included in the NCERT Mathematics Textbook Chapter 4. Students should practice the Ex 4.4 Class 12 Maths NCERT Solutions to have a good understanding of Determinants.

## Access NCERT Solutions for Class 12 Maths Chapter 4 – Determinants

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, which can be accessed by the NCERT Class 12 students anytime and anywhere. Basically, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 contains the content of the Chapter 4 Determinants. Determinants are studied as a subject of linear algebra and are used for linear equations. In this Chapter 4, Exercise 4.5 Class 12 Maths NCERT Solutions, the students will learn about: the Adjoint and Inverse of a Square Matrix, the condition for the Inverse of a Square Matrix to exist and the use of an Adjoint of a Matrix.

### What are Adjoint and Inverse of a Matrix?

The Inverse of a Matrix can be understood only if the student understands what the adjoint of a Matrix is, so they are clearly related to each other and students must give equal attention to both terms and their significance. Also, to clearly understand the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, students must first be clear with what both the Adjoint and Inverse of a Matrix mean and where these concepts apply specifically. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 particularly deals with problems related to Adjoint of a Matrix and Inverse of a Matrix. Extramarks has proper solutions and study material for even different Matrix problems for Class 12 students to go through and solve, which it provides in different chapters including NCERT solutions Maths for Matrix.

### Inverse of a Matrix

As numbers have reciprocals for them, similarly, Matrices too have reciprocals. This reciprocal of a Matrix is known as the Inverse of a Matrix. The Inverse of a Matrix is set by any other given Matrix, which when multiplying with the given Matrix results in the multiplicative identity. For instance, the Inverse of a Matrix M will be denoted by M-1. However, to get the Inverse of 3 × 3 matrices, students need to learn about the Determinant and Adjoint of the Matrix first. There are two ways by which the Inverse of the Matrix can be obtained. The Inverse of a Matrix can be computed by elementary operations and also by using the Adjoint of a Matrix. Therefore, it is said that knowing the Adjoint of a Matrix is important for solving further Matrix problems. The simple operations on Matrices can be calculated through transformations of rows or columns. In addition, the Inverse of a Matrix can be found by using the Inverse of the Matrix formula, that is by using the Determinant and the Adjoint of the Matrix. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, consists of solutions based most specifically on an Inverse of a Matrix along with solutions for Adjoint of a Matrix. These NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, clarifies the basic understanding of the Matrices so that the students can use them to solve the Determinant problems. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 is therefore a significant set of NCERT solutions for students of Class 12 to work on to get conceptualised knowledge on the NCERT Class 12 Mathematics Chapter 4.

### Maths NCERT Solutions Class 12 Chapter 4 Exercise 4.5

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 available at Extramarks are intensively and authentically solved, also curated by the Extramarks expert faculty. They are based on the NCERT textbook and give out easy solutions for quick concept clearance of the Class 12 NCERT students. Therefore, students are relieved from the worry of any faults or errors in these NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 and can use them for practice and reference, ideally.

NCERT Class 12 students need effective solutions and comprehension for every topic and subtopic of Mathematics to get high scores, as Mathematics as a subject is very high scoring if the solutions and results are on point. Therefore, every NCERT chapter of the subject of Mathematics for Class 12 must be given equal importance by students so that they have no hesitation while solving the problems related to the topics and can easily find solutions to them. NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, in this case, emphasises the Adjoint and Inverse of a Matrix, like how to find an Adjoint of a given Matrix or how to find an Inverse of a given Matrix. Based on these problems, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 also deals with the various theorems and definitions of both Adjoint and Inverse of a Matrix to understand both the elements of the Matrix distinctly. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 consists of solutions that are given by using those theorems, formulas, and definitions of the Adjoint and Inverse of a Matrix. The several subtopics determine all the functions and total significance of Determinants in Algebra, like the types of Determinants: the First Order Determinant, Second-Order Determinant, and the Third Order Determinant, many properties of Determinants, How do Minors and Cofactors work for Matrices and Determinants, what are the major applications of Matrices and Determinants, how are they correlated? And even the method of using the Inverse of a Matrix for Solving a System of Linear Equations. Students gain a perspective on how to solve any given question on Matrix or Determinants by going through the Class 12 maths chapter 4 Exercise 4.5 Solutions and the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5.

Out of these subtopics of Class 12 Mathematics Chapter 4, the Adjoint and Inverse of a Matrix, which is the main topic of explanation for the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, it mainly talks about 4 theorems related to them plus specific definitions. The main 4 theorems of Adjoint and Inverse of a Matrix are explained very briefly below:

• Theorem 1: If M is the given Square Matrix of order x, then M(adj M) = (adj M) M= |A| I , where I is the Identity Matrix of order x.
• Theorem 2: If M and N are Nonsingular Matrices of the same order, then MN and NM are also nonsingular Matrices of the same order.
• Theorem 3: The Determinant of the product of Matrices is equal to the product of their Respective Determinants, which means, |MN|= |M| |N|, where M and N are square matrices of the same order.
• Theorem 4: A Square Matrix M is Invertible if specifically, M is a Nonsingular Matrix.

Some important points that the students should note are:

1. A Square Matrix, say M is said to be singular only if |M| = 0.
2. Similarly, a Square Matrix M is said to be Nonsingular if |M|≠ 0

They will be able to solve any problem related to the questions included in Exercise 4.5 or any similar question mentioned in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 for that matter. Moreover, Extramarks prepares important questions and sets different difficulty levels for students to test every concept they have learned to improve their learning skills. The important questions of the practice series vary in levels from an easy level to the difficulty level and also have a medium level in order to not intimidate students. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 provide verifications and examples for all the theorems and definitions related to the Adjoint and Inverse of a Matrix so that Class 12 students understand why the theorems and solutions work the way they do in minute detail. Extramarks provides extra live lessons and discussions of the topics comprising the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. It also provides an analysis of all the solutions given in NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 by the Extramarks experts of the subject.

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### Chapter 4 – Determinants of Other Exercises

 Chapter 4 – Determinants of Other Exercises Exercise 4.1 8 Questions & Solutions (3 Short Answers, 5 Long Answers) Exercise 4.2 10 Questions & Solutions (4 Short Answers, 10 Long Answers) Exercise 4.3 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.4 5 Questions & Solutions (2 Short Answers, 3 Long Answers) Exercise 4.6 16 Questions & Solutions (3 Short Answers, 13 Long Answers)

Q.1 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.2 Find adjoint of each of the matrices

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

Ans

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

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

Ans

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

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

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}$

Q.5 Find the inverse of each of the matrices (if it exists)

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

Ans

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

Q.6 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.7 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.8 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.9 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.10 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.11 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.12

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

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.13

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

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.14

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

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.15

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

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.16

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

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.17 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.18

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

Ans

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

## 1. Are the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 available online?

Yes, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5  are available online at Extramarks which can be accessed by Class 12 students anytime anywhere. The solutions are skillfully curated by the Mathematics experts faculty of Extramarks and are prepared in a way that the students can easily comprehend and solve any problems that were referred into the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5.

## 2. Are the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 important for students to pass the Class 12 board examination?

Yes, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 is an important part of the Class 12 board examination. In fact, every exercise in the NCERT solutions is essential for the Class 12 students in their preparation for boards. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 are provided by Extramarks so that students can prepare for their examination by referring to these solutions and improving their concepts. Therefore, they can solve any problem that comes in the question paper of the Class 12 board examination. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 consist of a very important topic of what is Adjoint and Inverse of a Matrix and what it is used for, so the NCERT Mathematics Board Question Paper will consist of questions related to those in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. If students have no knowledge of the chapter and its exercises, they will struggle in solving the questions corresponding to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5. So, they will have more chances of getting fewer marks for not completing the concerned questions.

## 3. What do the questions of the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 deal with?

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 deals with basically the questions and solutions of the Adjoint and Inverse of a Matrix. Similar to finding either the Adjoint of a Matrix or the Inverse of a given Matrix or finding both. There are solutions for verifying certain equations using the Adjoint or the Inverse of a Matrix. Also, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 include all the theorems and formulas associated with the Adjoint and Inverse of Matrices which makes the students understand the topics in a detailed manner.

## 4. Does Extramarks have NCERT solutions for other chapters and exercises of NCERT Maths for Class 12 apart from the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5?

Yes, besides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5, Extramarks also provides solutions for all the chapters and exercises of NCERT Mathematics for Class 12. Students are offered the entire syllabus of the NCERT Mathematics textbook for Class 12 for practice and solving problems by Extramarks. Just like the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 which has solutions for their specific topic or subtopic, other NCERT solutions also have conceptualised solutions for the topics they are related to and deal with all the problems, evaluations, and verifications of the topic. For example the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.5 deal with Adjoint and Inverse of a Matrix and their Mathematical calculations of the main topic Determinants. Likewise, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 deal with Minors and Cofactors of Determinants. All other chapters dealing with different topics of the Class 12 NCERT Mathematics syllabus are available at Extramarks.