# NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 (Ex 4.4)

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4, in an easy format which is conveniently accessible by Class 12 students. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 are about the topic, “Determinants”, which is studied as a subject of linear algebra. Chapter 4, Exercise 4.4 Class 12 Maths NCERT Solutions, specifically deals with the use of Minors and Cofactors which are used to determine the adjoint and inverse of a matrix.

## NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 (Ex 4.4)

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 about Determinants are specially curated by the Extramarks Maths faculty with easy examples and detailed explanations. Students will be able to have a clear understanding of Chapter 4 Exercise 4.4 Class 12 Maths NCERT Solutions. This will help students solve any problem associated with the Exercise 4.4 Class 12 Maths NCERT Solutions. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 provides students with examples along with solutions that make it easy for the students to get their concepts right. The students can analyse the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 and practice more NCERT problems related to Exercise 4.4 Class 12 Maths NCERT Solutions.

Extramarks provides the NCERT Solutions for Class 12 Maths Chapter Exercise 4.4, along with NCERT Solutions for other classes 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 provides NCERT Solutions Class 5, NCERT Solutions Class 4, NCERT Solutions Class 3, NCERT Solutions Class 2 and even NCERT Solutions Class 1.  All these NCERT Solutions are easily accessible and explained in simple language with proper steps for solutions. All the Class 12 students can download the PDFs of the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 and analyse them until they solve all the problems present in the Ex4 4 Class 12 Maths NCERT Solutions.

## Access NCERT Solutions for Maths Chapter 4 – Determinants

Extramarks provides students with all the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 along with 5 more exercises from Exercise 4.1 to Exercise 4.6. NCERT Solutions for Class 12 Maths Chapter 4 explains to the students what Determinants are and how they can be applied, and also explains the use of Minors and Cofactors properly. Extramarks also recommends students to go through all other NCERT Solutions for Class 12 Maths Chapter 4 Exercises apart from the Exercise 4.4 Class 12 Maths NCERT Solutions. It explains how to evaluate Determinants, to use the properties of Determinants for proving sums, and to use Determinants for finding out Areas of Triangles with Vertices, Values, and Equations of Lines.

### Chapter 4 of Class 12th NCERT

Along with the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4, Extramarks provides all the other exercises included in the NCERT Mathematics Textbook Chapter 4: Determinants. After a precise summary of each topic, students can download the PDFs of all the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 along with other available Exercises of the Class 12 Mathematics Chapter 4.

### Introduction

A Determinant to be precise is the number that determines the uniqueness of a solution which is associated with a matrix. For example; if a1 b2- a2 b1 ≠ 0, the number a1 b2- a2 b1 determines the uniqueness of the solution associated with the matrix A= [a1 b1 a2  b2 ] and therefore is the Determinant of the Matrix. A Determinant in Mathematics is a scalar value which is the result of a calculation obtained from a Square Matrix. Determinants are used for various purposes like to give direct formulas for the solution of a system of X equations in X unknowns, and for the Inverse of what is known as an Invertible Matrix. Determinants can also be used particularly to give formulas for the area or volume of certain geometric figures.

NCERT Solutions for Class 12 Maths Chapter 4 provides a basic understanding of several properties of Determinants and their applications. Including finding the Area of a Triangle or the Adjoint and Inverse of a Square Matrix, and also both the consistency and inconsistency of the system of Linear Equations. Extramarks also provides solutions of linear equations in two or three variables using the Inverse of a Matrix, along with topics like Minors and Cofactors. Mainly, the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 available on Extramarks, help the students understand the use of Minors and Cofactors to write or evaluate Determinants in the compact. Extramarks Mathematics experts provide an adequate amount of examples attached along with the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 to make students understand the Exercise 4.4 Class 12 Maths NCERT Solutions more clearly and effectively. Problems related to, for instance, finding the Minor or Cofactor of an element of any Determinant or like evaluation of any Determinant using Minors or Cofactors are easily solved by the Class 12 students. It happens if the student is properly acquainted with the NCERT Solutions for Class 12 Maths Ex4 4

### Determinant

To every Square Matrix A = [aij] of order n. It implies that it is associated with a number called the Determinant of the Square Matrix A, where aij = (i, j)th element of A. Let it be considered a function which associates each Square Matrix with a unique number which might be real or complex. If we consider M to be the set of square matrices, and N to be the set of numbers, be it real or complex and f : M → N is defined by f (A) = k, where A ∈ M and k ∈ N, then f(A) is called the Determinant of A which can also be denoted by |A| or get A or Δ.

### Matrix

According to the NCERT Class 12 Textbook, “a matrix is an ordered rectangular array of numbers or functions. The numbers or functions are called the elements or the entries of the matrix” and it is denoted by capital letters. Matrices also consist of horizontal and vertical lines of elements which constitute rows and columns respectively. It is significant to note that a matrix can be in any format like in 2 × 2 order, 3 × 3 order, or even 5 × 5 order, and every matrix has its distinct location. For the relation between Determinants and Matrices, there are various equations considering them to be Determinants that vary with different Matrix orders. In the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4, different solutions are given to determine Minors and Cofactors or use Minors and Cofactors to find Determinants in the form of Matrix Equations. Extramarks has proper solutions and study materials for even different Matrix problems for Class 12 students to go through and solve.

### 2 x 2 Matrix Calculation

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 has problems with a Matrix of 2 x 2 order where either Minors and Cofactors are to be determined or Minors and Cofactors are used to determine Determinants. So the apprehension of 2 x 2 Matrix order and its calculation is extremely necessary for students to be able to understand the complex questions given in the  NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4. Students must be able to differentiate between different orders of Matrix calculation be it 2 x 2 or 3 x 3 to understand the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 and solve many other similar questions. A 2 x 2 Matrix calculation has two rows and 2 columns in a rectangular array like M = a1 b2  a2 b1 where |M| is the Determinant. Extramarks also provides solutions and study materials for analysis of the 2 x 2 Matrix calculation along with simple examples. Thus, students can easily understand the format and proceed to the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 to solve and practice problems related to Minors and Cofactors.

### 3 x 3 Matrices Calculation

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 available at Extramarks, also further proceeds to deal with Matrices with 3 x 3 order and their calculations. A 3 x 3 Matrix order consists of 3 rows and 3 columns formatted into a Rectangular Array and also is necessary for many Determinant problems. Extramarks, along with the content and study materials for other Matrix calculations have various problems and solutions related to 3 x 3 matrix calculations making it efficient for students to even understand the questions and solutions given in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4. Extramarks also recommends students to get the study materials of other NCERT Chapters of Class 12 available at Extramarks itself like Chapter 3 Exercises related to Matrix and its other uses and applications.

### Minor

The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 consists of solutions of problems majorly related to Minors and Cofactors. So knowing what a Minor is primarily necessary for Class 12 students to understand the solutions given in the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4. The NCERT Mathematics textbook defines Minor as “Minor of an element aij of a Determinant is the Determinant obtained by deleting its row and jth column in which element aij lies. Minor of an element aij is denoted by Mi”. A Minor in simple terms is a value calculated from the Determinant of a Square Matrix resulting from crossing out a row and a column corresponding to the element which is under consideration. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 has many questions relating to Minors and also explains the usage of Minor in order to evaluate a Determinant.

### Cofactor of a Minor

In the NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 provided by Extramarks many problems are about finding or using Cofactors in Determinant problems. Students must be familiar with the term and what it is used for, or what is its importance in solving any Determinant problem. For an element say ‘aij’, the Cofactor will be considered as ‘Aij’, where the mentioned Cofactor will be considered with the help of the formula A= (-1)i+j B, where B is Minor of the particular element ‘aij’. The NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4 explains the Cofactor calculations in a precise and comprehensible manner, and students can refer to them for practising and solving similar problems related to cofactor of a Minor.

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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 4.5 18 Questions & Solutions (4 Short Answers, 14 Long Answers) Exercise 4.6 16 Questions & Solutions (3 Short Answers, 13 Long Answers)

Q.1 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.2 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.3

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

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.4

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

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.5

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

Ans

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