NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4

NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4 Determinants are discussed here. These NCERT solutions are created by subject matter expert at Careers360 considering the latest syllabus and pattern of CBSE 2023-24. NCERT solutions for Class 12 Maths chapter 4 exercise 4.4 consists of questions related to finding minors and cofactors of determinants. The minor of an element of the determinant is obtained by deleting the row and the column in which the element lies. Minors and cofactors are useful to find the adjoint and inverse of the matrix that you will learn in the next exercise of the Class 12 Maths NCERT book.

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There are some examples given before Class 12 Maths ch 4 ex 4.4 that you can solve for better understanding. Exercise 4.4 Class 12 Maths is a very small exercise but useful for all the upcoming exercises. Class 12th Maths chapter 4 exercise 4.4 questions are very similar to the solved examples given before the NCERT exercise. First, try to solve these exercise questions by yourself. If you are finding difficulties while solving Class 12th Maths chapter 4 exercise 4.4, you can go through NCERT solutions for Class 12 Maths chapter 4 exercise 4.4. 12th class Maths exercise 4.1 answers are designed as per the students demand covering comprehensive, step by step solutions of every problem. Practice these questions and answers to command the concepts, boost confidence and in depth understanding of concepts. Students can find all exercise together using the link provided below.

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Assess NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4

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Determinants Exercise:4.4

Question:1(i) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix}2 &-4 \\0 &3 \end{vmatrix}$

Answer:

GIven determinant: $\begin{vmatrix}2 &-4 \\0 &3 \end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$ .

Therefore we have

$M_{11}$ = minor of element $a_{11}$ = 3

$M_{12}$ = minor of element $a_{12}$ = 0

$M_{21}$ = minor of element $a_{21}$ = -4

$M_{22}$ = minor of element $a_{22}$ = 2

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1)^{i+j}M_{ij}$ .

Therefore we have:

$A_{11} = (-1)^{1+1}M_{11} = (-1)^2(3) = 3$

$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(0) = 0$

$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(-4) = 4$

$A_{22} = (-1)^{2+2}M_{22} = (-1)^4(2) = 2$

Question:1(ii) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix} a &c \\ b &d \end{vmatrix}$

Answer:

GIven determinant: $\begin{vmatrix} a &c \\ b &d \end{vmatrix}$

Minor of element $a_{ij}$ is $M_{ij}$ .

Therefore we have

$M_{11}$ = minor of element $a_{11}$ = d

$M_{12}$ = minor of element $a_{12}$ = b

$M_{21}$ = minor of element $a_{21}$ = c

$M_{22}$ = minor of element $a_{22}$ = a

and finding cofactors of $a_{ij}$ is $A_{ij}$ = $(-1)^{i+j}M_{ij}$ .

Therefore we have:

$A_{11} = (-1)^{1+1}M_{11} = (-1)^2(d) = d$

$A_{12} = (-1)^{1+2}M_{12} = (-1)^3(b) = -b$

$A_{21} = (-1)^{2+1}M_{21} = (-1)^3(c) = -c$

$A_{22} = (-1)^{2+2}M_{22} = (-1)^4(a) = a$

Question:2(i) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{vmatrix}$

Answer:

Given determinant : $\begin{vmatrix} 1 & 0 &0 \\ 0 &1 &0 \\ 0 &0 &1 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 0 &1 \\ 0 &0 \end{vmatrix} = 0$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &0 \\ 0 &1 \end{vmatrix} = 0$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &0 \\ 1 &0 \end{vmatrix} = 0$ $M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &0 \\ 0 &0 \end{vmatrix} = 0$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 1$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = 0$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 0$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 0$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 1$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = 0$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = 0$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 0$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 1$ .

Question:2(ii) Write Minors and Cofactors of the elements of following determinants:

$\small \begin{vmatrix} 1 &0 &4 \\ 3 & 5 &-1 \\ 0 &1 &2 \end{vmatrix}$

Answer:

Given determinant : $\begin{vmatrix} 1 &0 &4 \\ 3 & 5 &-1 \\ 0 &1 &2 \end{vmatrix}$

Finding Minors: by the definition,

$M_{11} =$ minor of $a_{11} = \begin{vmatrix} 5 &-1 \\ 1 &2 \end{vmatrix} = 11$ $M_{12} =$ minor of $a_{12} = \begin{vmatrix} 3 &-1 \\ 0 &2 \end{vmatrix} = 6$

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 3 &5 \\ 0 &1 \end{vmatrix} = 3$ $M_{21} =$ minor of $a_{21} = \begin{vmatrix} 0 &4 \\ 1 &2 \end{vmatrix} = -4$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 1 &4 \\ 0 &2 \end{vmatrix} = 2$ $M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &0 \\ 0 &1 \end{vmatrix} = 1$

$M_{31} =$ minor of $a_{31} = \begin{vmatrix} 0 &4 \\ 5 &-1 \end{vmatrix} = -20$

$M_{32} =$ minor of $a_{32} = \begin{vmatrix} 1 &4 \\ 3 &-1 \end{vmatrix} = -1-12=-13$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &0 \\ 3 &5 \end{vmatrix} = 5$

Finding the cofactors:

$A_{11}=$ cofactor of $a_{11} = (-1)^{1+1}M_{11} = 11$

$A_{12}=$ cofactor of $a_{12} = (-1)^{1+2}M_{12} = -6$

$A_{13}=$ cofactor of $a_{13} = (-1)^{1+3}M_{13} = 3$

$A_{21}=$ cofactor of $a_{21} = (-1)^{2+1}M_{21} = 4$

$A_{22}=$ cofactor of $a_{22} = (-1)^{2+2}M_{22} = 2$

$A_{23}=$ cofactor of $a_{23} = (-1)^{2+3}M_{23} = -1$

$A_{31}=$ cofactor of $a_{31} = (-1)^{3+1}M_{31} = -20$

$A_{32}=$ cofactor of $a_{32} = (-1)^{3+2}M_{32} = 13$

$A_{33}=$ cofactor of $a_{33} = (-1)^{3+3}M_{33} = 5$ .

Question:3 Using Cofactors of elements of second row, evaluate . $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

Answer:

Given determinant : $\small \Delta =\begin{vmatrix} 5 &3 &8 \\ 2 & 0 & 1\\ 1 &2 &3 \end{vmatrix}$

First finding Minors of the second rows by the definition,

$M_{21} =$ minor of $a_{21} = \begin{vmatrix} 3 &8 \\ 2 &3 \end{vmatrix} =9-16 = -7$

$M_{22} =$ minor of $a_{22} = \begin{vmatrix} 5 &8 \\ 1 &3 \end{vmatrix} = 15-8=7$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 5 &3 \\ 1 &2 \end{vmatrix} = 10-3 =7$

Finding the Cofactors of the second row:

$A_{21}=$ Cofactor of $a_{21} = (-1)^{2+1}M_{21} = 7$

$A_{22}=$ Cofactor of $a_{22} = (-1)^{2+2}M_{22} = 7$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = -7$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the second row with their corresponding cofactors.

Therefore we have,

$\triangle = a_{21}A_{21} + a_{22}A_{22} + a_{23}A_{23} = 2(7) +0(7) +1(-7) =14-7=7$

Question:4 Using Cofactors of elements of third column, evaluate $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

Answer:

Given determinant : $\small \Delta =\begin{vmatrix} 1 &x &yz \\ 1 &y &zx \\ 1 &z &xy \end{vmatrix}$

First finding Minors of the third column by the definition,

$M_{13} =$ minor of $a_{13} = \begin{vmatrix} 1 &y \\ 1 &z \end{vmatrix} =z-y$

$M_{23} =$ minor of $a_{23} = \begin{vmatrix} 1 &x \\ 1 &z \end{vmatrix} = z-x$

$M_{33} =$ minor of $a_{33} = \begin{vmatrix} 1 &x \\ 1 &y \end{vmatrix} =y-x$

Finding the Cofactors of the second row:

$A_{13}=$ Cofactor of $a_{13} = (-1)^{1+3}M_{13} = z-y$

$A_{23}=$ Cofactor of $a_{23} = (-1)^{2+3}M_{23} = x-z$

$A_{33}=$ Cofactor of $a_{33} = (-1)^{3+3}M_{33} = y-x$

Therefore we can calculate $\triangle$ by sum of the product of the elements of the third column with their corresponding cofactors.

Therefore we have,

$\triangle = a_{13}A_{13} + a_{23}A_{23} + a_{33}A_{33}$

$= (z-y)yz + (x-z)zx +(y-x)xy$

$=yz^2-y^2z + zx^2-xz^2 + xy^2-x^2y$

$=z(x^2-y^2) + z^2(y-x) +xy(y-x)$

$= (x-y) \left [ zx+zy-z^2-xy \right ]$

$=(x-y)\left [ z(x-z) +y(z-x) \right ]$

$= (x-y)(z-x)[-z+y]$

$= (x-y)(y-z)(z-x)$

Thus, we have value of $\triangle = (x-y)(y-z)(z-x)$ .

Question:5 If $\small \Delta =\begin{vmatrix} a{_{11}} & a_1_2 & a_1_3\\ a_2_1 & a_2_2 & a_2_3\\ a_3_1 &a_3_2 &a_3_3 \end{vmatrix}$ and $\small A{_{ij}}$ is Cofactors of $\small a{_{ij}}$ , then the value of $\small \Delta$ is given by

(A) $\small a_1_1A_3_1+a_1_2A_3_2+a_1_3A_3_3$

(B) $\small a_1_1A_1_1+a_1_2A_2_1+a_1_3A_3_1$

(D) $\small a_1_1A_1_1+a_2_1A_2_1+a_3_1A_3_1$

Answer:

Answer is (D) $\small a_1_1A_1_1+a_2_1A_2_1+a_3_1A_3_1$ by the definition itself, $\small \Delta$ is equal to the product of the elements of the row/column with their corresponding cofactors.

More about NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4:-

In this article, you will get NCERT solutions for Class 12 Maths chapter 4 exercise 4.4 consists of five questions related to finding minor and cofactors of a determinant. There are four examples given before this NCERT exercise. After these examples, you will easily solve Class 12th Maths chapter 4 exercise 4.4 problems. This NCERT book exercise is very useful for all the upcoming exercises of this chapter. You are advised to be thorough with this exercise to understand the next exercise.

Also Read| Determinants Class 12 Chapter 4 Notes

Benefits of NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4:-

NCERT solutions for Class 12 Maths chapter 4 exercise 4.4 are helpful when you are not able to solve NCERT problems.
These questions are prepared by our experienced experts who will give the best content for the board exam based on NCERT syllabus.
Exercise 4.4 Class 12 Maths solutions are designed in a descriptive manner so you will understand them very easily.
If you have solved all the NCERT problems, you can try to solve previous years' papers to get familiar board exam patterns.
You can take Class 12 Maths chapter 4 exercise 4.4 solutions for quick revision before the exam.

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Key Features Of NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4

Comprehensive Coverage: The solutions encompass all the topics covered in ex 4.4 class 12, ensuring a thorough understanding of the concepts.
Step-by-Step Solutions: In this class 12 maths ex 4.4, each problem is solved systematically, providing a stepwise approach to aid in better comprehension for students.
Accuracy and Clarity: Solutions for class 12 ex 4.4 are presented accurately and concisely, using simple language to help students grasp the concepts easily.
Conceptual Clarity: In this 12th class maths exercise 4.4 answers, emphasis is placed on conceptual clarity, providing explanations that assist students in understanding the underlying principles behind each problem.
Inclusive Approach: Solutions for ex 4.4 class 12 cater to different learning styles and abilities, ensuring that students of various levels can grasp the concepts effectively.
Relevance to Curriculum: The solutions for class 12 maths ex 4.4 align closely with the NCERT curriculum, ensuring that students are prepared in line with the prescribed syllabus.

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Frequently Asked Questions (FAQs)

1. Where can I get CBSE Class 12 previous years paper ?

Here you will get CBSE Class 12 previous years paper.

2. Does CBSE provides previous papers solutions ?

No, CBSE doesn't provide previous papers solutions. Students can download the question papers and marking scheme from the CBSE website.

3. Can I get CBSE Class 12 Maths previous years paper with solutions ?

You can check here for CBSE Class 12 Maths previous years paper with solutions. The questions are based on the content of the NCERT syllabus. Refering to the previous year papers is helpful to understand the area from which more questions are asked.

4. what is a singular matrix ?

If the determinant of a square matrix A is zero, it is called a singular matrix.

5. what is a non-singular matrix ?

If the determinant of a square matrix A is not zero, it is called a non-singular matrix.

6. Can I get CBSE Class 10 Maths previous years paper with solutions ?

Here you will get CBSE Class 10 Maths previous years paper with solutions.

7. Does questions from miscellaneous exercises are asked in the CBSE board exams ?

Most of the questions are not asked from miscellaneous exercises but sometimes a few questions are asked from the miscellaneous exercises too.

8. Does miscellaneous exercises are important ?

As most of the questions are not asked from miscellaneous exercises in the board exams, so it is not important from the board exam point of view but very important for competitive exams.

Also Read

NCERT Books for Class 12 - Download All Subjects PDFs Here

NCERT Class 12 Physics Chapter 9 Notes Ray Optics and Optical Instruments - Download PDF

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i have taken admission in clg after my campartment exam in chemistry along with the neet 2025 preperation but unfortunately i again got campartment after doing all this preparation i got 15 marks only now I can't able to think what should I do next.

You can use them people also used problem

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Divyan 06 August,2024

i scored 84 percent in cbse 2 years ago.Am i eligible for medhavi scholarship if give 12th again after 2 drop years and will got 75+ in mpboard

Hi,

The Medhavi National Scholarship Program, under the Human Resources & Development Mission (HRDM), offers financial assistance to meritorious students through a scholarship exam. To be eligible, candidates must be between 16 and 40 years old as of the last date of registration and have at least passed the 10th grade from a recognized board. Higher qualifications, such as 11th/12th grade, graduation, post-graduation, or a diploma, are also acceptable.

To apply, download the Medhavi App from the Google Play Store, sign up, and read the detailed notification about the scholarship exam. Complete the registration within the app, take the exam from home using the app, and receive your results within two days. Following this, upload the necessary documents and bank account details for verification. Upon successful verification, the scholarship amount will be directly transferred to your bank account.

The scholarships are categorized based on the marks obtained in the exam: Type A for those scoring 60% or above, Type B for scores between 50% and 60%, and Type C for scores between 40% and 50%. The cash scholarships range from Rs. 2,000 to Rs. 18,000 per month, depending on the exam and the marks obtained.

Since you already have a 12th-grade qualification with 84%, you meet the eligibility criteria and can apply for the Medhavi Scholarship exam. Preparing well for the exam can increase your chances of receiving a higher scholarship.

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Yuvan 01 September,2024

i upload 12th board result screenshot in ncweb form from the cbse site so it is valid

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

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Ashvadha 12 July,2024

Kya mujhe class 12 ka important question milega jo exam mai per year puche jate hai

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

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kumardurgesh1802 10 July,2024

I have completed my 12th with arts cbse. I want to repeat my 12th with pcmb. How to become a private candidate and what are deadline to be able to appear in 2025 exam?

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
You have to appear for the 2025 12th board exams.
Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
Aim to register before late October to avoid extra fees.
Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

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NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4 - Determinants

NCERT Solutions For Class 12 Maths Chapter 4 Exercise 4.4

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Assess NCERT Solutions for Class 12 Maths Chapter 4 Exercise 4.4

Determinants Exercise:4.4

Key Features Of NCERT Solutions for Exercise 4.4 Class 12 Maths Chapter 4

JEE Main Important Mathematics Formulas

NCERT Solutions of Class 12 Subject Wise

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