IIT JAM Mathematics 2022: All you need to know to prepare for it

About IIT JAM Mathematics

IIT JAM Mathematics (MA) is considered one of the most sought-after master’s level competitive exams after BSc./B.Tech. Students can get direct admission into IITs and into IISC (upon clearing the interview). The IISER’s also take IIT JAM rank into account so all in all, it is a pretty important entrance if you are interested in higher mathematics.

IIT JAM Mathematics (MA) 2022: Important Details


Eligibility for IIT JAM MA 2022

To apply for admission to the Mathematics program, a candidate is required to qualify in the Mathematics Test Paper and also satisfy the Minimum Educational Qualifications (MEQs) and Eligibility Requirements (ERs) of the Academic Programme.

The Eligibility Requirements for IIT JAM 2022 is given below:

Check out the official website of IIT JAM for the Application procedure, detailed Eligibility Requirements (ER) and Minimum Educational Qualification (MEQ), application fee, etc: https://jam.iitr.ac.in/

Duration of the course

The duration of the M.Sc course for Mathematics course is 2 years.

Important Dates:

Admitting Institutes for IIT JAM Mathematics

Candidates can secure admission to the following colleges:

Cheenta is initiating an Open for All Math Camp for College Students and High School Passouts who really love Mathematics and want to pursue it for higher studies. 

To get free resources for your IIT JAM Mathematics 2022 Preparation, Register here.

Exam Pattern for IIT JAM Mathematics 2022

The JAM 2022 Examination for all the seven Test Papers will be carried out as ONLINE Computer Based Test (CBT) where the candidates will be shown the questions in a random sequence on a computer screen. This is how the Exam pattern for IIT JAM Mathematics 2020 will look like:

SectionType of QuestionsNumber of QuestionsMarks Distribution
AMultiple Choice Questions (MCQ)3010 questions of one mark each and 20 questions of two marks each
BMultiple Select Questions (MSQ)10Two marks each
CNumerical Answer Type (NAT)2010 questions of one mark each and 10 questions of two marks each
Total60100

The time duration of the Exam is 3 hours in total.

Marking scheme:

In all sections, questions not attempted will result in zero marks. In Section-A (MCQ), the wrong answer will result in negative marks. For each wrong answer to one-mark questions, the one-third mark will be deducted and similarly, for each wrong answer to two-mark questions, a two-third mark will be deducted. In Section-B (MSQ), there are no negative and no partial marking provisions. There is no negative marking in Section-C (NAT) as well.

Syllabus

The syllabus for IIT JAM MA includes:

Click here to get the full syllabus:- https://jam.iitr.ac.in/assets/syllabi/MA_Syllabi.pdf

The Open for All Math Camp gives an opportunity to students to learn live from the experts for free.

Don't miss this opportunity! Register here for free.

How to prepare for IIT JAM Mathematics Entrance 2021


Books for IIT JAM Mathematics (MA) Entrance

Books for Linear Algebra

Books for Abstract Algebra

Books for Real Analysis

Books for Vector Calculus and Differential Equation

Preparation Tips for IIT JAM MA 2022

Go through the syllabus:-

Learning about the syllabus is an essential part of preparation. This helps to have a better understanding of the preparation strategy and what topics to focus on. It saves you time as well.

Work on your concept and solve different questions:- 

Try to clear your concepts on each topic and practice questions based on that topic from different books. This way, you will be able to understand your understanding of the topic.

Solve Previous Papers:- 

This will help you know the exact difficulty level of the exam and prepare accordingly. Try to solve them in a time-bound manner. Check out some of the past year's problems available here.

Take Mock Tests:- 

Take Mock Tests to check your ability of problem-solving. If it is an online test, then it can help you know your preparation status in the competition. You will be able to know where you stand and what should be improved.

Track your progress:-

You should also track your progress. If something unimportant is stealing your time, recognize it and try to avoid it. Keep yourself surrounded by people who encourage you. Discuss your dreams and goals with them.

Take proper rest and believe in yourself:- 

It is the biggest myth that studying for long hours brings productivity to your work. There should be a routine with a proper work-rest balance and then you should ritually follow it. Also having low self-esteem can ruin your performance in exams.

150+ Students are already part of the Open for All Math Camp

Why are you leaving behind? Register here for free.

Prepare for IIT JAM Mathematics (MA) with Cheenta

Cheenta offers a full-year program for college students, willing to crack the IIT JAM Mathematics 2022 Entrance Exam. The program includes classes on Pre-College Mathematics, Real Analysis, Linear Algebra, Abstract Algebra & Vector Calculus. The classes for the full-year program includes:

  1. Two Group classes per week (conceptual and problem solving)
  2. Access to Cheenta Genius App
  3. Regular Quizzes and Assignments
  4. Access to the Doubt Clearing Group

Get a Trial Class now!

What is TIFR and how to prepare for it?

About TIFR

Tata Institute of Fundamental Research, TIFR is the foremost institution for advanced research in foundational sciences based in Mumbai, Maharashtra, India. The institute offers a master's course, an integrated M.Sc and Ph.D. course and a Ph.D. degree in different science fields. One can get admission into this institute by clearing the TIFR Entrance Exam. It is a perfect place for pioneering scientists to commence their careers.

WHAT IS TIFR AND HOW TO PREPARE FOR TIFR ENTRANCE

TIFR Entrance for Integrated M.Sc-Ph.D. Programme in Mathematics


Eligibility

The eligibility criterion to be considered for admission to the Integrated Ph.D. programme is a Bachelor's degree in any of Mathematics, Statistics, Science or Technology (B.A. / B.Sc. / B.Math. / B.Stat. / B.E. / B.Tech.) from any recognized university.

Duration of the course

The duration of integrated courses is 6 years.

Application Procedure

Candidates can fill the application form online through the university website. After filling the form, candidates can pay the application fee online through internet banking or Debit/Credit Card.

Generally, the admission process takes place in the month of December.

Candidates can secure the admission in the following colleges:

Selection Procedure

The selection of the candidates will be done on the basis of the Graduate School (GS) exam. A list of selected candidates will be published on the official website, on the basis of which, interview round will take place. This interview will be conducted by the university.

TIFR GS Entrance Exam Pattern

PartType of QuestionsNumber of Questions
AMCQ20
BObjective (True/ False)20
Total40

The time duration of the Exam is 3 hours in total.

This is the latest exam pattern followed in 2020 paper and patterns may change.

Syllabus

The syllabus includes:

Click here to get the full syllabus:- https://zurl.co/0kbX

How to prepare for TIFR Entrance

Books for TIFR Entrance

Books for Linear Algebra

Books for Abstract Algebra

Books for Real Analysis

Books for Topology

Books for Vector Calculus and Differential Equation

For more information, visit the website:- https://www.tifr.res.in/

Preparation Tips

Go through the syllabus:-

Learning about the syllabus is an essential part of preparation. This helps to have a better understanding of the preparation strategy and what topics to focus on. It saves you time as well.

Work on your concept and solve different questions:- 

Try to clear your concepts on each topic and practice questions based on that topic from different books. This way, you will be able to understand your understanding of the topic.

Solve Previous Papers:- 

This will help you know the exact difficulty level of the exam and prepare accordingly. Try to solve them in a time-bound manner.

Take Mock Tests:- 

Take Mock Tests to check your ability of problem-solving. If it is an online test, then it can help you know your preparation status in the competition. You will be able to know where you stand and what should be improved.

Track your progress:-

You should also track your progress. If something unimportant is stealing your time, recognise it and try to avoid it. Keep yourself surrounded by people who encourage you. Discuss your dreams and goals with them.

Take proper rest and believe in yourself:- 

It is the biggest myth that studying for long hours brings productivity at your work. There should be a routine with a proper work-rest balance and then you should ritually follow it. Also having low self-esteem can ruin your performance in exams.

All the best!

Prepare for TIFR Mathematics with Cheenta

Cheenta offers a full year program for college students, willing to crack the TIFR Entrance. The Program includes classes on Pre-College Mathematics, Real Analysis, Linear Algebra, Abstract Algebra & Vector Calculus. The classes for the full year program includes:

  1. Two Group classes per week (conceptual and problem solving)
  2. Access to Cheenta Genius App
  3. Regular Quizzes and Assignments
  4. Access to the Doubt Clearing Group

Get a Trial Class now!

Order of rings: TIFR GS 2018 Part B Problem 12

Understand the problem

The number of rings of order 4, up to isomorphism, is:
(a) 1
(b) 2
(c) 3
(d) 4.

Start with hints

Hint 1:

First, ask yourself how many groups are there of order 4.
the answer is simple => Z/4Z and Klein’s four group (K).

Hint 2 :
So intuitively there should be two rings with 4 elements.
  Okay, I will give you two rings or same order=> (R,+,.), (R,+,*) . see that order of the rings are same but I have changed the multiplication, and

I define a*b=0 for all a,b in R. [(R,+,*) is called zero ring]

Hint 3:
Question: Prove that (R,+,.) and (R,+,*) are not isomorphic! (easy)
So, we had (Z/4Z,+,.) and (K,+,.) as our answers. But if you change the multiplication 
to “*” then there will be 4 different rings
*upto isomorphism* right? Hence the answer is 4.
Hint 4:
Bonus Problem:
Prove that there are only two non-ismorphic p-rings(ring with p elements) upto isomorphism.

Watch the video

Connected Program at Cheenta

The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

Similar Problems

Last three digit of the last year: TIFR GS 2018 Part B Problem 9

[et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

Understand the problem

[/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]What are the last 3 digits of \(2^{2017}\)?
(a) 072
(b) 472
(c) 512
(d) 912.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR GS 2018 Part B Problem 9[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Number theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Burton[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

Start with hints

[/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

[/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]Pretty convenient problem for number theory lovers. I’m going to give some insight from Group Theory and solve with basic congruence techniques.
  1. Now, an obvious fact is \(2^{2017}\) ≅ \(0(mod8)\)
  2. so my idea is to find modul0 1000. so where does group theory lend you a hand?
  3. see that 2 is a generator of (Z/125Z)* (why?)
[/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]
So we get \(2^{2017}\) ≅ \(2^{2017}(mod125)\) ≅ \(2^{17}\) ≅ \(72 (mod 125)\)
4. Now, what is the most demanding step after this?
combining 1 and 3 we get \(2^{2017}\) ≅ \(072(mod 1000)\) [as \(125|2^{2017}-72, 8|2^{2017}-72 \)=> third digit is 0]
Hence the answer is 072.
[/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"]
Bonus Problems:
Q. find last two digits of 2^2016 like this process.

SOME NUMBER THEORIC PROBLEMS:
Q. Prove Wilson's theorem using basic group theory

Q. Prove Wilson's theorem by using Sylows theorems.

[/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

Watch the video

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Connected Program at Cheenta

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The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

Similar Problems

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Graphs in groups or Groups in graphs : TIFR GS 2018

Understand the problem

Let G be a finite group and g ∈ G an element of even order. Then we
can colour the elements of G with two colours in such a way that x and
gx have different colours for each x ∈ G.

Start with hints

Do you really need a hint? Try it first!

Hint 1
One needs to know the basics of Graph Theory to understand the solution.
Hint 2
Theorem 1 : A graph is 2-colourable iff it is bipartite.
Theorem 2: A graph is bipartite iff it has no odd-cycle.
Hint 3 : Hint 4 :
The answer is therefore True.

Watch the video

Connected Program at Cheenta

College Mathematics Program the higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

Similar Problems

Are juniors countable if seniors are?: TIFR GS 2018 Part A Problem 21

Understand the problem

A countable group can have only countably many distinct subgroups.

Start with hints

Hint 1
This is a really artistic problem.
Pre-Solution Thoughts:
Lemma: A group is finite iff the number of subgroups of a group is finite.(Check!)
The lemma is a way to understand that an infinite group will have infinite number of subgroups.

Hint 2
  • We are considering here set of all subsets of a countable group G, which is a subset of the Power Set of G.
  • Given G is countable the Power Set of G is uncountable.
  • Now we know that \(2^d\), where d is the set of natural numbers denotes the cardinality of the Power Set of G which is an uncountable set as it is bijective to the Real Numbers [Here d is not finite by the way, d is countably infinite].
  • So it is kinda intuitive that it may be uncountable.
  • First I took the group (Z,+), but we all know that the subgroups of  Z are nZ only. This doesn’t solve our purpose.
  • So naturally the next choice was the group (Q.+) whose subgroup is (Z,+).
  • While understanding the subgroups of (Q.+), the question is solved.
  • Hint 3
  • We need to understand the subgroups of (Q.+).
  • Consider any rational number q and the subgroup qZ of (Q.+) generated by q.
  • What if we take two rational numbers?
  • For simplicity check the subgroup generated by {1/2 , 1/3}.
  • Prove that the subgroup generated by {1/2 , 1/3} is (1/2.3)Z.{Observe that co-prime property of 2 and 3 is playing an important role}.
  • Hint 4
  • Now what if we take three mutually coprime natural numbers say a,b,c and see the subgroup generated by { 1/a , 1/b ,1/c}. In fact for simplicity take a,b,c to be primes.
  • Observe that it is of the form (1/a.b.c)Z.
  • Hence for every finite subset of primes, we generate a distinct subgroup of (Q,+).
  • Hence the total number of subgroups contained all the subsets of primes, which has a bijection with the Real Numbers as mentioned above.
  • So the answer is False.
  • Exercise

    Watch the video

    Connected Program at Cheenta

    The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

    Similar Problems

    Group with Quotient : TIFR GS 2018 Part A Problem 16

    Understand the problem

    Let G be a finite group with a normal subgroup H such that G/H has
    order 7. Then \(G \cong\) H × G/H.

    Start with hints

    Hint 1
    This is also an interesting question. First of all we need to understand something in general.
    If G is a finite group and H Δ G. So Consider the quotient group G/H.
    Observe the following!
    We will illustrate this by giving a simple example.
    Hint 2 Hint 3
    1. But we will give an example where the given statement is also False.
    Hint 4
    1. Prove that the finite subgroups of the group of rigid body motion are only

    Watch the video

    Connected Program at Cheenta

    The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.

    Similar Problems

    Symmetric groups of order 30: TIFR GS 2018 Part A Problem 23

    [et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

    Understand the problem

    [/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" _i="1" _address="0.0.0.1"]The permutation group \(S_{10}\) has an element of order 30.[/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 23[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Group Theory[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Easy[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Dummit and Foote[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

    Start with hints

    [/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

    [/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]
    Consider S={1,2,...,10}.\(S_{10}\) be the permutation group on S.
    What will you do if one asked for a subgroup of order 3!=6?
    [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"]
    What do you do if one asked for a subgroup of order 5?
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    The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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    Matrix additive-multiplicative :TIFR 2018 Part A Problem 19

    [et_pb_section fb_built="1" _builder_version="3.22.4" fb_built="1" _i="0" _address="0"][et_pb_row _builder_version="3.25" _i="0" _address="0.0"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.0.0"][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_padding="20px|20px|20px|20px" _i="0" _address="0.0.0.0"]

    Understand the problem

    [/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]Let A, B ∈ \(M_n(\Bbb R)\) be such that A + B = AB. Then AB = BA. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" hover_enabled="0" _i="0" _address="0.1.0.0.0"]TIFR 2018 Part A Problem 19[/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.0.1" open="off"]Linear Algebra[/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.0.2" open="off"]Medium[/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.0.3" open="off"]Linear Algebra, Hoffman and Kunze[/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

    Start with hints

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    [/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" hover_enabled="0" _i="1" _address="0.1.0.2.1"]
    We need to play around with symbols.
    Try to make (AB-BA) containing equation so that we can get the idea how to show it 0.This is a symmetric operation so remember if you use only B to get something you can also use A to get that so we have to take help of the symmetry.
    [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" hover_enabled="0" _i="2" _address="0.1.0.2.2"][/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" hover_enabled="0" _i="3" _address="0.1.0.2.3"][/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" hover_enabled="0" _i="4" _address="0.1.0.2.4"]
    U+T are linear operators on R^n and U+T=UT.Then Img(T)=Img(U) and Ker(T)=Ker(U)
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    The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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    Commutative does not commute in matrices: TIFR 2018 Part A, Problem 11

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    Understand the problem

    [/et_pb_text][et_pb_text _builder_version="3.27" text_font="Raleway||||||||" background_color="#f4f4f4" box_shadow_style="preset2" custom_margin="10px||10px" custom_padding="10px|20px|10px|20px" hover_enabled="0" _i="1" _address="0.0.0.1"]True or False: Let A, B, C ∈ M3(\(\mathbb{R}\)) be such that A commutes with B, B commutes with C and B is not a scalar matrix. Then A commutes with C. [/et_pb_text][/et_pb_column][/et_pb_row][et_pb_row _builder_version="3.25" _i="1" _address="0.1"][et_pb_column type="4_4" _builder_version="3.25" custom_padding="|||" custom_padding__hover="|||" _i="0" _address="0.1.0"][et_pb_accordion open_toggle_text_color="#0c71c3" _builder_version="3.27" toggle_font="||||||||" body_font="Raleway||||||||" text_orientation="center" custom_margin="10px||10px" hover_enabled="0" _i="0" _address="0.1.0.0"][et_pb_accordion_item title="Source of the problem" open="on" _builder_version="3.27" _i="0" _address="0.1.0.0.0" hover_enabled="0"]TATA INSTITUTE OF FUNDAMENTAL RESEARCH GS-2018 (Mathematics) [/et_pb_accordion_item][et_pb_accordion_item title="Topic" _builder_version="3.27" _i="1" _address="0.1.0.0.1" open="off" hover_enabled="0"]Linear Algebra [/et_pb_accordion_item][et_pb_accordion_item title="Difficulty Level" _builder_version="3.27" _i="2" _address="0.1.0.0.2" open="off" hover_enabled="0"]Medium [/et_pb_accordion_item][et_pb_accordion_item title="Suggested Book" _builder_version="3.27" _i="3" _address="0.1.0.0.3" open="off" hover_enabled="0"]Linear Algebra Hoffman and Kunze [/et_pb_accordion_item][/et_pb_accordion][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="1" _address="0.1.0.1"]

    Start with hints

    [/et_pb_text][et_pb_tabs active_tab_background_color="#0c71c3" inactive_tab_background_color="#000000" _builder_version="3.27" tab_text_color="#ffffff" tab_font="||||||||" background_color="#ffffff" hover_enabled="0" _i="2" _address="0.1.0.2"][et_pb_tab title="Hint 0" _builder_version="3.22.4" _i="0" _address="0.1.0.2.0"]Do you really need a hint? Try it first!

    [/et_pb_tab][et_pb_tab title="Hint 1" _builder_version="3.27" _i="1" _address="0.1.0.2.1" hover_enabled="0"]First we need to get some idea whether or not this may be true or false  . As a result we need to make some calculations. The first step to approach is always to build the expression of AC - CA & then see whether it is zero or not [/et_pb_tab][et_pb_tab title="Hint 2" _builder_version="3.27" _i="2" _address="0.1.0.2.2" hover_enabled="0"]Given AB = BA   &  BC = CB . Prove the following ! B(AC) = AC(B) & B(CA) = (CA)B and then subtract to get ( AC - CA )B =  B(AC -CA) Now this is not obvious if DB = BD & B being non- scalar matrix then D = 0 is too strong statement to be true  So this gives us idea  that it maybe false . Now to prove it false we need to construct a counter example [/et_pb_tab][et_pb_tab title="Hint 3" _builder_version="3.27" _i="3" _address="0.1.0.2.3" hover_enabled="0"]We can take approach using beautiful fact of matrices that they are transformation of spaces . Now given they are transformation of spaces and this is sort of abelian character showing up , we seek help from Groups     [/et_pb_tab][et_pb_tab title="Hint 4" _builder_version="3.27" _i="4" _address="0.1.0.2.4" hover_enabled="0"]You know why B is restricted to be non scalar because they form the centre of the \(GL_n\)(F) So we approach it in the following way If we can find a group of 3 x 3 matrices with non trivial centre . If we search for centre of groups then the only example available to us is Heisenberg Group In mathematics, the Heisenberg group    , named after Werner Heisenberg, is the group of 3×3 upper triangular matrices of the form
    under the operation of matrix multiplication. Elements a, b and c can be taken from any commutative ring with identity, often taken to be the ring of real numbers(resulting in the "continuous Heisenberg group") or the ring of integers (resulting in the "discrete Heisenberg group"). The continuous Heisenberg group arises in the description of one-dimensional quantum mechanical systems, especially in the context of the Stone–von Neumann theorem. More generally, one can consider Heisenberg groups associated to n-dimensional systems, and most generally, to any symplectic vector space. [/et_pb_tab][/et_pb_tabs][et_pb_text _builder_version="3.23.3" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="48px||48px" custom_padding="20px|20px|20px|20px" _i="3" _address="0.1.0.3"]

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    The higher mathematics program caters to advanced college and university students. It is useful for I.S.I. M.Math Entrance, GRE Math Subject Test, TIFR Ph.D. Entrance, I.I.T. JAM. The program is problem driven. We work with candidates who have a deep love for mathematics. This program is also useful for adults continuing who wish to rediscover the world of mathematics.[/et_pb_blurb][et_pb_button button_url="https://www.cheenta.com/collegeprogram/" button_text="Learn More" button_alignment="center" _builder_version="3.23.3" custom_button="on" button_bg_color="#0c71c3" button_border_color="#0c71c3" button_border_radius="0px" button_font="Raleway||||||||" button_icon="%%3%%" button_text_shadow_style="preset1" box_shadow_style="preset1" box_shadow_color="#0c71c3" background_layout="dark" _i="7" _address="0.1.0.7"][/et_pb_button][et_pb_text _builder_version="3.22.4" text_font="Raleway|300|||||||" text_text_color="#ffffff" header_font="Raleway|300|||||||" header_text_color="#e2e2e2" background_color="#0c71c3" border_radii="on|5px|5px|5px|5px" box_shadow_style="preset3" custom_margin="50px||50px" custom_padding="20px|20px|20px|20px" _i="8" _address="0.1.0.8"]

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