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Navjot Singh Class 12 Tuition trainer in Gurgaon

Navjot Singh

Mathematics Tutor - Postgraduate - IIT Delhi

Sector 54 Suncity, Gurgaon, India - 122003.

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Referral Discount: Get ₹ 250 off when you make a payment to start classes. Get started by Booking a Demo.

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Overview

I have 3 years of experience in teaching. I have great hold of all the core concepts of Mathematics and Computer Science and I'm able to make basics of my students extremely strong. I create a very good and an optimistic environment for the child to study in.

Languages Spoken

English

Hindi

Punjabi

Education

KIRORIMAL COLLEGE, DELHI UNIVERSITY 2014

Bachelor of Science (B.Sc.)

IIT DELHI 2016

Master of Science (M.Sc.)

Address

Sector 54 Suncity, Gurgaon, India - 122003

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Teaches

Class 12 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

I am postgraduate in Mathematics from IIT Delhi and keen to make students learn mathematics. My students love learning Mathematics with me.

Class 11 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 11 Tuition

I am postgraduate in Mathematics from IIT Delhi and keen to make students learn mathematics. My students love learning Mathematics with me.

Class 10 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 10 Tuition

I am postgraduate in Mathematics from IIT Delhi. I love to make students learn Mathematics. I've always received a positive feedback about my teaching and wish to continue the same.

Class 9 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 9 Tuition

I am postgraduate in Mathematics from IIT Delhi. I love to make students learn Mathematics. I've always received a positive feedback about my teaching and wish to continue the same.

Class 8 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Reviews

No Reviews yet!

FAQs

1. Which school boards of Class 12 do you teach for?

CBSE

2. Have you ever taught in any School or College?

No

3. Which classes do you teach?

I teach Class 10 Tuition, Class 11 Tuition, Class 12 Tuition, Class 8 Tuition and Class 9 Tuition Classes.

4. Do you provide a demo class?

Yes, I provide a free demo class.

5. How many years of experience do you have?

I have been teaching for 3 years.

Answers by Navjot Singh (3)

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

(i) Step 1 R1 = R1 + R2 + R3 Step 2 (5x+4) common in R1, bring that out, you are left with all 1s in R1 Applying C2 → C2 − C1, C3 → C3 − C1, we have: Step 3 R2 = R2 - 2x(R1) R3 = R3 - 2x(R1) Now matrix is : (5x+4) =(5x+4)(4-x)(4-x) Expanding along C3, we have: Hence, the given... ...more

(i) Step 1 

R1 = R1 + R2 + R3

Step 2

(5x+4) common in R1, bring that out, you are left with all 1s in R1

Applying C2 → C2 − C1, C3 → C3 − C1, we have:

Step 3

R2 = R2 - 2x(R1)

R3 = R3 - 2x(R1)

Now matrix is :

(5x+4)[(1,1,1),(0,4-x,0),(0,0,4-x)]

=(5x+4)(4-x)(4-x)

Expanding along C3, we have:

Hence, the given result is proved.

 

 

(ii) Here you need to subtract y(R1) from R2 and R3.

Applying R1 → R1 + R+ R3, we have:

Applying C2 → C2 − Cand C3 → C3 − C1, we have:

Expanding along C3, we have:

Answers 2 Comments
Dislike Bookmark

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

Teaches

Class 12 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 12 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 12 Tuition

I am postgraduate in Mathematics from IIT Delhi and keen to make students learn mathematics. My students love learning Mathematics with me.

Class 11 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 11 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 11 Tuition

I am postgraduate in Mathematics from IIT Delhi and keen to make students learn mathematics. My students love learning Mathematics with me.

Class 10 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 10 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 10 Tuition

I am postgraduate in Mathematics from IIT Delhi. I love to make students learn Mathematics. I've always received a positive feedback about my teaching and wish to continue the same.

Class 9 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 9 Tuition

3

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

Teaching Experience in detail in Class 9 Tuition

I am postgraduate in Mathematics from IIT Delhi. I love to make students learn Mathematics. I've always received a positive feedback about my teaching and wish to continue the same.

Class 8 Tuition
2 Students

Class Location

Online (video chat via skype, google hangout etc)

Student's Home

Tutor's Home

Years of Experience in Class 8 Tuition

4

Board

CBSE

CBSE Subjects taught

Mathematics

Taught in School or College

No

No Reviews yet!

Answers by Navjot Singh (3)

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

(i) Step 1 R1 = R1 + R2 + R3 Step 2 (5x+4) common in R1, bring that out, you are left with all 1s in R1 Applying C2 → C2 − C1, C3 → C3 − C1, we have: Step 3 R2 = R2 - 2x(R1) R3 = R3 - 2x(R1) Now matrix is : (5x+4) =(5x+4)(4-x)(4-x) Expanding along C3, we have: Hence, the given... ...more

(i) Step 1 

R1 = R1 + R2 + R3

Step 2

(5x+4) common in R1, bring that out, you are left with all 1s in R1

Applying C2 → C2 − C1, C3 → C3 − C1, we have:

Step 3

R2 = R2 - 2x(R1)

R3 = R3 - 2x(R1)

Now matrix is :

(5x+4)[(1,1,1),(0,4-x,0),(0,0,4-x)]

=(5x+4)(4-x)(4-x)

Expanding along C3, we have:

Hence, the given result is proved.

 

 

(ii) Here you need to subtract y(R1) from R2 and R3.

Applying R1 → R1 + R+ R3, we have:

Applying C2 → C2 − Cand C3 → C3 − C1, we have:

Expanding along C3, we have:

Answers 2 Comments
Dislike Bookmark

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

Answered on 03/09/2019 Learn CBSE/Class 12/Mathematics/Determinants/NCERT Solutions/Exercise 4.2

Navjot Singh describes himself as Mathematics Tutor - Postgraduate - IIT Delhi. He conducts classes in Class 10 Tuition, Class 11 Tuition and Class 12 Tuition. Navjot is located in Sector 54 Suncity, Gurgaon. Navjot takes at students Home and Regular Classes- at his Home. He has 4 years of teaching experience . Navjot has completed Bachelor of Science (B.Sc.) from KIRORIMAL COLLEGE, DELHI UNIVERSITY in 2014 and Master of Science (M.Sc.) from IIT DELHI in 2016. HeĀ is well versed in English, Hindi and Punjabi.

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