# Ex 4.2, 14

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 4.2, 14 By using properties of determinants, show that: a2+1 ab ac ab b2+1 bc ca cb c2+1 = 1 + a2 + b2 + c2 Taking L.H.S a2+1 ab ac ab b2+1 bc ca cb c2+1 Multiplying & Dividing by abc = a2+1 ab ac ab b2+1 bc ca cb c2+1 Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 aR1 , R2 bR3 , R3 bR3 ) = 1 (a2+1) (ab) (ac) (ab) (b2+1) (bc) (ca) (cb) (c2+1) = 1 a3+a 2b 2c ab2 b3+b 2c c2a 2b c3+c Applying R1 R1 + R2 + R3 = 1 a3+a+ 2+ 2 2b+b3+b+c2b 2c+b2c+c3+c ab2 b3+b 2c c2a 2b c3+c = 1 a( + + + ) ( + + + ) ( + + +1) ab2 b3+b 2c c2a 2b c3+c Taking (1+ 2+ 2+ 2) common from 1st Row = ( + + + ) a ab2 b3+b 2c c2a 2b c(c3+1) Taking a common from C1 ,b from C2 & c from C3 . = (1+ 2+ 2+ 2) 1 1 1 b2 b3+1 2 c2 2 c2+1 Applying C1 C1 C2 = (1+ 2+ 2+ 2) 1 1 b2 2 1 b2+1 2 c2 c2 2 c2+1 = (1+ 2+ 2+ 2) 1 1 1 b2+1 2 0 2 c2+1 Applying C2 C2 C3 = (1+ 2+ 2+ 2) 0 1 1 b2+1 2 2 0 2 2 1 c2+1 = (1+ 2+ 2+ 2) 0 1 1 1 2 0 1 c2+1 Expanding along R1 = (1+ 2+ 2+ 2) 1 1 2 1 2+1 0 1 2 0 2+1 + 0 1 1 0 1 = (1 + a2 + b1 + c2) (0 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S = Hence proved

Ex 4.1, 7
Important

Example 14 Important

Example 15 Important

Example 16 Important

Ex 4.2, 7 Important

Ex 4.2, 8 Important

Ex 4.2, 11 Important

Ex 4.2, 12 Important

Ex 4.2, 13 Important

Ex 4.2, 14 Important You are here

Ex 4.2, 15 Important

Example 18 Important

Ex 4.3, 2 Important

Ex 4.3, 3 Important

Example 24 Important

Example 26 Important

Ex 4.5, 10 Important

Ex 4.5, 15 Important

Ex 4.5, 18 Important

Ex 4.6, 13 Important

Ex 4.6, 15 Important

Ex 4.6, 16 Important

Example 32 Important

Example 34 Important

Misc. 2 Important

Misc 11 Important

Misc. 15 Important

Misc. 16 Important

Misc. 19 Important

Class 12

Important Question for exams Class 12

- Chapter 1 Class 12 Relation and Functions
- Chapter 2 Class 12 Inverse Trigonometric Functions
- Chapter 3 Class 12 Matrices
- Chapter 4 Class 12 Determinants
- Chapter 5 Class 12 Continuity and Differentiability
- Chapter 6 Class 12 Application of Derivatives
- Chapter 7 Class 12 Integrals
- Chapter 8 Class 12 Application of Integrals
- Chapter 9 Class 12 Differential Equations
- Chapter 10 Class 12 Vector Algebra
- Chapter 11 Class 12 Three Dimensional Geometry
- Chapter 12 Class 12 Linear Programming
- Chapter 13 Class 12 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.