# Ex 4.2, 8

Last updated at March 11, 2017 by Teachoo

Last updated at March 11, 2017 by Teachoo

Transcript

Ex 4.2, 8 By using properties of determinants, show that: (i) 1𝑎𝑎21𝑏𝑏21𝑐𝑐2 = (a - b) (b - c)(c – a) Taking L.H.S 1𝑎𝑎21𝑏𝑏21𝑐𝑐2 Applying R1 → R1 − R2 = 𝟏−𝟏𝑎−𝑏 𝑎2− 𝑏21𝑏𝑏21𝑐𝑐2 = 𝟎(𝑎−𝑏)(𝑎−𝑏)(𝑎+𝑏)1𝑏𝑏21𝑐𝑐2 = 0(𝐚−𝐛)(𝐚−𝐛) 𝐚−𝒃(a+b)1bb21cc2 Taking Common (a – b) from R1 = 𝐚−𝒃 01a+b1bb21cc2 Applying R2 → R2 − R3 = a−b 01a+b𝟏−𝟏b−cb2−c21cc2 = (a – b) 01a+𝑏𝟎b−c(b−c)(b+c)1cc2 Taking common (b – c) from R2 = (a – b) (b – c) 01a+b01b+c1cc2 Expanding Determinant along C1 = (a – b) (b – c) 0 1𝑏+𝑐𝑐𝑐2−0 1𝑎+𝑏𝑐𝑐2+1 1𝑎+𝑏1𝑏+𝑐 = (a – b) (b – c) 0−0+1 1𝑎+𝑏1𝑏+𝑐 = (a – b) (b – c) (1(b + c) – 1(a + b) ) = (a – b) (b – c) (b + c – a – b) = (a – b) (b – c)(c – a) = R.H.S Hence Proved Ex 4.2, 8 By using properties of determinants, show that: (ii) 111abca3b3c3 = (a – b) (b – c) (c – a) (a + b + c) Taking L.H.S 111abca3b3c3 Applying C1 → C1 − C2 = 𝟏−𝟏11a−bbc𝐚𝟑 −𝐛𝟑b3 c3 = 𝟎11a−bbc(𝐚 −𝐛)(𝐚𝟐+𝐛𝟐+𝐚𝐛) b3c3 = 011𝐚−𝐛bc(𝐚 −𝐛)(a2+b2+ab) b3c3 Taking Common (a – b) from C1 = (a – b) 0111bc(a2+b2+ab)b3c3 Applying C2 → C2 − C3 = (a – b) 0𝟏−𝟏11b−cc(a2+b2+ab)b3−c3c3 = (a – b) 0𝟎11b−cc(a2+b2+ab) b−c(b2+c2+bc)c3 Taking Common (b – c) from C2 = (a – b) (b – c) 00111ca2+b2+abb2+c2+bcc3 Expanding determinant along R1 = (a – b) (b – c) 0 1𝑐𝑏2+𝑐2+𝑏𝑐𝑐3−0 11𝑎2+𝑏2+𝑎𝑏𝑐3+1 11𝑎2+𝑏2+𝑎𝑏𝑏2+𝑐2+𝑏𝑐 = (a – b) (b – c) 0−0+1 11𝑎2+𝑏2+𝑎𝑏𝑏2+𝑐2+𝑏𝑐 = (a – b) (b – c) (1((b2 + c2 + bc) – (a2 + b2 + ab)) = (a – b) (b – c) (b2 + c2 + bc – a2 – b2 – ab) = (a – b) (b – c) (c2 – a2 + bc – ab) = (a – b) (b – c) ((c – a) (c + a) + b (c – a)) = (a – b) (b – c) ((c – a) (c + a + b)) = (a – b) (b – c) ((c – a) (a + b + c)) = 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 You are here

Ex 4.2, 11 Important

Ex 4.2, 12 Important

Ex 4.2, 13 Important

Ex 4.2, 14 Important

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

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.