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Ex 4.2, 5 - Prove using property of determinants |b+c q+r

Ex 4.2, 5 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 5 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.2, 5 - Chapter 4 Class 12 Determinants - Part 4

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Ex 4.2, 5 Using the property of determinants and without expanding, prove that: |■8(b+c&q+r&[email protected]+a&r+p&[email protected]+b&p+q&x+y)| = 2 |■8(a&p&[email protected]&q&[email protected]&r&z)| Solving L.H.S |■8(b+c&q+r&[email protected]+a&r+p&[email protected]+b&p+q&x+y)| Applying R3 → R3 + R2 + R1 = |■8(𝑏+𝑐&𝑞+𝑟&𝑦+𝑧@𝑐+𝑎&𝑟+𝑝&𝑧+𝑥@𝑎+𝑏+𝑐+𝑎+𝑏+𝑐&𝑝+𝑞+𝑟+𝑞+𝑞+𝑟&𝑥+𝑦+𝑧+𝑥+𝑦+𝑧)| = |■8(b+𝑐&𝑞+𝑟&𝑦+𝑧@c+a&r+𝑝&[email protected]𝟐(a+b+c)&𝟐(𝑝+𝑞+𝑟)&𝟐(x+𝑦+𝑧))| Taking 2 common from R3 = 2 |■8(b+𝑐&𝑞+𝑟&𝑦+𝑧@c+a&r+𝑝&[email protected](a+b+c)&(𝑝+𝑞+𝑟)&(x+𝑦+𝑧))| Applying R1 → R1 – R3 = 2 |■8(b+𝑐 −(a+b+c)&𝑞+𝑟 −(𝑝+𝑞+𝑟)&𝑦+𝑧 −(x+𝑦+𝑧)@c+a&r+𝑝&[email protected](a+b+c)&(𝑝+𝑞+𝑟)&(x+𝑦+𝑧))| = 2 |■8(b+𝑐−𝑎−𝑏−𝑐&𝑞+𝑟−𝑝−𝑞−𝑟&𝑦+𝑧−𝑥−𝑦−𝑧@c+a&r+𝑝&[email protected]+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| = 2 |■8(−𝑎&−𝑝&−𝑥@c+a&r+𝑝&[email protected]+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| Applying R2 → R2 – R3 = 2 |■8(−𝑎&−𝑝&−𝑥@c+a−a−b−c&r+𝑝−𝑝−𝑞−𝑟&z+x−x−y−[email protected]+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−[email protected]+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| Applying R3 → R3 + R1 + R2 = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−[email protected]+b+c−𝑎−𝑏&𝑝+𝑞+𝑟−𝑝−𝑞&x+𝑦+𝑧−𝑥−𝑦)| = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−[email protected]𝑐&𝑟&𝑧)| Taking –1 Common from R1 & R3 = 2 ( –1) ( –1)|■8(𝑎&𝑝&𝑥@b&𝑞&[email protected]𝑐&𝑟&𝑧)| = 2 |■8(𝑎&𝑝&𝑥@b&𝑞&[email protected]𝑐&𝑟&𝑧)| = R.H.S Hence Proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.