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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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Transcript

Ex 4.2, 5 Using the property of determinants and without expanding, prove that: |■8(b+c&q+r&y+z@c+a&r+p&z+x@a+b&p+q&x+y)| = 2 |■8(a&p&x@b&q&y@c&r&z)| Solving L.H.S |■8(b+c&q+r&y+z@c+a&r+p&z+x@a+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&z+x@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&z+x@(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&z+x@(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&z+x@a+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| = 2 |■8(−𝑎&−𝑝&−𝑥@c+a&r+𝑝&z+x@a+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−z@a+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−y@a+b+c&𝑝+𝑞+𝑟&x+𝑦+𝑧)| Applying R3 → R3 + R1 + R2 = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−y@a+b+c−𝑎−𝑏&𝑝+𝑞+𝑟−𝑝−𝑞&x+𝑦+𝑧−𝑥−𝑦)| = 2 |■8(−𝑎&−𝑝&−𝑥@−b&−𝑞&−y@𝑐&𝑟&𝑧)| Taking –1 Common from R1 & R3 = 2 ( –1) ( –1)|■8(𝑎&𝑝&𝑥@b&𝑞&y@𝑐&𝑟&𝑧)| = 2 |■8(𝑎&𝑝&𝑥@b&𝑞&y@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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.