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Ex 4.2

Ex 4.2, 1
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Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 3 Deleted for CBSE Board 2023 Exams

Ex 4.2, 4 Deleted for CBSE Board 2023 Exams

Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams You are here

Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams

Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at March 30, 2023 by Teachoo

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