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Ex 4.2

Ex 4.2, 1
Deleted for CBSE Board 2023 Exams

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 Jan. 22, 2020 by Teachoo

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