Solve all your doubts with Teachoo Black (new monthly pack available now!)

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

Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams You are here

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, 6 (Method 1) By using properties of determinants, show that: |■8(0&a&−b@−a&0&−c@b&c&0)| = 0 Let Δ = |■8(0&a&−b@−a&0&−c@b&c&0)| Multiply & Divide by ab = 𝑎𝑏/𝑎𝑏 |■8(0&a&−b@−a&0&−c@b&c&0)| = 1/𝑎𝑏 a . b |■8(0&a&−b@−a&0&−c@b&c&0)| Multiplying C2 by b & C3 by a = 1/𝑎𝑏 |■8(0&𝒃(𝑎)&𝒂(−𝑏)@−𝑎&𝒃(0)&𝒂(−𝑐)@𝑏&𝒃(𝑐)&𝒂(0))| = 1/𝑎𝑏 |■8(0&𝑎𝑏&−𝑏𝑎@−𝑎&0&−𝑐𝑎@𝑏&𝑏𝑐&0)| Applying C2 → C2 + C3 = 1/𝑎𝑏 |■8(0&𝑎𝑏+(−𝒃𝒂)&−𝑏𝑎@−𝑎&0+(−𝑐𝑎)&−𝑐𝑎@𝑏&𝑏𝑐+0&0)| = 1/𝑎𝑏 |■8(0&𝟎&−𝑏𝑎@−a&−ca&−ca@b&bc&0)| Expanding Determinant along R1 = 1/𝑎𝑏 (0|■8(−𝑎𝑐&−𝑐𝑎@𝑏𝑐&0)|−0|■8(−𝑎&−𝑐𝑎@𝑏&0)|−ba|■8(−𝑎&−𝑎𝑐@𝑏&𝑏𝑐)|) = 1/𝑎𝑏 (0−0−ba|■8(−𝑎&−𝑎𝑐@𝑏&𝑏𝑐)|) = 1/𝑎𝑏 (– ba ( (–a)bc – b(–ac)) ) = 1/𝑎𝑏 (–ba ( – abc + abc)) = 1/𝑎𝑏 (–ba × 0) = 1/𝑎𝑏 × 0 = 0 = R.H.S Hence proved Ex 4.2, 6 (Method 2) By using properties of determinants, show that: |■8(0&a&−b@−a&0&−c@b&c&0)| = 0 Let Δ = |■8(0&a&−b@−a&0&−c@b&c&0)| Taking (–1) common from each row Δ = (–1) (–1) (–1) |■8(0&−a&b@a&0&c@−b&−c&0)| Δ = (–1)3 |■8(0&−a&b@a&0&c@−b&−c&0)| Δ = –1 × |■8(0&−a&b@a&0&c@−b&−c&0)| Now, From (1) Δ = |■8(0&a&−b@−a&0&−c@b&c&0)| Interchanging rows & columns Δ = |■8(0&−a&b@a&0&c@−b&−c&0)| Adding (2) & (3) Δ + ∆ = –1 × |■8(0&−a&b@a&0&c@−b&−c&0)| + |■8(0&−a&b@a&0&c@−b&−c&0)| 2∆ = |■8(0&a&−b@−a&0&−c@b&c&0)| + |■8(0&−a&b@a&0&c@−b&−c&0)| 2∆ = |■8(0&0&0@0&0&0@0&0&0)| = 0 2∆ = 0 Δ = 0 Hence proved