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Ex 4.2
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
Last updated at March 16, 2023 by Teachoo
Ex 4.2, 6 (Method 1) By using properties of determinants, show that: |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| = 0 Let Δ = |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| Multiply & Divide by ab = 𝑎𝑏/𝑎𝑏 |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| = 1/𝑎𝑏 a . b |■8(0&a&−[email protected]−a&0&−[email protected]&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&−[email protected]&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&−[email protected]−a&0&−[email protected]&c&0)| = 0 Let Δ = |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| Taking (–1) common from each row Δ = (–1) (–1) (–1) |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| Δ = (–1)3 |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| Δ = –1 × |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| Now, From (1) Δ = |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| Interchanging rows & columns Δ = |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| Adding (2) & (3) Δ + ∆ = –1 × |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| + |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| 2∆ = |■8(0&a&−[email protected]−a&0&−[email protected]&c&0)| + |■8(0&−a&[email protected]&0&[email protected]−b&−c&0)| 2∆ = |■8(0&0&[email protected]&0&[email protected]&0&0)| = 0 2∆ = 0 Δ = 0 Hence proved