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Ex 4.2, 6  - Show that |0 a -b -a 0 -c b c 0| = 0 - Chapter 4 Class 12

Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 3 Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 4 Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 5 Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 6 Ex 4.2, 6 - Chapter 4 Class 12 Determinants - Part 7

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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

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.