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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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Transcript

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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.