Question 6 - Solving by simplifying det. - Chapter 4 Class 12 Determinants

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Question 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 Question 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