# Ex 4.2, 6 - Chapter 4 Class 12 Determinants

Last updated at June 15, 2018 by Teachoo

Last updated at June 15, 2018 by Teachoo

Transcript

Ex 4.2, 6 (Method 1) By using properties of determinants, show that: 0a−b−a0−cbc0 = 0 Let Δ = 0a−b−a0−cbc0 Multiply & Divide by ab = 𝑎𝑏𝑎𝑏 0a−b−a0−cbc0 = 1𝑎𝑏 a . b 0a−b−a0−cbc0 Multiplying C2 by b & C3 by a = 1𝑎𝑏 0𝐛(a)𝐚(−b)−a𝐛(0)𝐚(−c)b𝐛(c)𝐚(0) = 1𝑎𝑏 0ab−𝑏𝑎−a0−cabbc0 Applying C2 → C2 + C3 = 1𝑎𝑏 0ab+(−𝐛𝐚)−𝑏𝑎−a0+(−ca)−cabbc+00 = 1𝑎𝑏 0𝟎−𝑏𝑎−a−ca−cabbc0 Expanding Determinant along R1 = 1𝑎𝑏 0 −𝑎𝑐−𝑐𝑎𝑏𝑐0−0 −𝑎−𝑐𝑎𝑏0−ba −𝑎−𝑎𝑐𝑏𝑏𝑐 = 1𝑎𝑏 0−0−ba −𝑎−𝑎𝑐𝑏𝑏𝑐 = 1𝑎𝑏 (– ba ( (–a)bc – b(–ac)) ) = 1𝑎𝑏 ( – ba ( – abc + abc)) = 1𝑎𝑏 ( – ba (0)) = 1𝑎𝑏 × 0 = 0 = R.H.S Hence proved Ex 4.2, 6 (Method 2) By using properties of determinants, show that: 0a−b−a0−cbc0 = 0 Let Δ = 0a−b−a0−cbc0 Taking ( – 1) common from each row Δ = ( –1) ( –1) ( –1) 0−aba0c−b−c0 Δ = ( – 1)3 0−aba0c−b−c0 Δ = – 0−aba0c−b−c0 From (1) Δ = 0a−b−a0−cbc0 Interchanging rows & columns Δ = 0−aba0c−b−c0 Adding (2) & (3) Δ + ∆ = – 0−aba0c−b−c0 + 0−aba0c−b−c0 2∆ = 0a−b−a0−cbc0 + 0−aba0c−b−c0 = 000000000 = 0 2∆ = 0 ⇒ Δ = 0 Hence proved

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.