Ex 4.2, 8 - Show using property (i) |1 a a2 1 b b2 1 c ca|

Ex 4.2, 8 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 8 - Chapter 4 Class 12 Determinants - Part 3

 

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Transcript

Question 8 By using properties of determinants, show that: (i) |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| = (a - b) (b - c)(c – a) Solving L.H.S |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| Applying R1 → R1 − R2 = |■8(𝟏−𝟏&𝑎−𝑏&𝑎^2−𝑏^2@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(𝟎&(𝑎−𝑏)&(𝑎−𝑏)(𝑎+𝑏)@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(0(𝐚−𝐛)&(𝐚−𝐛)&(𝐚−𝒃)(a+b)@1&b&b2@1&c&c2 ) | Taking Common (a – b) from R1 = (𝐚−𝒃) |■8(0&1&a+b@1&b&b2@1&c&c2 ) | Applying R2 → R2 − R3 = (a−b) |■8(0&1&a+b@𝟏−𝟏&b−c&b2−c2@1&c&c2 ) | = (a – b) |■8(0&1&a+𝑏@𝟎&b−c&(b−c)(b+c)@1&c&c2 ) | Taking common (b – c) from R2 = (a – b) (b – c) |■8(0&1&a+b@0&1&b+c@1&c&c2 ) | Expanding Determinant along C1 = (a – b) (b – c) ( 0|■8(1&𝑏+𝑐@𝑐&𝑐2)|−0|■8(1&𝑎+𝑏@𝑐&𝑐2)|+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) ( 0−0+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) (1(b + c) – 1(a + b) ) = (a – b) (b – c) (b + c – a – b) = (a – b) (b – c)(c – a) = R.H.S 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.