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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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Ex 4.2, 8 By using properties of determinants, show that: (i) |■8(1&𝑎&𝑎[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2)| = (a - b) (b - c)(c – a) Solving L.H.S |■8(1&𝑎&𝑎[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2)| Applying R1 → R1 − R2 = |■8(𝟏−𝟏&𝑎−𝑏&𝑎^2−𝑏^[email protected]&𝑏&𝑏[email protected]&𝑐&𝑐2 ) | = |■8(𝟎&(𝑎−𝑏)&(𝑎−𝑏)(𝑎+𝑏)@1&𝑏&𝑏[email protected]&𝑐&𝑐2 ) | = |■8(0(𝐚−𝐛)&(𝐚−𝐛)&(𝐚−𝒃)(a+b)@1&b&[email protected]&c&c2 ) | Taking Common (a – b) from R1 = (𝐚−𝒃) |■8(0&1&[email protected]&b&[email protected]&c&c2 ) | Applying R2 → R2 − R3 = (a−b) |■8(0&1&[email protected]𝟏−𝟏&b−c&b2−[email protected]&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&[email protected]&1&[email protected]&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.