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

Ex 4.2, 8 - Chapter 4 Class 12 Determinants - Part 5
Ex 4.2, 8 - Chapter 4 Class 12 Determinants - Part 6 Ex 4.2, 8 - Chapter 4 Class 12 Determinants - Part 7

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Ex 4.2, 8 By using properties of determinants, show that: (ii) |■8(1&1&[email protected]&b&[email protected]&b3&c3)| = (a – b) (b – c) (c – a) (a + b + c) Solving L.H.S |■8(1&1&[email protected]&b&[email protected]&b3&c3)| Applying C1 → C1 − C2 = |■8(𝟏−𝟏&1&[email protected]−b&b&[email protected]𝐚𝟑 −𝐛𝟑&b3 &c3)| = |■8(𝟎&1&[email protected]−b&b&[email protected](𝐚 −𝐛)(𝐚𝟐+𝐛𝟐+𝐚𝐛) &b3&c3)| = |■8(0&1&[email protected]𝐚−𝐛&b&[email protected](𝐚 −𝐛)(a2+b2+ab) &b3&c3)| Taking Common (a – b) from C1 = (a – b) |■8(0&1&[email protected]&b&[email protected](a2+b2+ab)&b3&c3)| Applying C2 → C2 − C3 = (a – b) |■8(0&𝟏−𝟏&[email protected]&b−c&[email protected](a2+b2+ab)&b3−c3&c3)| (x3 – y3 = (x – y)(x2 + y2 +xy)) = (a – b) |■8(0&𝟎&[email protected]&b−c&[email protected](a2+b2+ab)&(b−c)(b2+c2+bc)&c3)| Taking Common (b – c) from C2 = (a – b) (b – c) |■8(0&0&[email protected]&1&[email protected]+b2+ab&b2+c2+bc&c3)| Expanding determinant along R1 = (a – b) (b – c) (0|■8(1&𝑐@𝑏2+𝑐2+𝑏𝑐&𝑐3)|−0|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑐3)|+1|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑏2+𝑐2+𝑏𝑐)|) = (a – b) (b – c) (0−0+1|■8(1&[email protected]𝑎2+𝑏2+𝑎𝑏&𝑏2+𝑐2+𝑏𝑐)|) = (a – b) (b – c) (1((b2 + c2 + bc) – (a2 + b2 + ab)) = (a – b) (b – c) (b2 + c2 + bc – a2 – b2 – ab) = (a – b) (b – c) (c2 – a2 + bc – ab) = (a – b) (b – c) ((c – a) (c + a) + b (c – a)) = (a – b) (b – c) ((c – a) (c + a + b)) = (a – b) (b – c) ((c – a) (a + b + c)) = 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.