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Ex 4.2, 13 - Using properties of determinants - Class 12

Ex 4.2, 13 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 13 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.2, 13 - Chapter 4 Class 12 Determinants - Part 4

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Ex 4.2, 13 By using properties of determinants, show that: |■8(1+a2−b2&2ab&−2b@2ab&1−a2+b2&2a@2b&−2a&1−a2−b2)| = (1 + a2+b2)3 Solving L.H.S |■8(1+a2−b2&2ab&−2b@2ab&1−a2+b2&2a@2b&−2a&1−a2−b2)| Applying R1 → R1 + bR3 = |■8(1+a2−b2+𝑏(2𝑏)&2ab+b(−2a)&−2b+𝑏(1−𝑎2−𝑏2)@2ab&1−a2−b2&2a@2b&−2a&1−a2−b2)| = |■8(1+a2−b2+2𝑏^2&2ab−2ab&−2b+𝑏−𝑏𝑎^2−𝑏^3@2ab&1−a2−b2&2a@2b&−2a&1−a2−b2)| = |■8(𝟏+𝐚𝟐+𝐛𝟐&0&−𝑏(𝟏+𝐚𝟐+𝐛𝟐)@2ab&1−a2+b2&2a@2b&−2a&1−a2−b2)| Taking Common (1+𝑎2+𝑏2) from R1 = (1+𝑎2+𝑏2) |■8(1&0&−b@2ab&1−a2+b2&2a@2b&−2a&1−a2−b2)| Applying R2 → R2 − aR3 = (1+𝑎2+𝑏2) |■8(1&0&−b@2ab−𝑎(2𝑏)&1−a2+b2−𝑎(−2𝑎) &2a−a(1−a2−b2) @2b&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−b@2ab−2𝑎𝑏&1−a2+b2+2𝑎2&2a−a+a3+ab2@2b&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−b@0&1+b2+𝑎2&a+a3+ab2@2b&−2a&1−a2−b2)| = (1+𝑎2+𝑏2) |■8(1&0&−b@0&𝟏+𝒂𝟐+𝒃𝟐&a(𝟏+𝐚𝟐+𝒃𝟐) @2b&−2a&1−a2−b2)| Taking Common (1+𝑎2+𝑏2) from R2 = (1+𝑎2+𝑏2)2 |■8(1&0&−b@0&1&a@2b&−2a&1−a2−b2)| = (1+𝑎2+𝑏2)2 ( 1|■8(1&𝑎@−2𝑎&1−a2−b2)|−0|■8(0&x@−2𝑎&1−a2−b2)|−𝑏|■8(0&1@2b&−2𝑏)|) = (1+𝑎2+𝑏2)2 ( 1|■8(1&𝑎@−2𝑎&1−a2−b2)|−0−𝑏|■8(0&1@2b&−2𝑏)|) = (1+𝑎2+𝑏2)2 (1(1 – a2 – b2) + 2a2) – b (0 – 2b)) = (1+𝑎2+𝑏2)2 (1 – a2 – b2 + 2a2 + 2b2) = (1+𝑎2+𝑏2)2 (1 + a2 + b2) = (1+𝑎2+𝑏2)3 = R.H.S Hence proved

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Davneet Singh

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.