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Ex 4.2

Ex 4.2, 1
Deleted for CBSE Board 2023 Exams

Ex 4.2, 2 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 3 Deleted for CBSE Board 2023 Exams

Ex 4.2, 4 Deleted for CBSE Board 2023 Exams

Ex 4.2, 5 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 6 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 7 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 8 (i) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 8 (ii) Deleted for CBSE Board 2023 Exams

Ex 4.2, 9 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (i) Deleted for CBSE Board 2023 Exams

Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 12 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 13 Important Deleted for CBSE Board 2023 Exams You are here

Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2023 Exams

Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2023 Exams

Chapter 4 Class 12 Determinants

Serial order wise

Last updated at Jan. 22, 2020 by Teachoo

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