


Ex 4.2
Ex 4.2, 2 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 3 Deleted for CBSE Board 2022 Exams
Ex 4.2, 4 Deleted for CBSE Board 2022 Exams
Ex 4.2, 5 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 6 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 7 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 8 (i) Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 8 (ii) Deleted for CBSE Board 2022 Exams
Ex 4.2, 9 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 10 (i) Deleted for CBSE Board 2022 Exams
Ex 4.2, 10 (ii) Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 11 (i) Deleted for CBSE Board 2022 Exams
Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 12 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 13 Important Deleted for CBSE Board 2022 Exams You are here
Ex 4.2, 14 Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams
Ex 4.2, 16 (MCQ) Deleted for CBSE Board 2022 Exams
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