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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

Ex 4.2, 14 Important Deleted for CBSE Board 2023 Exams You are here

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, 14 By using properties of determinants, show that: |■8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |■8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying & Dividing by abc = 𝒂𝒃𝒄/𝒂𝒃𝒄 |■8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 → aR1 , R2 → bR3 , R3 → bR3 ) = 1/𝑎𝑏𝑐 |■8(𝒂(a2+1)&𝒂(ab)&𝒂(ac)@𝒃(ab)&𝐛(b2+1)&𝒃(bc)@𝐜(ca)&𝒄(cb)&𝐜(c2+1))| = 1/𝑎𝑏𝑐 |■8(a3+a&𝑎2b&𝑎2c@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| Applying R1 → R1 + R2 + R3 = 1/𝑎𝑏𝑐 |■8(a3+a+𝑎𝑏2+𝑐2𝑎&𝑎2b+b3+b+c2b&𝑎2c+b2c+c3+c@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| = 1/𝑎𝑏𝑐 |■8(a(𝐚𝟐+𝟏+𝒃𝟐+𝒄𝟐)&𝑏(𝒂𝟐+𝐛𝟐+𝟏+𝐜𝟐)&𝑐(𝒂𝟐+𝐛𝟐+𝐜𝟐"+1" )@ab2&b3+b&𝑏2c@c2a&𝑐2b&c3+c)| Taking (1+𝑎2+𝑏2+𝑐2) common from 1st Row = ((𝟏 + 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐))/𝑎𝑏𝑐 |■8(a&𝑏&𝑐@ab2&b3+b&𝑏2c@c2a&𝑐2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = 𝒂𝒃𝒄 ( (1 + 𝑎2 + 𝑏2 + 𝑐2))/𝑎𝑏𝑐 |■8(1&1&1@b2&b3+1&𝑏2@c2&𝑐2&c2+1)| Applying C1 → C1 − C2 = (1+𝑎2+𝑏2+𝑐2) |■8(𝟏−𝟏&1&1@b2−𝑏2−1&b2+1&𝑏2@c2−c2&𝑐2&c2+1)| = (1+𝑎2+𝑏2+𝑐2) |■8(𝟎&1&1@−1&b2+1&𝑏2@0&𝑐2&c2+1)| Applying C2 → C2 − C3 = (1+𝑎2+𝑏2+𝑐2) |■8(0&𝟏−𝟏&1@−1&b2+1−𝑐2&𝑏2@0&𝑐2−𝑐2−1&c2+1)| = (1+𝑎2+𝑏2+𝑐2) |■8(0&𝟎&1@−1&1&𝑏2@0&−1&c2+1)| Expanding along R1 = (1+𝑎2+𝑏2+𝑐2)(1|■8(1&𝑏2@−1&𝑐2+1)|" – 0" |■8(−1&𝐶2@0&𝑐2+1)|" + 0" |■8(1&1@0&−1)|) = (1 + a2 + b1 + c2) (0 – 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S Hence proved