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Ex 4.2

Ex 4.2, 1
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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 March 30, 2023 by Teachoo

Ex 4.2, 14 By using properties of determinants, show that: |■8(a2+1&ab&[email protected]&b2+1&[email protected]&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |■8(a2+1&ab&[email protected]&b2+1&[email protected]&cb&c2+1)| Multiplying & Dividing by abc = 𝒂𝒃𝒄/𝒂𝒃𝒄 |■8(a2+1&ab&[email protected]&b2+1&[email protected]&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&𝑎[email protected]&b3+b&𝑏[email protected]&𝑐2b&c3+c)| Applying R1 → R1 + R2 + R3 = 1/𝑎𝑏𝑐 |■8(a3+a+𝑎𝑏2+𝑐2𝑎&𝑎2b+b3+b+c2b&𝑎[email protected]&b3+b&𝑏[email protected]&𝑐2b&c3+c)| = 1/𝑎𝑏𝑐 |■8(a(𝐚𝟐+𝟏+𝒃𝟐+𝒄𝟐)&𝑏(𝒂𝟐+𝐛𝟐+𝟏+𝐜𝟐)&𝑐(𝒂𝟐+𝐛𝟐+𝐜𝟐"+1" )@ab2&b3+b&𝑏[email protected]&𝑐2b&c3+c)| Taking (1+𝑎2+𝑏2+𝑐2) common from 1st Row = ((𝟏 + 𝒂𝟐 + 𝒃𝟐 + 𝒄𝟐))/𝑎𝑏𝑐 |■8(a&𝑏&𝑐@ab2&b3+b&𝑏[email protected]&𝑐2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = 𝒂𝒃𝒄 ( (1 + 𝑎2 + 𝑏2 + 𝑐2))/𝑎𝑏𝑐 |■8(1&1&[email protected]&b3+1&𝑏[email protected]&𝑐2&c2+1)| Applying C1 → C1 − C2 = (1+𝑎2+𝑏2+𝑐2) |■8(𝟏−𝟏&1&[email protected]−𝑏2−1&b2+1&𝑏[email protected]−c2&𝑐2&c2+1)| = (1+𝑎2+𝑏2+𝑐2) |■8(𝟎&1&[email protected]−1&b2+1&𝑏[email protected]&𝑐2&c2+1)| Applying C2 → C2 − C3 = (1+𝑎2+𝑏2+𝑐2) |■8(0&𝟏−𝟏&[email protected]−1&b2+1−𝑐2&𝑏[email protected]&𝑐2−𝑐2−1&c2+1)| = (1+𝑎2+𝑏2+𝑐2) |■8(0&𝟎&[email protected]−1&1&𝑏[email protected]&−1&c2+1)| Expanding along R1 = (1+𝑎2+𝑏2+𝑐2)(1|■8(1&𝑏[email protected]−1&𝑐2+1)|" – 0" |■8(−1&𝐶[email protected]&𝑐2+1)|" + 0" |■8(1&[email protected]&−1)|) = (1 + a2 + b1 + c2) (0 – 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = R.H.S Hence proved