Ex 4.2

Ex 4.2, 1
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Ex 4.2, 2 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 3 Deleted for CBSE Board 2022 Exams

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Ex 4.2, 8 (i) Important Deleted for CBSE Board 2022 Exams You are here

Ex 4.2, 8 (ii) Deleted for CBSE Board 2022 Exams

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Ex 4.2, 10 (i) Deleted for CBSE Board 2022 Exams

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Ex 4.2, 11 (ii) Important Deleted for CBSE Board 2022 Exams

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Ex 4.2, 15 (MCQ) Important Deleted for CBSE Board 2022 Exams

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Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

Last updated at Aug. 18, 2021 by Teachoo

Ex 4.2, 8 By using properties of determinants, show that: (i) |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| = (a - b) (b - c)(c – a) Solving L.H.S |■8(1&𝑎&𝑎2@1&𝑏&𝑏2@1&𝑐&𝑐2)| Applying R1 → R1 − R2 = |■8(𝟏−𝟏&𝑎−𝑏&𝑎^2−𝑏^2@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(𝟎&(𝑎−𝑏)&(𝑎−𝑏)(𝑎+𝑏)@1&𝑏&𝑏2@1&𝑐&𝑐2 ) | = |■8(0(𝐚−𝐛)&(𝐚−𝐛)&(𝐚−𝒃)(a+b)@1&b&b2@1&c&c2 ) | Taking Common (a – b) from R1 = (𝐚−𝒃) |■8(0&1&a+b@1&b&b2@1&c&c2 ) | Applying R2 → R2 − R3 = (a−b) |■8(0&1&a+b@𝟏−𝟏&b−c&b2−c2@1&c&c2 ) | = (a – b) |■8(0&1&a+𝑏@𝟎&b−c&(b−c)(b+c)@1&c&c2 ) | Taking common (b – c) from R2 = (a – b) (b – c) |■8(0&1&a+b@0&1&b+c@1&c&c2 ) | Expanding Determinant along C1 = (a – b) (b – c) ( 0|■8(1&𝑏+𝑐@𝑐&𝑐2)|−0|■8(1&𝑎+𝑏@𝑐&𝑐2)|+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) ( 0−0+1|■8(1&𝑎+𝑏@1&𝑏+𝑐)|) = (a – b) (b – c) (1(b + c) – 1(a + b) ) = (a – b) (b – c) (b + c – a – b) = (a – b) (b – c)(c – a) = R.H.S Hence Proved