Ex 4.2

Ex 4.2, 1
Deleted for CBSE Board 2022 Exams

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 You are here

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

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

Chapter 4 Class 12 Determinants (Term 1)

Serial order wise

Ex 4.2, 11 By using properties of determinants, show that: (i) |■8(a−b−c&2a&2a@2b&b−c−a&2b@2c&2c&c−a−b)| = (a + b + c)3 Solving L.H.S |■8(a−b−c&2a&2a@2b&b−c−a&2b@2c&2c&c−a−b)| Applying R1 → R1 + R2 + R3 = |■8(a−b−c+2𝑏+2𝑐&2a+b−c−a+2c&2a+2𝑏+𝑐−𝑎−𝑏@2b&b−c−a&2b@2c&2c&c−a−b)| = |■8(𝐚+𝐛+𝐜&𝐚+𝐛+𝐜&𝐚+𝒃+𝒄@2b&b−c−a&2b@2c&2c&c−a−b)| Taking common (a + b + c) from R1 = (a+b+c) |■8(1&1&1@2b&𝑏−𝑐−𝑎&2𝑏@2c&2𝑐&c−a−b)| Applying C1 → C1 – C2 = (a+b+c) |■8(𝟏−𝟏&1&1@2b−(b−c−a)&b−c−a&2𝑏@2c−2c&2𝑐&𝑐−𝑎−𝑏)| = (a+b+c) |■8(𝟎&1&1@b+c+a&b−c−a&2𝑏@0&2𝑐&𝑐−𝑎−𝑏)| Applying C2 → C2 – C3 = (𝑎+𝑏+𝑐)|■8(0&𝟏−𝟏&1@𝑏+𝑐+𝑎&𝑏−𝑐−𝑎−2𝑏&2𝑏@0&2𝑐−(𝑐−𝑎−𝑏)&𝑐−𝑎−𝑏)| = (𝑎+𝑏+𝑐)|■8(0&𝟎&1@𝑏+𝑐+𝑎&−𝑏−𝑐−𝑎&2𝑏@0&𝑎+𝑏+𝑐&𝑐−𝑎−𝑏)| Taking (a + b + c) common from C1 = (𝑎+𝑏+𝑐)(𝑎+𝑏+𝑐)|■8(0&0&1@1&−𝑏−𝑐−𝑎&2𝑏@0&(𝑎+𝑏+𝑐)&𝑐−𝑎−𝑏)| = (𝑎+𝑏+𝑐)2 |■8(0&0&1@1&−(𝒂+𝒃+𝒄)&2𝑏@0&(𝒂+𝒃+𝒄)&𝑐−𝑎−𝑏)| Taking common (a + b + c) from C2 = (𝑎+𝑏+𝑐)2(a+b+c) |■8(0&0&1@1&−1&2𝑏@0&1&𝑐−𝑎−𝑏)| Expanding determinant along R1 = (a + b + c)3 ( 0|■8(−1&2b@1&c−a−b)|−0|■8(1&2b@0&c−a−b)|+1|■8(1&−1@0&1)|) = (a + b + c)3 (0−0+1(1−0)) = (a + b + c)3 (1) = (a + b + c)3 Hence proved