# Ex 4.2, 11 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 22, 2020 by Teachoo

Last updated at Jan. 22, 2020 by Teachoo

Transcript

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 Ex 4.2, 11 By using properties of determinants, show that: (ii) |■8(x+y+2z&x&y@z&y+z+2x&y@z&x&z+x+2y)| = 2(x + y + z)3 Solving L.H.S |■8(x+y+2z&x&y@z&y+z+2x&y@z&x&z+x+2y)| Applying C1 → C1 + C2 + C3 = |■8(𝑥+𝑦+2𝑧+𝑥+𝑦&𝑥&𝑦@𝑧+𝑦+𝑧+2𝑥+𝑦&𝑦+𝑧+2𝑥&𝑦@𝑧+𝑥+𝑧+𝑥+2𝑦&𝑥&𝑧+𝑥+2𝑦)| = |■8(𝟐(𝒙+𝒚+𝒛)&𝑥&𝑦@𝟐(𝒙+𝒚+𝒛)&y+𝑧+2𝑥&y@𝟐(𝒙+𝒚+𝒛)&x&z+x+2y)| Taking common 2(𝑥+𝑦+𝑧) from C1 = 𝟐(𝐱+𝐲+𝐳) |■8(1&𝑥&𝑦@1&y+𝑧+2𝑥&y@1&x&z+x+2y)| Applying R2 → R2 – R3 = 2(x+y+z)|■8(1&𝑥&𝑦@𝟏−𝟏&y+𝑧+2𝑥−𝑥&y−(𝑧+𝑥+2𝑦)@1&x&z+x+2y)| = 2(x+y+z)|■8(1&𝑥&𝑦@𝟎&𝑥+𝑦+𝑧&−𝑥−𝑦−𝑧@1&x&z+x+2y)| = 2(x+y+z)|■8(1&𝑥&𝑦@0&(𝒙+𝒚+𝒛)&−(𝒙+𝒚+𝒛)@1&x&z+x+2y)| Taking common (𝑥+𝑦+𝑧) from 2nd Row = 2(x+y+z)(x+y+z)|■8(1&𝑥&𝑦@0&1&−1@1&x&z+x+2y)| Applying R3 → R3 – R1 = 2(x+y+z)2|■8(1&𝑥&𝑦@0&1&−1@𝟏−𝟏&x−𝑥&z+x+2y−y)| = 2(x+y+z)2|■8(1&𝑥&𝑦@0&1&−1@𝟎&0&x+y+z)| Taking common (𝑥+𝑦+𝑧) Common from 3rd Row = 2(x+y+z)2(x+y+z)|■8(1&𝑥&𝑦@0&1&−1@0&0&1)| Expanding Determinant along C1 = 2(x+y+z)3 ( 1|■8(1&−1@0&1)|−0|■8(𝑥&𝑦@0&1)|+0|■8(x&y@1&−1)|) = 2(x+y+z)3 ( 1|■8(1&−1@0&1)|−0+0) = 2(x+y+z)3 (1(1−0)−𝑥(0)+𝑦(0)) = 2(x+y+z)3 (1) = 2(x+y+z)3 = R.H.S Hence proved

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 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 9 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 10 Deleted for CBSE Board 2022 Exams

Ex 4.2, 11 Important Deleted for CBSE Board 2022 Exams You are here

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 Important Deleted for CBSE Board 2022 Exams

Ex 4.2, 16 Important Deleted for CBSE Board 2022 Exams

Chapter 4 Class 12 Determinants (Term 1)

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.