Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

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 You are here

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

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 Aug. 18, 2021 by Teachoo

Maths Crash Course - Live lectures + all videos + Real time Doubt solving!

Ex 4.2, 11 By using properties of determinants, show that: (ii) |■8(x+y+2z&x&[email protected]&y+z+2x&[email protected]&x&z+x+2y)| = 2(x + y + z)3 Solving L.H.S |■8(x+y+2z&x&[email protected]&y+z+2x&[email protected]&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&[email protected]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&[email protected]&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&−[email protected]&x&z+x+2y)| Applying R3 → R3 – R1 = 2(x+y+z)2|■8(1&𝑥&𝑦@0&1&−[email protected]𝟏−𝟏&x−𝑥&z+x+2y−y)| = 2(x+y+z)2|■8(1&𝑥&𝑦@0&1&−[email protected]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&−[email protected]&0&1)| Expanding Determinant along C1 = 2(x+y+z)3 ( 1|■8(1&−[email protected]&1)|−0|■8(𝑥&𝑦@0&1)|+0|■8(x&[email protected]&−1)|) = 2(x+y+z)3 ( 1|■8(1&−[email protected]&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