Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Properties of Determinant
Question 2 Important Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Important Deleted for CBSE Board 2024 Exams
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 (i) Important Deleted for CBSE Board 2024 Exams
Question 8 (ii) Deleted for CBSE Board 2024 Exams
Question 9 Important Deleted for CBSE Board 2024 Exams
Question 10 (i) Deleted for CBSE Board 2024 Exams
Question 10 (ii) Important Deleted for CBSE Board 2024 Exams
Question 11 (i) Deleted for CBSE Board 2024 Exams
Question 11 (ii) Important Deleted for CBSE Board 2024 Exams You are here
Question 12 Important Deleted for CBSE Board 2024 Exams
Question 13 Important Deleted for CBSE Board 2024 Exams
Question 14 Important Deleted for CBSE Board 2024 Exams
Question 15 (MCQ) Important Deleted for CBSE Board 2024 Exams
Question 16 (MCQ) Deleted for CBSE Board 2024 Exams
Properties of Determinant
Last updated at May 29, 2023 by Teachoo
Question 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