Properties of Determinant
Question 2 Important Deleted for CBSE Board 2025 Exams
Question 3 Deleted for CBSE Board 2025 Exams
Question 4 Deleted for CBSE Board 2025 Exams
Question 5 Important Deleted for CBSE Board 2025 Exams
Question 6 Important Deleted for CBSE Board 2025 Exams
Question 7 Important Deleted for CBSE Board 2025 Exams
Question 8 (i) Important Deleted for CBSE Board 2025 Exams You are here
Question 8 (ii) Deleted for CBSE Board 2025 Exams
Question 9 Important Deleted for CBSE Board 2025 Exams
Question 10 (i) Deleted for CBSE Board 2025 Exams
Question 10 (ii) Important Deleted for CBSE Board 2025 Exams
Question 11 (i) Deleted for CBSE Board 2025 Exams
Question 11 (ii) Important Deleted for CBSE Board 2025 Exams
Question 12 Important Deleted for CBSE Board 2025 Exams
Question 13 Important Deleted for CBSE Board 2025 Exams
Question 14 Important Deleted for CBSE Board 2025 Exams
Question 15 (MCQ) Important Deleted for CBSE Board 2025 Exams
Question 16 (MCQ) Deleted for CBSE Board 2025 Exams
Properties of Determinant
Last updated at April 16, 2024 by Teachoo
Question 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