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 You are here
Question 11 (ii) Important Deleted for CBSE Board 2024 Exams
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: (i) |■8(a−b−c&2a&[email protected]&b−c−a&[email protected]&2c&c−a−b)| = (a + b + c)3 Solving L.H.S |■8(a−b−c&2a&[email protected]&b−c−a&[email protected]&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&[email protected]&2c&c−a−b)| = |■8(𝐚+𝐛+𝐜&𝐚+𝐛+𝐜&𝐚+𝒃+𝒄@2b&b−c−a&[email protected]&2c&c−a−b)| Taking common (a + b + c) from R1 = (a+b+c) |■8(1&1&[email protected]&𝑏−𝑐−𝑎&2𝑏@2c&2𝑐&c−a−b)| Applying C1 → C1 – C2 = (a+b+c) |■8(𝟏−𝟏&1&[email protected]−(b−c−a)&b−c−a&2𝑏@2c−2c&2𝑐&𝑐−𝑎−𝑏)| = (a+b+c) |■8(𝟎&1&[email protected]+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&[email protected]&−𝑏−𝑐−𝑎&2𝑏@0&(𝑎+𝑏+𝑐)&𝑐−𝑎−𝑏)| = (𝑎+𝑏+𝑐)2 |■8(0&0&[email protected]&−(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&[email protected]&−1&2𝑏@0&1&𝑐−𝑎−𝑏)| Expanding determinant along R1 = (a + b + c)3 ( 0|■8(−1&[email protected]&c−a−b)|−0|■8(1&[email protected]&c−a−b)|+1|■8(1&−[email protected]&1)|) = (a + b + c)3 (0−0+1(1−0)) = (a + b + c)3 (1) = (a + b + c)3 Hence proved