Ex 4.2, 11 - By using properties of determinants, show that: (i)

Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 2
Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 3
Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 4

 

 

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Question 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

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.