Check sibling questions

Example 14 - Prove that |b+c a a b c+a b c c a+b| = 4abc

Example 14 - Chapter 4 Class 12 Determinants - Part 2


Transcript

Example 14 Prove that |■8(b+c&a&[email protected]&c+a&[email protected]&c&a+b)| = 4abc Solving L.H.S Let ∆ = |■8(b+c&a&[email protected]&c+a&[email protected]&c&a+b)| Applying R1 → R1 – R2 – R3 = |■8(b+c−𝐛−𝐜&a−c−a−c&a−b−a−[email protected]&c+a&[email protected]&c&a+b)| = |■8(𝟎&−2c&−[email protected]&c+a&[email protected]&c&a+b)| Expanding determinant along R1 = 0|■8(c+a&[email protected]&a+b)|−(−2𝑐)|■8(b&[email protected]&a+b)|−2𝑏|■8(b&c+𝑎@c&c)| = 0 −(−2𝑐)|■8(b&[email protected]&a+b)|−2𝑏|■8(b&[email protected]&c)| = 0 + 2c (b (a + b) – cb) – 2b (cb – c(c + a)) = 2c (ab + b2 – cb) – 2b (cb – c2 – ca) = 2abc + 2cb2 – 2bc2 – 2b2c + 2bc2 + 2abc = 2abc + 2abc + 2cb2 – 2cb2 – 2bc2 + 2bc2 = 4abc + 0 + 0 = 4abc = R.H.S 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 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.