Question 9 - Chapter 4 Class 12 Determinants (Important Question)

Last updated at Dec. 16, 2024 by

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

Go Ad-free

Transcript

Question 9 Prove that |■8(b+c&a&a@b&c+a&b@c&c&a+b)| = 4abc Solving L.H.S Let ∆ = |■8(b+c&a&a@b&c+a&b@c&c&a+b)| Applying R1 → R1 – R2 – R3 = |■8(b+c−𝐛−𝐜&a−c−a−c&a−b−a−b@b&c+a&b@c&c&a+b)| = |■8(𝟎&−2c&−2b@b&c+a&b@c&c&a+b)| Expanding determinant along R1 = 0|■8(c+a&b@c&a+b)|−(−2𝑐)|■8(b&b@c&a+b)|−2𝑏|■8(b&c+𝑎@c&c)| = 0 −(−2𝑐)|■8(b&b@c&a+b)|−2𝑏|■8(b&c+a@c&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

Davneet Singh's photo - Co-founder, Teachoo

Made by

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 and Computer Science at Teachoo