Question 9 - Examples - Chapter 4 Class 12 Determinants
Last updated at April 16, 2024 by Teachoo
Examples
Example 2
Example 3
Example 4
Example 5 Important
Example 6
Example 7 Important
Example 8
Example 9
Example 10
Example 11 Important
Example 12
Example 13 Important
Example 14
Example 15 Important
Example 16
Example 17 Important
Example 18
Example 19 Important
Question 1
Question 2
Question 3
Question 4 Important
Question 5 Important
Question 6
Question 7
Question 8
Question 9 Important You are here
Question 10 Important
Question 11 Important
Question 12
Question 13 Important
Question 14 Important
Question 15 Important
Last updated at April 16, 2024 by Teachoo
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