Question 6 - Solving by simplifying det. - Chapter 4 Class 12 Determinants

Last updated at April 16, 2024 by

Example 11 - Prove that |a a+b a+b+c 2a 3a+2b 4a+3b+2c 3a 6a+3b 10a+6b

Example 11 - Chapter 4 Class 12 Determinants - Part 2
Example 11 - Chapter 4 Class 12 Determinants - Part 3

Transcript

Question 6 Prove that |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@2𝑎&3𝑎+2𝑏&4𝑎+3𝑏+2𝑐@3𝑎&6𝑎+3𝑏&10𝑎+6𝑏+3𝑐)| = a3 Let Δ = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@2𝑎&3𝑎+2𝑏&4𝑎+3𝑏+2𝑐@3𝑎&6𝑎+3𝑏&10𝑎+6𝑏+3𝑐)| Applying R2 → R2 – 2R1 = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@𝟐𝒂−𝟐(𝒂)&3𝑎+3𝑏−2(𝑎+𝑏)&4𝑎+3𝑏+2𝑐−2(𝑎+𝑏+𝑐)@3𝑎&6𝑎+3𝑏&10𝑎+6𝑏+3𝑐)| = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@𝟎&3𝑎+3𝑏−2𝑎−2𝑏&4𝑎+3𝑏+2𝑐−2𝑎−2𝑏−2𝑐@3𝑎&6𝑎+3𝑏&10𝑎+6𝑏+3𝑐)| = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@𝟎&𝑎&2𝑎+𝑏@3𝑎&6𝑎+3𝑏&10𝑎+6𝑏+3𝑐)| Applying R3 → R3 – 3R1 = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@0&𝑎&2𝑎+𝑏@𝟑𝒂−𝟑(𝒂)&6𝑎+3𝑏−3(𝑎+𝑏)&10𝑎+6𝑏+3𝑐−3𝑎+𝑏+𝑐)| = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@0&𝑎&2𝑎+𝑏@𝟎&6𝑎+3𝑏−3𝑎−3𝑏&10𝑎+6𝑏+3𝑐−3𝑎+𝑏+𝑐)| = |■8(𝑎&𝑎+𝑏&𝑎+𝑏+𝑐@0&𝑎&2𝑎+𝑏@0&3𝑎&7𝑎+3𝑏)| Expanding along C1 = = a |■8(𝑎&2𝑎+𝑏@3𝑎&7𝑎+3𝑏)| – 0 + 0 = a (a(7a + 3b) – 3a (2a + b) = a (7a2 + 3ab – 6a2 – 3ab) = a(a2) = a3 = R.H.S Hence proved |■8(𝑎&2𝑎+𝑏@3𝑎&7𝑎+3𝑏)| |■8(𝑎+𝑏&𝑎+𝑏+𝑐@3𝑎&7𝑎+3𝑏)|

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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.