Question 6 - Examples - Chapter 4 Class 12 Determinants
Last updated at Dec. 16, 2024 by Teachoo
Last updated at Dec. 16, 2024 by Teachoo
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 and Computer Science at Teachoo