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Question 6 - Examples - Chapter 4 Class 12 Determinants

Last updated at May 29, 2023 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

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Question 6 Prove that |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@2π‘Ž&3π‘Ž+2𝑏&4π‘Ž+3𝑏+2𝑐@3π‘Ž&6π‘Ž+3𝑏&10π‘Ž+6𝑏+3𝑐)| = a3 Let Ξ” = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@2π‘Ž&3π‘Ž+2𝑏&4π‘Ž+3𝑏+2𝑐@3π‘Ž&6π‘Ž+3𝑏&10π‘Ž+6𝑏+3𝑐)| Applying R2 β†’ R2 – 2R1 = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@πŸπ’‚βˆ’πŸ(𝒂)&3π‘Ž+3π‘βˆ’2(π‘Ž+𝑏)&4π‘Ž+3𝑏+2π‘βˆ’2(π‘Ž+𝑏+𝑐)@3π‘Ž&6π‘Ž+3𝑏&10π‘Ž+6𝑏+3𝑐)| = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@𝟎&3π‘Ž+3π‘βˆ’2π‘Žβˆ’2𝑏&4π‘Ž+3𝑏+2π‘βˆ’2π‘Žβˆ’2π‘βˆ’2𝑐@3π‘Ž&6π‘Ž+3𝑏&10π‘Ž+6𝑏+3𝑐)| = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@𝟎&π‘Ž&2π‘Ž+𝑏@3π‘Ž&6π‘Ž+3𝑏&10π‘Ž+6𝑏+3𝑐)| Applying R3 β†’ R3 – 3R1 = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@0&π‘Ž&2π‘Ž+𝑏@πŸ‘π’‚βˆ’πŸ‘(𝒂)&6π‘Ž+3π‘βˆ’3(π‘Ž+𝑏)&10π‘Ž+6𝑏+3π‘βˆ’3π‘Ž+𝑏+𝑐)| = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@0&π‘Ž&2π‘Ž+𝑏@𝟎&6π‘Ž+3π‘βˆ’3π‘Žβˆ’3𝑏&10π‘Ž+6𝑏+3π‘βˆ’3π‘Ž+𝑏+𝑐)| = |β– 8(π‘Ž&π‘Ž+𝑏&π‘Ž+𝑏+𝑐@0&π‘Ž&2π‘Ž+𝑏@0&3π‘Ž&7π‘Ž+3𝑏)| Expanding along C1 = = a |β– 8(π‘Ž&2π‘Ž+𝑏@3π‘Ž&7π‘Ž+3𝑏)| – 0 + 0 = a (a(7a + 3b) – 3a (2a + b) = a (7a2 + 3ab – 6a2 – 3ab) = a(a2) = a3 = R.H.S Hence proved |β– 8(π‘Ž&2π‘Ž+𝑏@3π‘Ž&7π‘Ž+3𝑏)| |β– 8(π‘Ž+𝑏&π‘Ž+𝑏+𝑐@3π‘Ž&7π‘Ž+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 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.