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Example 11 - Chapter 4 Class 12 Determinants (Term 1)

Last updated at Jan. 23, 2020 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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Example 11 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

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.