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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

Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.