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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

Transcript

Example 11 Prove that |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@2๐‘Ž&3๐‘Ž+2๐‘&4๐‘Ž+3๐‘+2๐‘@3๐‘Ž&6๐‘Ž+3๐‘&10๐‘Ž+6๐‘+3๐‘)| = a3 Let ฮ” = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@2๐‘Ž&3๐‘Ž+2๐‘&4๐‘Ž+3๐‘+2๐‘@3๐‘Ž&6๐‘Ž+3๐‘&10๐‘Ž+6๐‘+3๐‘)| Applying R2 โ†’ R2 โ€“ 2R1 = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@๐Ÿ๐’‚โˆ’๐Ÿ(๐’‚)&3๐‘Ž+3๐‘โˆ’2(๐‘Ž+๐‘)&4๐‘Ž+3๐‘+2๐‘โˆ’2(๐‘Ž+๐‘+๐‘)@3๐‘Ž&6๐‘Ž+3๐‘&10๐‘Ž+6๐‘+3๐‘)| = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@๐ŸŽ&3๐‘Ž+3๐‘โˆ’2๐‘Žโˆ’2๐‘&4๐‘Ž+3๐‘+2๐‘โˆ’2๐‘Žโˆ’2๐‘โˆ’2๐‘@3๐‘Ž&6๐‘Ž+3๐‘&10๐‘Ž+6๐‘+3๐‘)| = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@๐ŸŽ&๐‘Ž&2๐‘Ž+๐‘@3๐‘Ž&6๐‘Ž+3๐‘&10๐‘Ž+6๐‘+3๐‘)| Applying R3 โ†’ R3 โ€“ 3R1 = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@0&๐‘Ž&2๐‘Ž+๐‘@๐Ÿ‘๐’‚โˆ’๐Ÿ‘(๐’‚)&6๐‘Ž+3๐‘โˆ’3(๐‘Ž+๐‘)&10๐‘Ž+6๐‘+3๐‘โˆ’3๐‘Ž+๐‘+๐‘)| = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@0&๐‘Ž&2๐‘Ž+๐‘@๐ŸŽ&6๐‘Ž+3๐‘โˆ’3๐‘Žโˆ’3๐‘&10๐‘Ž+6๐‘+3๐‘โˆ’3๐‘Ž+๐‘+๐‘)| = |โ– 8(๐‘Ž&๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@0&๐‘Ž&2๐‘Ž+๐‘@0&3๐‘Ž&7๐‘Ž+3๐‘)| Expanding along C1 = = a |โ– 8(๐‘Ž&2๐‘Ž+๐‘@3๐‘Ž&7๐‘Ž+3๐‘)| โ€“ 0 + 0 = a (a(7a + 3b) โ€“ 3a (2a + b) = a (7a2 + 3ab โ€“ 6a2 โ€“ 3ab) = a(a2) = a3 = R.H.S Hence proved |โ– 8(๐‘Ž&2๐‘Ž+๐‘@3๐‘Ž&7๐‘Ž+3๐‘)| |โ– 8(๐‘Ž+๐‘&๐‘Ž+๐‘+๐‘@3๐‘Ž&7๐‘Ž+3๐‘)|

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.