Example 11 - Prove |a a+b a+b+c 2a 3a+2b - Chapter 4 NCERT - Solving by simplifying det.

Slide20.JPG
Slide21.JPG

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

Transcript

Example 11 Prove that ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ2๐‘Ž๏ทฎ3๐‘Ž+2๐‘๏ทฎ4๐‘Ž+3๐‘+2๐‘๏ทฎ3๐‘Ž๏ทฎ6๐‘Ž+3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘๏ทฏ๏ทฏ = a3 Let ฮ” = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ2๐‘Ž๏ทฎ3๐‘Ž+2๐‘๏ทฎ4๐‘Ž+3๐‘+2๐‘๏ทฎ3๐‘Ž๏ทฎ6๐‘Ž+3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘๏ทฏ๏ทฏ Applying R2 โ†’ R2 โ€“ 2R1 = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ๐Ÿ๐’‚โˆ’๐Ÿ(๐’‚)๏ทฎ3๐‘Ž+3๐‘โˆ’2(๐‘Ž+๐‘)๏ทฎ4๐‘Ž+3๐‘+2๐‘โˆ’2(๐‘Ž+๐‘+๐‘)๏ทฎ3๐‘Ž๏ทฎ6๐‘Ž+3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘๏ทฏ๏ทฏ = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ๐ŸŽ๏ทฎ3๐‘Ž+3๐‘โˆ’2๐‘Žโˆ’2๐‘๏ทฎ4๐‘Ž+3๐‘+2๐‘โˆ’2๐‘Žโˆ’2๐‘โˆ’2๐‘๏ทฎ3๐‘Ž๏ทฎ6๐‘Ž+3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘๏ทฏ๏ทฏ = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ๐ŸŽ๏ทฎ๐‘Ž๏ทฎ2๐‘+๐‘๏ทฎ3๐‘Ž๏ทฎ6๐‘Ž+3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘๏ทฏ๏ทฏ Applying R3 โ†’ R3 โ€“ 3R1 = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ0๏ทฎ๐‘Ž๏ทฎ2๐‘+๐‘๏ทฎ๐Ÿ‘๐’‚โˆ’๐Ÿ‘(๐’‚)๏ทฎ6๐‘Ž+3๐‘โˆ’3(๐‘Ž+๐‘)๏ทฎ10๐‘Ž+6๐‘+3๐‘โˆ’3๐‘Ž+๐‘+๐‘๏ทฏ๏ทฏ = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ0๏ทฎ๐‘Ž๏ทฎ2๐‘+๐‘๏ทฎ๐ŸŽ๏ทฎ6๐‘Ž+3๐‘โˆ’3๐‘Žโˆ’3๐‘๏ทฎ10๐‘Ž+6๐‘+3๐‘โˆ’3๐‘Ž+๐‘+๐‘๏ทฏ๏ทฏ = ๐‘Ž๏ทฎ๐‘Ž+๐‘๏ทฎ๐‘Ž+๐‘+๐‘๏ทฎ0๏ทฎ๐‘Ž๏ทฎ2๐‘+๐‘๏ทฎ0๏ทฎ3๐‘Ž๏ทฎ7๐‘Ž+3๐‘๏ทฏ๏ทฏ Expanding along C1 = = a ๐‘Ž๏ทฎ2๐‘Ž+๐‘๏ทฎ3๐‘Ž๏ทฎ7๐‘Ž+3๐‘๏ทฏ๏ทฏ 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

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.
Jail