Example 11
Last updated at Dec. 8, 2016 by Teachoo
Transcript
Example 11 Prove that ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ2๐๏ทฎ3๐+2๐๏ทฎ4๐+3๐+2๐๏ทฎ3๐๏ทฎ6๐+3๐๏ทฎ10๐+6๐+3๐๏ทฏ๏ทฏ = a3 Let ฮ = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ2๐๏ทฎ3๐+2๐๏ทฎ4๐+3๐+2๐๏ทฎ3๐๏ทฎ6๐+3๐๏ทฎ10๐+6๐+3๐๏ทฏ๏ทฏ Applying R2 โ R2 โ 2R1 = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ๐๐โ๐(๐)๏ทฎ3๐+3๐โ2(๐+๐)๏ทฎ4๐+3๐+2๐โ2(๐+๐+๐)๏ทฎ3๐๏ทฎ6๐+3๐๏ทฎ10๐+6๐+3๐๏ทฏ๏ทฏ = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ๐๏ทฎ3๐+3๐โ2๐โ2๐๏ทฎ4๐+3๐+2๐โ2๐โ2๐โ2๐๏ทฎ3๐๏ทฎ6๐+3๐๏ทฎ10๐+6๐+3๐๏ทฏ๏ทฏ = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ๐๏ทฎ๐๏ทฎ2๐+๐๏ทฎ3๐๏ทฎ6๐+3๐๏ทฎ10๐+6๐+3๐๏ทฏ๏ทฏ Applying R3 โ R3 โ 3R1 = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ0๏ทฎ๐๏ทฎ2๐+๐๏ทฎ๐๐โ๐(๐)๏ทฎ6๐+3๐โ3(๐+๐)๏ทฎ10๐+6๐+3๐โ3๐+๐+๐๏ทฏ๏ทฏ = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ0๏ทฎ๐๏ทฎ2๐+๐๏ทฎ๐๏ทฎ6๐+3๐โ3๐โ3๐๏ทฎ10๐+6๐+3๐โ3๐+๐+๐๏ทฏ๏ทฏ = ๐๏ทฎ๐+๐๏ทฎ๐+๐+๐๏ทฎ0๏ทฎ๐๏ทฎ2๐+๐๏ทฎ0๏ทฎ3๐๏ทฎ7๐+3๐๏ทฏ๏ทฏ Expanding along C1 = = a ๐๏ทฎ2๐+๐๏ทฎ3๐๏ทฎ7๐+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
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9
Example 10
Example 11 You are here
Example 12
Example 13
Example 14 Important
Example 15 Important
Example 16 Important
Example 17
Example 18 Important
Example 19
Example 20
Example 21
Example 22
Example 23
Example 24 Important
Example 25
Example 26 Important
Example 27
Example 28
Example 29
Example 30
Example 31
Example 32 Important
Example 33
Example 34 Important
Chapter 4 Class 12 Determinants
Serial order wise
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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/.