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Example 11 - Chapter 4 Class 12 Determinants
Last updated at Jan. 23, 2020 by Teachoo
Transcript
Example 11 Prove that |โ 8(๐&๐+๐&๐+๐+๐@2๐&3๐+2๐&4๐+3๐+2๐@3๐&6๐+3๐&10๐+6๐+3๐)| = a3 Let ฮ = |โ 8(๐&๐+๐&๐+๐+๐@2๐&3๐+2๐&4๐+3๐+2๐@3๐&6๐+3๐&10๐+6๐+3๐)| Applying R2 โ R2 โ 2R1 = |โ 8(๐&๐+๐&๐+๐+๐@๐๐โ๐(๐)&3๐+3๐โ2(๐+๐)&4๐+3๐+2๐โ2(๐+๐+๐)@3๐&6๐+3๐&10๐+6๐+3๐)| = |โ 8(๐&๐+๐&๐+๐+๐@๐&3๐+3๐โ2๐โ2๐&4๐+3๐+2๐โ2๐โ2๐โ2๐@3๐&6๐+3๐&10๐+6๐+3๐)| = |โ 8(๐&๐+๐&๐+๐+๐@๐&๐&2๐+๐@3๐&6๐+3๐&10๐+6๐+3๐)| Applying R3 โ R3 โ 3R1 = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@๐๐โ๐(๐)&6๐+3๐โ3(๐+๐)&10๐+6๐+3๐โ3๐+๐+๐)| = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@๐&6๐+3๐โ3๐โ3๐&10๐+6๐+3๐โ3๐+๐+๐)| = |โ 8(๐&๐+๐&๐+๐+๐@0&๐&2๐+๐@0&3๐&7๐+3๐)| Expanding along C1 = = a |โ 8(๐&2๐+๐@3๐&7๐+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&2๐+๐@3๐&7๐+3๐)| |โ 8(๐+๐&๐+๐+๐@3๐&7๐+3๐)|
Examples
Example 1
Example 2
Example 3
Example 4
Example 5
Example 6
Example 7
Example 8
Example 9 Important
Example 10 Important
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 Important
Example 29
Example 30
Example 31 Important
Example 32 Important
Example 33 Important
Example 34 Important
Chapter 4 Class 12 Determinants
Serial order wise
About the Author
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.