Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 5

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Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 6

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Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 7 Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 8

  1. Chapter 4 Class 12 Determinants (Term 1)
  2. Serial order wise

Transcript

Ex 4.2, 11 By using properties of determinants, show that: (ii) |โ– 8(x+y+2z&x&y@z&y+z+2x&y@z&x&z+x+2y)| = 2(x + y + z)3 Solving L.H.S |โ– 8(x+y+2z&x&y@z&y+z+2x&y@z&x&z+x+2y)| Applying C1 โ†’ C1 + C2 + C3 = |โ– 8(๐‘ฅ+๐‘ฆ+2๐‘ง+๐‘ฅ+๐‘ฆ&๐‘ฅ&๐‘ฆ@๐‘ง+๐‘ฆ+๐‘ง+2๐‘ฅ+๐‘ฆ&๐‘ฆ+๐‘ง+2๐‘ฅ&๐‘ฆ@๐‘ง+๐‘ฅ+๐‘ง+๐‘ฅ+2๐‘ฆ&๐‘ฅ&๐‘ง+๐‘ฅ+2๐‘ฆ)| = |โ– 8(๐Ÿ(๐’™+๐’š+๐’›)&๐‘ฅ&๐‘ฆ@๐Ÿ(๐’™+๐’š+๐’›)&y+๐‘ง+2๐‘ฅ&y@๐Ÿ(๐’™+๐’š+๐’›)&x&z+x+2y)| Taking common 2(๐‘ฅ+๐‘ฆ+๐‘ง) from C1 = ๐Ÿ(๐ฑ+๐ฒ+๐ณ) |โ– 8(1&๐‘ฅ&๐‘ฆ@1&y+๐‘ง+2๐‘ฅ&y@1&x&z+x+2y)| Applying R2 โ†’ R2 โ€“ R3 = 2(x+y+z)|โ– 8(1&๐‘ฅ&๐‘ฆ@๐Ÿโˆ’๐Ÿ&y+๐‘ง+2๐‘ฅโˆ’๐‘ฅ&yโˆ’(๐‘ง+๐‘ฅ+2๐‘ฆ)@1&x&z+x+2y)| = 2(x+y+z)|โ– 8(1&๐‘ฅ&๐‘ฆ@๐ŸŽ&๐‘ฅ+๐‘ฆ+๐‘ง&โˆ’๐‘ฅโˆ’๐‘ฆโˆ’๐‘ง@1&x&z+x+2y)| = 2(x+y+z)|โ– 8(1&๐‘ฅ&๐‘ฆ@0&(๐’™+๐’š+๐’›)&โˆ’(๐’™+๐’š+๐’›)@1&x&z+x+2y)| Taking common (๐‘ฅ+๐‘ฆ+๐‘ง) from 2nd Row = 2(x+y+z)(x+y+z)|โ– 8(1&๐‘ฅ&๐‘ฆ@0&1&โˆ’1@1&x&z+x+2y)| Applying R3 โ†’ R3 โ€“ R1 = 2(x+y+z)2|โ– 8(1&๐‘ฅ&๐‘ฆ@0&1&โˆ’1@๐Ÿโˆ’๐Ÿ&xโˆ’๐‘ฅ&z+x+2yโˆ’y)| = 2(x+y+z)2|โ– 8(1&๐‘ฅ&๐‘ฆ@0&1&โˆ’1@๐ŸŽ&0&x+y+z)| Taking common (๐‘ฅ+๐‘ฆ+๐‘ง) Common from 3rd Row = 2(x+y+z)2(x+y+z)|โ– 8(1&๐‘ฅ&๐‘ฆ@0&1&โˆ’1@0&0&1)| Expanding Determinant along C1 = 2(x+y+z)3 ( 1|โ– 8(1&โˆ’1@0&1)|โˆ’0|โ– 8(๐‘ฅ&๐‘ฆ@0&1)|+0|โ– 8(x&y@1&โˆ’1)|) = 2(x+y+z)3 ( 1|โ– 8(1&โˆ’1@0&1)|โˆ’0+0) = 2(x+y+z)3 (1(1โˆ’0)โˆ’๐‘ฅ(0)+๐‘ฆ(0)) = 2(x+y+z)3 (1) = 2(x+y+z)3 = R.H.S Hence proved

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.