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

Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 6
Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 7 Ex 4.2, 11 - Chapter 4 Class 12 Determinants - Part 8

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Question 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 has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.