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  1. Chapter 4 Class 12 Determinants
  2. Concept wise

Transcript

Ex 4.2, 14 By using properties of determinants, show that: |โ– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| = 1 + a2 + b2 + c2 Solving L.H.S |โ– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying & Dividing by abc = ๐’‚๐’ƒ๐’„/๐’‚๐’ƒ๐’„ |โ– 8(a2+1&ab&ac@ab&b2+1&bc@ca&cb&c2+1)| Multiplying 1st row by a, 2nd row by b & 3rd row by c ( R1 โ†’ aR1 , R2 โ†’ bR3 , R3 โ†’ bR3 ) = 1/๐‘Ž๐‘๐‘ |โ– 8(๐’‚(a2+1)&๐’‚(ab)&๐’‚(ac)@๐’ƒ(ab)&๐›(b2+1)&๐’ƒ(bc)@๐œ(ca)&๐’„(cb)&๐œ(c2+1))| = 1/๐‘Ž๐‘๐‘ |โ– 8(a3+a&๐‘Ž2b&๐‘Ž2c@ab2&b3+b&๐‘2c@c2a&๐‘2b&c3+c)| Applying R1 โ†’ R1 + R2 + R3 = 1/๐‘Ž๐‘๐‘ |โ– 8(a3+a+๐‘Ž๐‘2+๐‘2๐‘Ž&๐‘Ž2b+b3+b+c2b&๐‘Ž2c+b2c+c3+c@ab2&b3+b&๐‘2c@c2a&๐‘2b&c3+c)| = 1/๐‘Ž๐‘๐‘ |โ– 8(a(๐š๐Ÿ+๐Ÿ+๐’ƒ๐Ÿ+๐’„๐Ÿ)&๐‘(๐’‚๐Ÿ+๐›๐Ÿ+๐Ÿ+๐œ๐Ÿ)&๐‘(๐’‚๐Ÿ+๐›๐Ÿ+๐œ๐Ÿ"+1" )@ab2&b3+b&๐‘2c@c2a&๐‘2b&c3+c)| Taking (1+๐‘Ž2+๐‘2+๐‘2) common from 1st Row = ((๐Ÿ + ๐’‚๐Ÿ + ๐’ƒ๐Ÿ + ๐’„๐Ÿ))/๐‘Ž๐‘๐‘ |โ– 8(a&๐‘&๐‘@ab2&b3+b&๐‘2c@c2a&๐‘2b&c(c3+1))| Taking a common from C1 ,b from C2 & c from C3 . = ๐’‚๐’ƒ๐’„ ( (1 + ๐‘Ž2 + ๐‘2 + ๐‘2))/๐‘Ž๐‘๐‘ |โ– 8(1&1&1@b2&b3+1&๐‘2@c2&๐‘2&c2+1)| Applying C1 โ†’ C1 โˆ’ C2 = (1+๐‘Ž2+๐‘2+๐‘2) |โ– 8(๐Ÿโˆ’๐Ÿ&1&1@b2โˆ’๐‘2โˆ’1&b2+1&๐‘2@c2โˆ’c2&๐‘2&c2+1)| = (1+๐‘Ž2+๐‘2+๐‘2) |โ– 8(๐ŸŽ&1&1@โˆ’1&b2+1&๐‘2@0&๐‘2&c2+1)| Applying C2 โ†’ C2 โˆ’ C3 = (1+๐‘Ž2+๐‘2+๐‘2) |โ– 8(0&๐Ÿโˆ’๐Ÿ&1@โˆ’1&b2+1โˆ’๐‘2&๐‘2@0&๐‘2โˆ’๐‘2โˆ’1&c2+1)| = (1+๐‘Ž2+๐‘2+๐‘2) |โ– 8(0&๐ŸŽ&1@โˆ’1&1&๐‘2@0&โˆ’1&c2+1)| Expanding along R1 = (1+๐‘Ž2+๐‘2+๐‘2)(1|โ– 8(1&๐‘2@โˆ’1&๐‘2+1)|" โ€“ 0" |โ– 8(โˆ’1&๐ถ2@0&๐‘2+1)|" + 0" |โ– 8(1&1@0&โˆ’1)|) = (1 + a2 + b1 + c2) (0 โ€“ 0 + 1) = (1 + a2 + b1 + c2) (1) = (1 + a2 + b1 + c2) = 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 9 years. He provides courses for Maths and Science at Teachoo.