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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise

Transcript

Ex 4.2, 8 By using properties of determinants, show that: (i) |โ– 8(1&๐‘Ž&๐‘Ž2@1&๐‘&๐‘2@1&๐‘&๐‘2)| = (a - b) (b - c)(c โ€“ a) Solving L.H.S |โ– 8(1&๐‘Ž&๐‘Ž2@1&๐‘&๐‘2@1&๐‘&๐‘2)| Applying R1 โ†’ R1 โˆ’ R2 = |โ– 8(๐Ÿโˆ’๐Ÿ&๐‘Žโˆ’๐‘&๐‘Ž^2โˆ’๐‘^2@1&๐‘&๐‘2@1&๐‘&๐‘2 ) | = |โ– 8(๐ŸŽ&(๐‘Žโˆ’๐‘)&(๐‘Žโˆ’๐‘)(๐‘Ž+๐‘)@1&๐‘&๐‘2@1&๐‘&๐‘2 ) | = |โ– 8(0(๐šโˆ’๐›)&(๐šโˆ’๐›)&(๐šโˆ’๐’ƒ)(a+b)@1&b&b2@1&c&c2 ) | Taking Common (a โ€“ b) from R1 = (๐šโˆ’๐’ƒ) |โ– 8(0&1&a+b@1&b&b2@1&c&c2 ) | Applying R2 โ†’ R2 โˆ’ R3 = (aโˆ’b) |โ– 8(0&1&a+b@๐Ÿโˆ’๐Ÿ&bโˆ’c&b2โˆ’c2@1&c&c2 ) | = (a โ€“ b) |โ– 8(0&1&a+๐‘@๐ŸŽ&bโˆ’c&(bโˆ’c)(b+c)@1&c&c2 ) | Taking common (b โ€“ c) from R2 = (a โ€“ b) (b โ€“ c) |โ– 8(0&1&a+b@0&1&b+c@1&c&c2 ) | Expanding Determinant along C1 = (a โ€“ b) (b โ€“ c) ( 0|โ– 8(1&๐‘+๐‘@๐‘&๐‘2)|โˆ’0|โ– 8(1&๐‘Ž+๐‘@๐‘&๐‘2)|+1|โ– 8(1&๐‘Ž+๐‘@1&๐‘+๐‘)|) = (a โ€“ b) (b โ€“ c) ( 0โˆ’0+1|โ– 8(1&๐‘Ž+๐‘@1&๐‘+๐‘)|) = (a โ€“ b) (b โ€“ c) (1(b + c) โ€“ 1(a + b) ) = (a โ€“ b) (b โ€“ c) (b + c โ€“ a โ€“ b) = (a โ€“ b) (b โ€“ c)(c โ€“ a) = R.H.S Hence Proved Ex 4.2, 8 By using properties of determinants, show that: (ii) |โ– 8(1&1&1@a&b&c@a3&b3&c3)| = (a โ€“ b) (b โ€“ c) (c โ€“ a) (a + b + c) Solving L.H.S |โ– 8(1&1&1@a&b&c@a3&b3&c3)| Applying C1 โ†’ C1 โˆ’ C2 = |โ– 8(๐Ÿโˆ’๐Ÿ&1&1@aโˆ’b&b&c@๐š๐Ÿ‘ โˆ’๐›๐Ÿ‘&b3 &c3)| = |โ– 8(๐ŸŽ&1&1@aโˆ’b&b&c@(๐š โˆ’๐›)(๐š๐Ÿ+๐›๐Ÿ+๐š๐›) &b3&c3)| = |โ– 8(0&1&1@๐šโˆ’๐›&b&c@(๐š โˆ’๐›)(a2+b2+ab) &b3&c3)| Taking Common (a โ€“ b) from C1 = (a โ€“ b) |โ– 8(0&1&1@1&b&c@(a2+b2+ab)&b3&c3)| Applying C2 โ†’ C2 โˆ’ C3 = (a โ€“ b) |โ– 8(0&๐Ÿโˆ’๐Ÿ&1@1&bโˆ’c&c@(a2+b2+ab)&b3โˆ’c3&c3)| (x3 โ€“ y3 = (x โ€“ y)(x2 + y2 +xy)) = (a โ€“ b) |โ– 8(0&๐ŸŽ&1@1&bโˆ’c&c@(a2+b2+ab)&(bโˆ’c)(b2+c2+bc)&c3)| Taking Common (b โ€“ c) from C2 = (a โ€“ b) (b โ€“ c) |โ– 8(0&0&1@1&1&c@a2+b2+ab&b2+c2+bc&c3)| Expanding determinant along R1 = (a โ€“ b) (b โ€“ c) (0|โ– 8(1&๐‘@๐‘2+๐‘2+๐‘๐‘&๐‘3)|โˆ’0|โ– 8(1&1@๐‘Ž2+๐‘2+๐‘Ž๐‘&๐‘3)|+1|โ– 8(1&1@๐‘Ž2+๐‘2+๐‘Ž๐‘&๐‘2+๐‘2+๐‘๐‘)|) = (a โ€“ b) (b โ€“ c) (0โˆ’0+1|โ– 8(1&1@๐‘Ž2+๐‘2+๐‘Ž๐‘&๐‘2+๐‘2+๐‘๐‘)|) = (a โ€“ b) (b โ€“ c) (1((b2 + c2 + bc) โ€“ (a2 + b2 + ab)) = (a โ€“ b) (b โ€“ c) (b2 + c2 + bc โ€“ a2 โ€“ b2 โ€“ ab) = (a โ€“ b) (b โ€“ c) (c2 โ€“ a2 + bc โ€“ ab) = (a โ€“ b) (b โ€“ c) ((c โ€“ a) (c + a) + b (c โ€“ a)) = (a โ€“ b) (b โ€“ c) ((c โ€“ a) (c + a + b)) = (a โ€“ b) (b โ€“ c) ((c โ€“ a) (a + b + c)) = 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.