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


Misc 6 Prove that |โ– 8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = 4a2b2c2 Solving L.H.S |โ– 8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = |โ– 8(๐š(๐‘Ž)&๐›(c)&๐œ(a+c)@๐š(๐‘Ž+๐‘)&๐›(b)&๐œ(a)@๐š(๐‘)&๐›(b+c)&๐œ(c))| Taking out a common from C1 ,b common from C2 & c common from C3 = abc |โ– 8(a&c&a+c@๐‘Ž+๐‘&b&a@๐‘&b+c&c)| Applying C1โ†’ C1 + C2 + C3 = abc |โ– 8(a+๐‘+๐‘Ž+๐‘&c&a+c@๐‘Ž+๐‘+๐‘+๐‘Ž&b&a@๐‘+๐‘+๐‘+๐‘&b+c&c)| = abc |โ– 8(2a+2๐‘&c&a+c@2a+2๐‘&b&a@2๐‘+2๐‘&b+c&c)| = abc |โ– 8(๐Ÿ(๐‘Ž+๐‘)&c&a+c@๐Ÿ(๐‘Ž+๐‘)&b&a@๐Ÿ(๐‘+๐‘)&b+c&c)| Taking out 2 Common from C2 = 2a๐‘๐‘|โ– 8((๐‘Ž+๐‘)&c&a+c@(๐‘Ž+๐‘)&b&a@(๐‘+๐‘)&b+c&c)| Applying C3โ†’ C3 โ€“ C1 = 2a๐‘๐‘|โ– 8((๐‘Ž+๐‘)&c&a+cโˆ’(๐š+๐œ)@(๐‘Ž+๐‘)&b&aโˆ’(a+b)@(๐‘+๐‘)&b+c&cโˆ’(b+c))| = 2a๐‘๐‘|โ– 8(๐‘Ž+๐‘&c&๐ŸŽ@๐‘Ž+๐‘&b&โˆ’๐‘@๐‘+๐‘&b+c&โˆ’b)| Applying C2โ†’ C2 โ€“ C1 = 2abc |โ– 8(๐‘Ž+๐‘&cโˆ’(๐‘Ž+๐‘)&0@๐‘Ž+๐‘&bโˆ’(๐‘Ž+๐‘)&โˆ’b@๐‘+๐‘&b+cโˆ’(๐‘+๐‘)&โˆ’b)| = 2abc |โ– 8(๐‘Ž+๐‘&โˆ’๐’‚&0@๐‘Ž+๐‘&โˆ’๐’‚&โˆ’๐›@๐‘+๐‘&0&โˆ’๐›)| Taking out โ€“a common from C2, โ€“b common from C3 = 2abc (โ€“a) (โ€“b) |โ– 8(๐‘Ž+๐‘&1&0@๐‘Ž+๐‘&1&1@๐‘+๐‘&0&1)| Expanding Determinant along R1 = 2a2b2c ((a+c)|โ– 8(1&1@0&1)|โˆ’1|โ– 8(๐‘Ž+๐‘&1@๐‘+๐‘&1)|+0|โ– 8(๐‘Ž+๐‘&1@๐‘+๐‘&0)|) = 2a2b2c ((a+c)|โ– 8(1&1@0&1)|โˆ’1|โ– 8(๐‘Ž+๐‘&1@๐‘+๐‘&1)|+0) = 2a2b2c ((a + c) (1 โ€“ 0) โ€“ 1(a + b โ€“ (b + c))) = 2a2b2c ((a + c) (1) โ€“ 1 (a + b โ€“ b โ€“ c)) = 2a2b2c (a + c โ€“ a + c) = 2a2b2c (2c) = 4a2b2c = 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.