# Misc 6

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc 6 Prove that |โ 8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = 4a2b2c2 Taking L.H.S |โ 8(a2&bc&ac+c2@a2+ab&b2&ac@ab&b2+bc&c2)| = |โ 8(๐(๐)&๐(c)&๐(a+c)@๐(๐+๐)&๐(b)&๐(a)@๐(๐)&๐(b+c)&๐(c))| Taking out a common from C1 ,b common from C2 & c common from C3 = abc |โ 8(a&c&a+c@๐+๐&b&a@๐&b+c&c)| Applying C1โ C1 + C2 + C3 = abc |โ 8(a+๐+๐+๐&c&a+c@๐+๐+๐+๐&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(๐+๐)&c&a+c@๐(๐+๐)&b&a@๐(๐+๐)&b+c&c)| Taking out 2 Common from C2 = 2a๐๐|โ 8((๐+๐)&c&a+c@(๐+๐)&b&a@(๐+๐)&b+c&c)| Applying C3โ C3 โ C1 = 2a๐๐|โ 8((๐+๐)&c&a+cโ(๐+๐)@(๐+๐)&b&aโ(a+b)@(๐+๐)&b+c&cโ(b+c))| = 2a๐๐|โ 8(๐+๐&c&๐@๐+๐&b&โ๐@๐+๐&b+c&โb)| Applying C2โ C2 โ C1 = 2abc |โ 8(๐+๐&cโ(๐+๐)&0@๐+๐&bโ(๐+๐)&โb@๐+๐&b+cโ(๐+๐)&โb)| = 2abc |โ 8(๐+๐&โ๐&0@๐+๐&โ๐&โ๐@๐+๐&0&โ๐)| Taking out โa common from C2, โb common from C3 = 2abc (โa) (โb) |โ 8(๐+๐&1&0@๐+๐&1&1@๐+๐&0&1)| Expanding Determinant along R1 = 2a2b2c ((a+c)|โ 8(1&1@0&1)|โ1|โ 8(๐+๐&1@๐+๐&1)|+0|โ 8(๐+๐&1@๐+๐&0)|) = 2a2b2c ((a+c)|โ 8(1&1@0&1)|โ1|โ 8(๐+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

Chapter 4 Class 12 Determinants

Serial order wise

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Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 7 years. He provides courses for Mathematics and Science from Class 6 to 12. You can learn personally from here https://www.teachoo.com/premium/maths-and-science-classes/.