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Misc 13 - Using determinants 3 (a + b + c) (ab + bc + ac) - Making whole row/column one and simplifying

Misc. 13 - Chapter 4 Class 12 Determinants - Part 2
Misc. 13 - Chapter 4 Class 12 Determinants - Part 3
Misc. 13 - Chapter 4 Class 12 Determinants - Part 4

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Misc. 13 Using properties of determinants, prove that: 3a a+b a+c b+a 3b b+c c+a c+b 3c = 3 ( a + b + c) (ab + bc + ac) Taking L.H. S 3a a+b a+c b+a 3b b+c c+a c+b 3c Applying C1 C1 + C2 + C3 = 3a a+b a+c a+b a+c b+a 3b b+c 3b b+c c+a c+b 3c c+b 3c = + + a+b a+c + + 3b b+c + + c+b 3c Taking (a + b + c) common from C1 = ( + + ) 1 a+b a+c 1 3b b+c 1 c+b 3c Applying R1 R1 R2 = (a+b+c) a+b 3b a+c ( b+c) 1 3b b+c 1 c+b 3c = (a+b+c) a 2b a+c+b c 1 3b b+c 1 c+b 3c = (a+b+c) 0 a 2b a+b 1 3b b+c 1 c+b 3c Applying R2 R2 R3 =(a+b+c) 0 a 2b a+b 3b b+c 1 c+b 3c 3c =(a+b+c) 0 a 2b a+b 2b+c b 2c 1 c+b 3c Expanding determinant along C1 = (a + b + c ) 0 2b+c b 2c c+b 3c 0 a 2b a+b c+b 3c +1 a 2b a+b 2b+c b 2c = (a + b + c ) b+2c a+2b (2b+c)( a+b) = (a + b + c ) b+2c a+2b (2b+c)( a+b) = (a + b + c ) ab+2b2+2ca+4ab (2b( a+b) c( a+b) = (a + b + c ) ab+2b2+2ca+4ab ( 2ba+2b2+ac bc) = (a + b + c) ab+2b2+2ca+4ab ( 2ba+2b2+ac bc) = (a + b + c ) ab+2b2+2ca+4ab+2ba 2b2+ca cb = (a + b + c ) ab+2ba+2ca+ +4 +2 2 2 2 = (a + b + c ) (3ab + 3ac + 3bc + 0) = (a + b + c ) (3ab + 3ac + 3bc) = (a + b + c ). 3 (ab + ac + bc) = 3 (a + b + c ) (ab + ac + bc) = R.H.S Hence Proved

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Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.