# Misc. 13

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Misc. 13 Using properties of determinants, prove that: 3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c = 3 ( a + b + c) (ab + bc + ac) Taking L.H. S 3a−a+b−a+c−b+a3b−b+c−c+a−c+b3c Applying C1→ C1 + C2 + C3 = 3a−a+b−a+c−a+b−a+c−b+a−3b−b+c3b−b+c−c+a−c+b−3c−c+b3c = 𝐚+𝐛+𝐜−a+b−a+c𝐚+𝐛+𝐜3b−b+c𝐚+𝐛+𝐜−c+b3c Taking (a + b + c) common from C1 = (𝐚+𝐛+𝐜) 1−a+b−a+c13b−b+c1−c+b3c Applying R1→ R1 – R2 = (a+b+c) 𝟏−𝟏−a+b−3b−a+c−(−b+c)13b−b+c1−c+b3c = (a+b+c) 𝟎−a−2b−a+c+b−c13b−b+c1−c+b3c = (a+b+c) 0−a−2b−a+b13b−b+c1−c+b3c Applying R2→ R2 – R3 =(a+b+c) 0−a−2b−a+b𝟏−𝟏3b−b+c1−c+b3c−3c =(a+b+c) 0−a−2b−a+b𝟎2b+c−b−2c1−c+b3c Expanding determinant along C1 = (a + b + c ) 0 2b+c−b−2c−c+b3c−0 −a−2b−a+b−c+b3c+1 −a−2b−a+b2b+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+𝑐𝑎+4𝑐𝑏−𝑐𝑏+2𝑏2−2𝑏2 = (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 is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.