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

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  1. Chapter 4 Class 12 Determinants
  2. Serial order wise
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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+𝐜﷮−a+b﷮−a+c﷮𝐚+𝐛+𝐜﷮3b﷮−b+c﷮𝐚+𝐛+𝐜﷮−c+b﷮3c﷯﷯ Taking (a + b + c) common from C1 = (𝐚+𝐛+𝐜) 1﷮−a+b﷮−a+c﷮1﷮3b﷮−b+c﷮1﷮−c+b﷮3c﷯﷯ Applying R1→ R1 – R2 = (a+b+c) 𝟏−𝟏﷮−a+b−3b﷮−a+c−(−b+c)﷮1﷮3b﷮−b+c﷮1﷮−c+b﷮3c﷯﷯ = (a+b+c) 𝟎﷮−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﷮𝟏−𝟏﷮3b﷮−b+c﷮1﷮−c+b﷮3c﷯−3c﷯ =(a+b+c) 0﷮−a−2b﷮−a+b﷮𝟎﷮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+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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