Using properties of Determinants, prove that:- = 2(a+b)(b+c)(c+a) Share with your friends Share 9 Varun.Rawat answered this Let ∆ = a+b+c-c-b-ca+b+c-a-b-aa+b+cApplying R1 = R1 + R3∆ = a+c-a+ca+c-ca+b+c-a-b-aa+b+c∆ = a+c1-11-ca+b+c-a-b-aa+b+capplying C1 = C1 + C2 and C2 = C2 + C3∆ =a+c001a+bb+c-a-a+bb+ca+b+c⇒∆ = a+bb+cc+a00111-a-11a+b+c⇒∆ = a+bb+cc+a 11+1⇒∆ = 2a+bb+cc+a 35 View Full Answer Shrutika Pedamkar answered this First do C1+C2, you get (a+b) common in 1st column . Then do C2+C3, you get (b+c) common in 2nd column. Then solve the determinant.:):) 4
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= 2(a+b)(b+c)(c+a)