Question

# Using properties of determinants, prove that ∣∣ ∣ ∣ ∣ ∣∣(a+b)2ccca(b+c)2aabb(c+a)2b∣∣ ∣ ∣ ∣ ∣∣=2(a+b+c)3 OR If p≠0,q≠0 and ∣∣ ∣∣pqpα+qqrpα+rpα+qqα+r0∣∣ ∣∣=0, then, using properties of determinants. Prove that at least one of the following statements is true: (a) p, q, r are in G.P. (b) α is a root of the equation px2+2qx+r=0

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Solution

## LHS : Let Δ=∣∣ ∣ ∣ ∣ ∣∣(a+b)2ccca(b+c)2aabb(c+a)2b∣∣ ∣ ∣ ∣ ∣∣=2(a+b+c)3 By R1→cR1,R2→aR2,R3→bR3⇒Δ=1abc∣∣ ∣ ∣∣(a+b)2c2c2a2(b+c)2a2b2b2(c+a)2∣∣ ∣ ∣∣ By C1→C1−C3 and C2→C2−C3,⇒Δ=1abc∣∣ ∣ ∣∣(a+b+c)(a+b−c)0c20(b+c+a)(b+c−a)a2(b+c+a)(b−c−a)(b+c+a)(b−c−a)(c+a)2∣∣ ∣ ∣∣ Taking (a+b+c) common form C1 and C2 both ⇒Δ=(a+b+c)2abc∣∣ ∣ ∣∣(a+b−c)0c20(b+c−a)a2(b−c−a)(b−c−a)(c+a)2∣∣ ∣ ∣∣ By R3→R3−(R1+R2), ⇒Δ=(a+b+c)2abc∣∣ ∣ ∣∣(a+b−c)0c20(b+c−a)a2−2a−2c2ca∣∣ ∣ ∣∣ By C1→cC1,C2→aC2 ⇒Δ=(a+b+c)2abcac∣∣ ∣ ∣∣(ac+bc−c2)0c20(ba+ca−a2)a2−2ca−2ca2ca∣∣ ∣ ∣∣ By C1→C1+C3,C2→C2+C3⇒Δ=(a+b+c)2abcac∣∣ ∣ ∣∣ac+bcc2c2a2ba+caa2002ca∣∣ ∣ ∣∣ Take c,a, 2ca common from R1,R2,andR3, ⇒Δ=2c2a2(a+b+c)2abcac∣∣ ∣∣a+bccab+ca001∣∣ ∣∣ Expanding along R3, ⇒Δ=2(a+b+c)2b[0−0+1(ab+ac+b2+bc−ac)] ∴Δ=2(a+b+c)2b(ab+b2+bc)=2(a+b+c)2=RHS. OR We have ∣∣ ∣∣pqpα+qqrqα+rpα+qqα+r0∣∣ ∣∣=0 By R1→αR1,1α∣∣ ∣ ∣∣pαqαpα2+qαqrqα+rpα+qqα+r0∣∣ ∣ ∣∣=0 By R3→R3−(R1+R2),1α∣∣ ∣ ∣∣pαqαpα2+qαqrqα+r00−pα2−2qα−r∣∣ ∣ ∣∣=0 Expanding along R1,1α[0−0+(−pα2−2qα−r)(pαr−q2α)]=0 ⇒−1α(−α(q2−pr))(2qα+r+pα2)=0 Either (q2−pr)=0 or (2qα+r+pα2)=0. That implies q2=pr or p(α)2+2q(α)+r=0 Therefore either p,q,r are G.P. or α satisfies the equation px2+2qx+r=0 Hence either p,q,r are in G.P. or, α is a root of the equation px2+2qx+r=0

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