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Question

Using properties of determinants, prove that ∣ ∣ ∣ ∣ ∣(a+b)2ccca(b+c)2aabb(c+a)2b∣ ∣ ∣ ∣ ∣=2(a+b+c)3

OR

If p0,q0 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


Solution

LHS : Let Δ=∣ ∣ ∣ ∣ ∣(a+b)2ccca(b+c)2aabb(c+a)2b∣ ∣ ∣ ∣ ∣=2(a+b+c)3

By R1cR1,R2aR2,R3bR3Δ=1abc∣ ∣ ∣(a+b)2c2c2a2(b+c)2a2b2b2(c+a)2∣ ∣ ∣

By C1C1C3 and C2C2C3,Δ=1abc∣ ∣ ∣(a+b+c)(a+bc)0c20(b+c+a)(b+ca)a2(b+c+a)(bca)(b+c+a)(bca)(c+a)2∣ ∣ ∣

Taking (a+b+c) common form C1 and C2 both

Δ=(a+b+c)2abc∣ ∣ ∣(a+bc)0c20(b+ca)a2(bca)(bca)(c+a)2∣ ∣ ∣

By R3R3(R1+R2),
Δ=(a+b+c)2abc∣ ∣ ∣(a+bc)0c20(b+ca)a22a2c2ca∣ ∣ ∣

By C1cC1,C2aC2
Δ=(a+b+c)2abcac∣ ∣ ∣(ac+bcc2)0c20(ba+caa2)a22ca2ca2ca∣ ∣ ∣

By C1C1+C3,C2C2+C3Δ=(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[00+1(ab+ac+b2+bcac)]

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 R1αR1,1α∣ ∣ ∣pαqαpα2+qαqrqα+rpα+qqα+r0∣ ∣ ∣=0

By R3R3(R1+R2),1α∣ ∣ ∣pαqαpα2+qαqrqα+r00pα22qαr∣ ∣ ∣=0

Expanding along R1,1α[00+(pα22qαr)(pαrq2α)]=0

1α(α(q2pr))(2qα+r+pα2)=0

Either (q2pr)=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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