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Question

Let Δ=∣ ∣ ∣a2abababa2abababa2∣ ∣ ∣
Prove that Δ is non-negative and establish the relation between a,b and c if Δ=0.

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Solution

Apply C1+C2+C3 and a2+2ab=(a)2 be
taken out from new C1.
Δ=(a)2∣ ∣ ∣1abab1a2ab1aba2∣ ∣ ∣
Now making two zeros by R2R1 and R3R1
Δ=(a)2[a2ab]2
Δ being perfect square of real quantities is always +ive.
Δ=0 if either a=0 or a2ab=0
i.e. 12[(ab)2+(bc)2+(ca)2]=0
Above will hold only when ab=0, bc=0, ca=0 or a=b=c.

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