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Question

By using properties of determination, show that
∣ ∣ ∣1+a2b22ab2b2ab1a2+b22b2b2a1a2b2∣ ∣ ∣=(1+a2+b2)3

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Solution

∣ ∣ ∣1+a2b22ab2b2ab1a2+b22b2b2a1a2b2∣ ∣ ∣

c1c1bc3; c2c2+ac3
∣ ∣ ∣1+a2+b202b01+a2+b22a2bb(1a2b2)2a+a(1a2b2)1a2b2∣ ∣ ∣=∣ ∣ ∣1+a2+b202b01+a2+b22a2bb+b(a2+b2)2a+aa(a2+b2)1a2b2∣ ∣ ∣

=∣ ∣ ∣1+a2+b202b01+a2+b22ab(1+a2+b2)a(1+a2+b2)1a2b2∣ ∣ ∣=(1+a2+b2)2∣ ∣ ∣102b012aba1a2b2∣ ∣ ∣

=(1+a2+b2)2(1+a2+b2)=(1+a2+b2)3

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