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Question
If Δ is an operation such that for integers a and b we have a Δ b = a × b – 2 × a × b + b × b (–a) × b + b × b then find 4 Δ (–3) Also show that 4 ∆ (–3) ≠ (–3) ∆ 4.
Solution
We have a Δ b = a × b – 2 × a × b + b × (b) (–a) × b + b × b
Now, put a = 4 and b = (–3)
4 Δ (–3) = 4 × (–3) – 2 × 4 × (–3) + (–3) × (–3)(– 4) × (–3) + (–3) × (–3)
= –12 – 2 × (–12) + (9)(12) + 9
= –12 + 24 + 108 + 9
= –12 + 141
= 129
Now, put a = –3 and b = 4
⇒ (–3) Δ 4 = (–3) × 4 –2 × (–3) × (4) + 4 × 4{–(–3)} × 4 + 4 × 4
= (–12) + 24 + 16(3) × 4 + 16
= (–12) + 24 + 192 + 16
= 220
Clearly, 4 Δ (–3) ≠ (–3) Δ 4
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