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Question

Evaluate the following :
∣ ∣1abc1bca1cab∣ ∣

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Solution

Let Δ=∣ ∣1abc1bca1cab∣ ∣.

Applying R2R2R1,R3R3R1, we get

Δ=∣ ∣1abc0bacabc0caabbc∣ ∣ =∣ ∣ ∣1abc0bac(ab)0cab(ac)∣ ∣ ∣

Δ=(ab)(ac)∣ ∣1abc01c01b∣ ∣ [Taking (a-b) and (a-c)common from R1 and R3 respectively.]

Δ=(ab)(ac)(b+c) [Expanding along first column]

Δ=(ab)(bc)(ca)

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