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Question

What is the value of the determinant ∣ ∣ ∣1bca(b+c)1cab(c+a)1abc(a+b)∣ ∣ ∣?

A
0
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B
abc
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C
ab+bc+ca
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D
abc(a+b+c)
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Solution

The correct option is A 0
Now, in the matrix
∣ ∣1bcab+ac1acbc+ba1abca+cb∣ ∣

C3C3+C2

∣ ∣1bcab+ac+bc1acab+ac+bc1abab+ac+bc∣ ∣ taking (ab+bc+ac) common in C3

(ab+bc+ac)∣ ∣1bc11ac11ab1∣ ∣
Now two columns of the matrix are same so it's determinant will be 0.
(a)

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