Let a,b,c∈R be such that a+b+c>0 and abc=2. Let A=⎡⎢⎣abcbcacab⎤⎥⎦ If A2=I, then value of a3+b3+c3 is
A
7
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B
2
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C
0
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D
-1
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Solution
The correct option is D 7 A2=I ⇒a2+b2+c2=1 and bc+ca+ab=0 Also, |A2|=1⇒|A|=±1. But A|=3abc−(a3+b3+c3) =−(a+b+c)(a2+b2+c2−bc−ca−ab) =−(a+b+c) As a+b+c>0, we get |A|=−1. ∴a3+b3+c3=7