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Question

If matrix A is an circulant matrix whose elements of first row are a,b,c>0 such that abc=1 and AτA=1 then a3+b3+c3 equals to,

A
0
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B
4
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C
1
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D
3
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Solution

The correct option is B 4
Given ATA=1
abcbcacababcbcacab=100010001
⎢ ⎢a2ababcbb2bcacacc2⎥ ⎥=100010001
a2=b2=c2=1 and ab=bc=ca=0 ...(1)
Now, a3+b3+c3=(a+b+c)(a2+b2+c2abbcca)+3abc
a3+b3+c3=(a+b+c)[10]+3(1)
a3+b3+c3=(a+b+c)+3 ....(2)
Now, (a+b+c)2=a2+2ab=1
a+b+c=1.
Substitute this value in (2)
a3+b3+c3=1+3
a3+b3+c3=4

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