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Question

For x>0, let A=⎢ ⎢ ⎢x+1x000x00016⎥ ⎥ ⎥ and B=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢5xx2+10003x00014⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ be two matrices.
Two other matrices X and Y are defined as
X=(AB)1+(AB)2+AB3+(AB)n and Y=limnX. If tr(P) denotes the trace of matrix P, then the least positive integral value of tr(AY) is

A
5
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B
6
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C
7
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D
8
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Solution

The correct option is B 6
AB=⎢ ⎢ ⎢x2+1x000x00016⎥ ⎥ ⎥⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢5xx2+10003x00014⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥AB=500030004

(AB)1=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢150001300014⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

(AB)2=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢152000132000142⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥
and so on.

X=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢15+152++15n00013+132++13n00014+142++14n⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥Y=limnX=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢151150001311300014114⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥Y=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢140001200013⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥

AY=⎢ ⎢ ⎢x+1x000x00016⎥ ⎥ ⎥⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢140001200013⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥AY=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢14(x+1x)000x2000163⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥tr(AY)=x4+14x+x2+163tr(AY)=3x4+14x+163
As x>0, using A.M.G.M., we get
3x4+14x23163x4+14x32

So, tr(AY)32+163
Least positive integral value of tr(AY)=7

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