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Question

For x>0, let A=⎢ ⎢ ⎢x+1x000x00016⎥ ⎥ ⎥ and B=⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢5xx2+10003x00014⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ be two matrices.
Three other matrices X, Y and Z are defined as
X=(AB)1+(AB)2+AB3+(AB)n, Y=limnX and Z=Y12I, where I is the identity matrix of order 3. If tr(P) denotes the trace of matrix P and adj(P) denotes the adjoint of matrix P, then which of the following is/are CORRECT?

A
adj(5Y1)=5(5!)2
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B
adj(5Y1)=(5!)2
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C
The least positive integral value of tr(AY) is 7
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D
tr(Z+Z2+Z3++Z10)=211+8
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Solution

The correct options are
A adj(5Y1)=5(5!)2
C The least positive integral value of tr(AY) is 7
D tr(Z+Z2+Z3++Z10)=211+8
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⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥Y1=400020003

adj(5Y1)=(5)2 adj(Y1)=53|Y1|2=53242=5(5!)2

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

Z=Y12I=200000001Zn=2n00000001n

tr(Z+Z2+Z3++Z10)=2+22+23++210+10=2(210121)+10=211+8

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