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Question

Let A=207010121 and B=x14x7x010x4x2x be two matrices such that AB=(AB)1 and ABI, where I is an identity matrix of order 3×3. Then the value of tr(AB+(AB)2+(AB)3++(AB)100) is
( Here, tr(A) denotes the trace of matrix A, i.e., sum of diagonal elements of A.)

A
150
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B
300
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C
200
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D
100
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Solution

The correct option is D 100
AB=207010121x14x7x010x4x2x

AB=5x00010010x25x

(AB)2=25x200010010x2+50x210x25x2
(AB)2=I
25x2=1 and 50x22=0
x=±15

At x=15,AB=I but ABI
x=15
and AB=100010041
(AB)2=I (AB)3=AB
(AB)4=(AB)2=I
(AB)2n=I and (AB)2n1=AB, nN
AB+(AB)2+(AB)3++(AB)100
=AB+I+AB+I++I
=50AB+50I
Hence, tr(AB+(AB)2+(AB)3++(AB)100)
=tr(50AB+50I)
=tr(50AB)+tr(50I)
=50(1+11)+50(1+1+1)
=100

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