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Question

Let A=207010121 and α14α7α010α4α2α. If AB=I, where I is an identity matrix of order 3 then trace B has value equal to

A
0
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B
25
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C
15
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D
5
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Solution

The correct option is A 0
x=0.47777
100x=47.777
10x=4.77790x=43
x=4390



x=0.001
x=0.001001001
100x=.100100100
100000x=100.10010099900x=99
x=9999900
x=1111100


A=207010121
B=α14α7α010α4α2α
AB=2α+07α0+0+02α+02α0+0+00+1+08α009α
α=19
trace (B)=0

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