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Let `omega`be a complex cube root of unity with `omega!=1a n dP=[p_(i j)]`be a `nxxn`matrix withe `p_(i j)=omega^(i+j)dot`Then `p^2!=O ,w h e nn=`a.`57`b. `55`c. `58`d. `56`A. 57B. 55C. 58D. 56 |
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Answer» Correct Answer - A We have, `P=[P_(ij)]_(nxxx)` `:. (P^2)_(ij)=sum_(r=1)^n p ir" " prj=sum_(r=1)^nomega^(i+r)omega^(r+j)=sum_(r-1)^nomega^(i+j+2r)` `=omega^(i+j)sum_(r=1)^n(omega^2)^r=omega^(i+j)[omega^2{((omega^2)^n-1)/(omega^2-1)}]` `=(omega^(i+j+2))/(omega^2-1)(omega^(2n)-1)` Now, `P^2=O` `rArr (P^2)_(ij)=0" for all " o,j` `rArr (omega^(i+j+2))/(omega^2-1)(omega^(2n)-1)=0" for all "i,j` `rArr omega^(2n)=1` `rArr " n is a multiple of 3"` Clearly, 57 is a multiple of 3, So, option (a) is correct. |
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