InterviewSolution
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InterviewSolution
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let `P_(1)=[(1,0,0),(0,1,0),(0,0,1)],P_(2)=[(1,0,0),(0,0,1),(0,1,0)],P_(3)=[(0,1,0),(1,0,0),(0,0,1)],P_(4)=[(0,1,0),(0,0,1),(1,0,0)],P_(5)=[(0,0,1),(1,0,0),(0,1,0)],P_(6)=[(0,0,1),(0,1,0),(1,0,0)]` and `X=sum_(k=1)^(6) P_(k)[(2,1,3),(1,0,2),(3,2,1)]P_(k)^(T)` Where `P_(k)^(T)` is transpose of matrix `P_(k)`. Then which of the following options is/are correct?A. X is a symmetric matrixB. if `X=[(1),(1),(1)]=alpha[(1),(1),(1)]`, then `alpha=30`C. X-30I is an invertible matrixD. The sum of diagonal entries of X is 18. |
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Answer» Correct Answer - A::B::D Clearly `P_(1)=P_(1)^(T)=P_(1)^(-1)` `P_(2)=P_(2)^(T)=P_(2)^(-1)` . . . `P_(6)=P_(6)^(T)=P_(6)^(-1)` And `A^(T)=A` where `A=[(2,1,3),(1,0,2),(3,2,1)]` Using formula `(A+B)^(T)=A^(T)+B^(T)` `X^(T)=(P_(1)AP_(1)^(T)+ . . . .+P_(6)AP_(6)T)^(T)=P_(1)A^(T)P_(1)^(T)+ . . . . . +P_(6)A^(T)P_(6)^(T)=XimpliesX` is symmetric Let `B=[(1),(1),(1)]` `XB=P_(1)AP_(1)^(T)B+P_(2)AP_(2)^(T)B+. . . .+P_(6)AP_(6)^(T)B=P_(1)AB+PAB+. . . . +P_(6)AB` `XB=(P_(1)+P_(2)+. . . .+P_(6))[(6),(3),(6)]` `=[(6xx2+3xx2+6xx2),(6xx2+3xx2+6xx2),(6xx2+3xx2+6xx2)]=[(30),(30),(30)]=30Bimpliesalpha=30` since X`[(1),(1),(1)=30[(1),(1),(1)]` `implies(X-30I)B=0` has a non trivial solution `B=[(1),(1),(1)]` `implies|X-30I|=0` `X=P_(1)P_(1)AP_(1)^(T)+. . . .+P_(6)AP_(6)^(T)` trace `(x)=t_(r)(P_(1)AP_(1)^(T))+. . . . .+t_(r)(P_(6)AP_(6)^(T))=(2+0+1)+. . . . +(2+0+1)=3xx6=18` |
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