InterviewSolution
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InterviewSolution
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Let P_1=I=[{:(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)denotes the transpose of the matrix P_(X). then which of the following option is/are correct ? |
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Answer» X is a symmertic matrix `P_(1) = I = [{:(,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),(,0,0,1),(,0,1,0):}], P_(6)=[{:(,0,0,1),(,0,1,0),(,0,1,0):}]` and `X = underset(k-1)overset(6)sumP_(k) [{:(,2,1,3),(,1,0,2),(,3,2,1):}] P_(K)^(T)` `because P_(1)^(T) = P_(1),P_(2)^(T)= P_(2),P_(3)^(T) = P_(3), P_(4)^(T) = P_(5), P_(5)^(T) = P_(4)` and `P_(6)^(T) = P_(6)` and Let Q`= [{:(,2,1,3),(,1,0,2),(,3,2,1):}] ANDBECAUSE Q^(T) = Q` Now,`X= (P_(1)QP_(1)^(T)) + (P_(2)QP_(2)^(T))^(T) + (P_(3)QP_(3)^(T)) + (P_(4)QP_(4)^(T))+(P_(5)QP_(5)^(T))+(P_(6)QP_(6)^(T))` So, `X^(T) = (P_(1)QP_(1)^(T))^(T) + (P_(2)QP_(2)^(T))^(T)+(P_(3)QP_(3)^(T))^(T) + (P_(4)QP_(4)^(T))^(T) + (P_(5)QP_(5)^(T))^(T)+(P_(6)QP_(6)^(T))^(T)` `P_(1)QP_(1)^(T) + P_(2)QP_(2)^(T)+P_(3)QP_(3)^(T) + P_(4)QP_(4)^(T) + P_(5)QP_(5)^(T)+P_(6)QP_(6)^(T)` `[because(ABC)^(T) = C^(T)B^(T)A^(T) and (A^(T))^(T) = A and Q^(T) = Q]` `rArr X^(T) = X` `rArr` X is a symmetric matrix. The sum ofdiagonal entries of `X = Tr(x)` `=sum_(i=1)^(6) T_(R)(P_(i)QP_(i)^(T))` `= sum_(i-1)^(6) T_(r)(QP_(i)^(T)P_(i))""[becauseT_(r)(ABC) = T_(r)(BCA)]` `= sum_(i=l)^(6) T_(r)(Ql) ""[because P_(i)`'s "are othogonal matrices"]` `=sum_(i=1)^(6) T_(r)(Q)= 6T_(r)(Q)= 6xx3 = 18` `Now Let `R= [{:(,1),(,1),(,1):}]`, then `XR = sum_(k=1)^(6) (P_(k)QP_(k)^(T)) R = sum_(k=1)^(6) (P_(k) QP_(k)^(T)R)` `= sum_(k=1)^(6)(P_(k) QR)""[becauseP_(k)^(T)R = R]` `= sum_(k=1)^(6) P_(k)[{:(6),(3),(6):}] = sum_(k=1)^(6) P_(k) [{:(6),(3),(6):}] = [{:(2,2,2),(2,2,2),(2,2,2):}][{:(6),(3),(6):}]` `rArr XR = [{:(30),(30),(30):}]rArr XR = 30 R rArr X [{:(1),(1),(1):}]= 30[{:(1),(1),(1):}]` `rArr (X-30I0 R = 0 rArr |X-30I|=0` So, `(X - 30I)` is notinvertible and value of `alpha = 30`. Hence, options (a),(b) and (c) are correct. |
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