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If `{:P=[(sqrt3/2,1/2),(1/2,sqrt3/2)],A=[(1,1),(0,1)]:}and Q=PAP^T," then" P^TQ^2015P`, isA. `{:[(1,2005),(0,1)]:}`B. `{:[(1,2005),(2005,1)]:}`C. `{:[(1,0),(2005,1)]:}`D. `{:[(1,0),(0,1)]:}` |
Answer» Correct Answer - A We have, `P P^T={:[(sqrt3/2,1/2),(-1/2,sqrt3/2)][(sqrt3/2,-1/sqrt2),(1/2,sqrt3/2)]=[(1,0),(0,1)]=I:}` `:. P^T=P^(-1)` Now, `Q=PAP^T` `rArr P^TQP=P^T(PAP^T)P` `rArr P^TQP=(P^TP)A(P^TP)=IAI=A={:[(1,1),(0,1)]:}` `and, Q_2=(PAP^T)(PAP^T)=PA(P^TP)AP^T=PA^2P^T` `:. P^TQ^2P=P^T(PA^2P^T)P=(P^TP)A^2(P^TP)=IA^2` `rArr P^TQ^2P=A^2={:[(1,2),(0,1)]:}` Continuing in this manner, we have `P^TQ^2005P=A^2005=`{:[(1,2005),(0,1)]:}` |
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