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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. `{:[(2015,1),(1,2015)]:}`B. `{:[(1,2015),(0,1)]:}`C. `{:[(0,2015),(0,0)]:}`D. `{:[(2015,0),(1,2015)]:}` |
Answer» Correct Answer - B We have, `{:P=[(sqrt3/2,1/2),(-1/2,sqrt3/2)]rArr P^T=[(sqrt3/2,-1/2),(1/2,sqrt3/2)]:}` Clearly, `P^T=P^(-1)` `:. Q=PAP^T=PAP^(-1)` `rArr Q^2015=(PAP^(-1))^2015=PA^2015P^(-1)` `:. P^TQ^2015P=P^(-1)(PA^2015P^(-1))P=A^2015` Now, `A={:[(1,1),(0,1)]:}` `:.{:A^2=[(1,1),(0,1)][(1,1),(0,1)]=[(1,2),(0,1)]:}` `{:A^3=A^2A=[(1,2),(0,1)][(1,1),(0,1)]=[(1,3),(0,1)]:}`and so no. In general `{:A^2015=[(1,2015),(0,1)]:}` Hence, `P^TQ^2015P=A^2015={:[(1,2015),(0,1)]:}`. |
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