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If `P=[[sqrt(3)/2,1/2],[-1/2,sqrt(3)/2]], A=[[1,1],[0,1]]` and `Q=PAP^T`, then `P^TQ^(2015)P` is |
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Answer» `P = [[sqrt3/2,1/2],[-1/2, sqrt3/2]]` `:. P^T = [[sqrt3/2,-1/2],[1/2, sqrt3/2]]` Now, `P^TP = [[1,0],[0,1]] = I` Now, `P^TQ^2015P = P^T(PAP^T)^2015P = P^T(PAP^T)(PAP^T)(PAP^T)...P = A^2015` Here, `A = [[1,1],[0,1]]` `A^2 = A.A = [[1,1],[0,1]][[1,1],[0,1]] = [[1,2],[0,1]]` `A^3 = A^2.A = [[1,2],[0,1]][[1,1],[0,1]] = [[1,3],[0,1]]` `:. A^2015 = [[1,2015],[0,1]]` `:. P^TQ^2015P = [[1,2015],[0,1]]` |
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