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Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Jurassic World Alive From The Cutting Room Floor Jump to: navigation, search Jurassic World Alive Developer : Ludia Publisher : Ludia Platforms : Android , iOS Released internationally : May 24, 2018 This game has unused areas. This game has unused playable characters. This game has unused code. This game has revisional differences. This game is still under active development. Be aware that any unused content you find may become used or removed in the future. Please only add things to the article that are unlikely to ever be used, or went unused for some time. If they do get used, please remove them from the page and specify in the edit summary! To do: Intro More test files exist Find out what the test files do or if they still work Contents ...

.mw-parser-output .nota-disambigua{clear:both;margin-bottom:.5em;border:1px solid #CCC;padding-left:4px}.mw-parser-output .nota-disambigua i{vertical-align:middle} Disambiguazione – Se stai cercando il comune del Massachusetts, vedi Sherborn . Questa voce sull'argomento centri abitati dell'Inghilterra è solo un abbozzo . Contribuisci a migliorarla secondo le convenzioni di Wikipedia. Sherborne parrocchia civile Localizzazione Stato   Regno Unito       Inghilterra Regione Sud Ovest Contea Dorset Distretto West Dorset Territorio Coordinate 50°57′N 2°31′W  /  50.95°N 2.516667°W 50.95; -2.516667  ( Sherborne ) Coordinate: 50°57′N 2°31′W  /  50.95°N 2.516667°W 50.95; -2.516667  ( Sherborne ) Abitanti 9 350 (2001) Altre informazioni Cod. postale DT9 Prefisso 01935 Fuso orario UTC+0 Cartografia Sherborne Sito istituzionale Modifica dati su Wikidata  · Manuale Sherborne è un paese di 9.350 abita...

0 1 $begingroup$ Let $$A= begin{bmatrix} 2&2&3\ 1&3&3\ -1&-2&-2 end{bmatrix} . $$ Find the Jordan Form, $J$ , of this matrix, and an invertible matrix $Q$ such that $A = QJQ^{-1}$ . I have already found the Jordan Form of this matrix, that is, $$J = begin{bmatrix}1&0&0\0&1&1\0&0&1end{bmatrix}.$$ The part that I am confused about is finding the matrix $Q$ . I know that the columns of $Q$ will consist of the eigenvectors, and generalized eigenvectors of $A - lambda I$ . The characteristic polynomial of $A$ is $$p_A(lambda) = lambda^3 - 3lambda^2 + 3lambda + 1 = (lambda - 1)^3.$$ I have found the eigenvector associated with $lambda = 1$ to be $$v = (-5, 1, 1).$$ However, $(A - I)^2 = 0$ , so I am confused on how to find the generalized eigenvec...