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A-1-1-4-2-A-2016-




Question Number 11035 by suci last updated on 08/Mar/17
A= [(1,(−1)),((−4),(−2)) ]  A^(2016) =.....?
A=[1142]A2016=..?
Answered by diofanto last updated on 19/Jul/17
Finding the eigenvalues of A:   determinant (((1−λ),(−1)),((−4),(−2−λ))) = 0  (1−λ)(−2−λ) − 4 = 0  λ^2  + λ − 6 = 0  λ_1  = 2, λ_2  = −3  Finding an eigenvector for each eigenvalue:  Ax = λ_1 x ⇔ (A−λ_1 I)x = 0  ⇔  [((−1),(−1)),((−4),(−4)) ] [(x_1 ),(x_2 ) ] =  [(0),(0) ]   ⇔ x_1  = −x_2 , e.g.  [((−1)),(1) ]  Ay = λ_2 y ⇔ (A−λ_2 I)y = 0  ⇔  [(4,(−1)),((−4),1) ] [(y_1 ),(y_2 ) ] =  [(0),(0) ]   ⇔ y_1  = y_2 /4, e.g.  [(1),(4) ]  Diagonalizing:  A =  [((−1),1),(1,4) ] [(2,0),(0,(−3)) ] [((−1),1),(1,4) ]^(−1)   Finding  [((−1),1),(1,4) ]^(−1) :   (((−1),1,1,0),(1,4,0,1) ) ∼  (((−1),1,1,0),(0,5,1,1) )  ∼  (((−1),0,(4/5),(−1/5)),(0,5,1,1) ) ∼  ((1,0,(−4/5),(1/5)),(0,1,(1/5),(1/5)) )  ⇔  [((−1),1),(1,4) ]^(−1)  =  [((−4/5),(1/5)),((1/5),(1/5)) ]  A^(2016)  =  [((−1),1),(1,4) ] [(2^(2016) ,0),(0,((−3)^(2016) )) ] [((−4/5),(1/5)),((1/5),(1/5)) ]  …
FindingtheeigenvaluesofA:|1λ142λ|=0(1λ)(2λ)4=0λ2+λ6=0λ1=2,λ2=3Findinganeigenvectorforeacheigenvalue:Ax=λ1x(Aλ1I)x=0[1144][x1x2]=[00]x1=x2,e.g.[11]Ay=λ2y(Aλ2I)y=0[4141][y1y2]=[00]y1=y2/4,e.g.[14]Diagonalizing:A=[1114][2003][1114]1Finding[1114]1:(11101401)(11100511)(104/51/50511)(104/51/5011/51/5)[1114]1=[4/51/51/51/5]A2016=[1114][2201600(3)2016][4/51/51/51/5]

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