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Question Number 123868 by benjo_mathlover last updated on 28/Nov/20

	Answered by liberty last updated on 29/Nov/20
	$we begin from A^3 =A.A^2 and A^2 =8A−13I gives A^3 =A(8A−13I)=8A^2 −13A = 8(8A−13I)−13A = 64A −104I−13A = 51A−104I xA+yI = 51A−104I we conclude that { ((x=51)),((y=−104)) :}$
	$w e b e g i n f r o m A^{3} = A . A^{2} a n d A^{2} = 8 A - 13 I$ $g i v e s A^{3} = A (8 A - 13 I) = 8 A^{2} - 13 A$ $= 8 (8 A - 13 I) - 13 A$ $= 64 A - 104 I - 13 A = 51 A - 104 I$ $x A + y I = 51 A - 104 I$ $w e c o n c l u d e t h a t {\begin{cases} x = 51 \\ y = - 104 \end{cases}$
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