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Calcule A^2 e A^3 . O que devem ser A^((2 n)) e A^((2 n + 1))

Passo 1

Saaaalve, tranquilo? Vamos lá resolver mais uma questão juntos?! 😊

A = ([[1,//,2,-,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,-,1,//,2,1,//,2]])

Então

A = ([[1,//,2,-,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,-,1,//,2,1,//,2]]) ([[1,//,2,-,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,1,//,2,-,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,1,//,2,-,1,//,2],[-,1,//,2,-,1,//,2,-,1,//,2,1,//,2]])

= ([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]])

Passo 2

Já o A^3

([[(1)/(2),-,(1)/(2),-,(1)/(2),-,(1)/(2)],[-,(1)/(2),(1)/(2),-,(1)/(2),-,(1)/(2)],[-,(1)/(2),-,(1)/(2),(1)/(2),-,(1)/(2)],[-,(1)/(2),-,(1)/(2),-,(1)/(2),(1)/(2)]])

Então podemos interpretar que A^n = ([[1,0,0,0],[0,1,0,0],[0,0,1,0],[0,0,0,1]])

E

A^((2 n + 1)) = ([[(1)/(2),-,(1)/(2),-,(1)/(2),-,(1)/(2)],[-,(1)/(2),(1)/(2),-,(1)/(2),-,(1)/(2)],[-,(1)/(2),-,(1)/(2),(1)/(2),-,(1)/(2)],[-,(1)/(2),-,(1)/(2),-,(1)/(2),(1)/(2)]])

Resposta

Resposta nos passos.