Enunciado

A matriz A do Exercício 21.

Passo 1

A matriz do Exercício 21 é dada por:

A = [[[3,0,0],[0,-,1,3],[0,-,3,-,1]]]

Como a gente quer testar a validade de p_1 (A) = p_2 (A) p_3 ( A ) . Vamos encontrar os termos separados e depois comparar.

Passo 2

Como p_1 (x) = x^2 - 9 , adaptando pra forma matricial ficamos com p_1 (A) = A^2 - 9 I .

A^2 = [[[9,0,0],[0,-,8,-,6],[0,6,-,8]]]

9 I = [[[9,0,0],[0,9,0],[0,0,9]]]

p_1 (A) = A^2 - 9 I => [[[9,0,0],[0,-,8,-,6],[0,6,-,8]]] - [[[9,0,0],[0,9,0],[0,0,9]]] = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

Passo 3

Belezinha, agora falta encontrar a outra parte da igualdade, p_2 (A) p_3 ( A ) , como p_2 (x) = x + 3 e p_3 (x) = x - 3 , adaptando a equação pra forma matricial vamos ficar com p_2 (A) = A + 3 e p_3 (A) = A - 3 .

p_2 (A) = A + 3 => [[[3,0,0],[0,-,1,3],[0,-,3,-,1]]] - [[[3,0,0],[0,3,0],[0,0,3]]] = [[[0,0,0],[0,-,4,3],[0,-,3,-,4]]]

p_3 (A) = A - 3 => [[[3,0,0],[0,-,1,3],[0,-,3,-,1]]] + [[[3,0,0],[0,3,0],[0,0,3]]] = [[[6,0,0],[0,2,3],[0,-,3,2]]]

p_2 (A) p_3 (A) = [[[0,0,0],[0,-,4,3],[0,-,3,-,4]]] [[[6,0,0],[0,2,3],[0,-,3,2]]] = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

Passo 4

Agora que descobrimos os elementos dos dois lados, podemos comparar

p_1 (A) = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

p_2 (A) p_3 (A) = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

p_1 (A) = p_2 (A) p_3 (A) <=> [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]] = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

São EXATAMENTE iguais!!!

Resposta

p_1 (A) = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

p_2 (A) p_3 (A) = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

p_1 (A) = p_2 (A) p_3 (A) <=> [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]] = [[[0,0,0],[0,-,17,-,6],[0,6,-,17]]]

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