Enunciado

Verifique que det (A B) = d e t ( B A ) e determine se vale a igualdade det (A + B) = det (A) + d e t ( B ) .

[[A,=,([[2,1,0],[3,4,0],[0,0,2]]),B,=,([[1,-,1,3],[7,1,2],[5,0,1]])]]

Passo 1

Dadas as matrizes A = ([[2,1,0],[3,4,0],[0,0,2]]) e B = ([[1,-,1,3],[7,1,2],[5,0,1]]) vamos primeiro determinar os produtos A B e B A .

  • Resolvendo A B
  • A B = ([[2,1,0],[3,4,0],[0,0,2]]) ([[1,-,1,3],[7,1,2],[5,0,1]])

    A B = ([[2,+,7,+,0,-,2,+,1,+,0,6,+,2,+,0],[3,+,28,+,0,-,3,+,4,+,0,9,+,8,+,0],[0,+,0,+,10,0,+,0,+,0,0,+,0,+,2]])

    A B = ([[9,-,1,8],[31,1,17],[10,0,2]])

  • Resolvendo B A

B A = ([[1,-,1,3],[7,1,2],[5,0,1]]) ([[2,1,0],[3,4,0],[0,0,2]])

B A = ([[2,-,3,+,0,1,-,4,+,0,0,+,0,+,6],[14,+,3,+,0,7,+,4,+,0,0,+,0,+,4],[10,+,0,+,0,5,+,0,+,0,0,+,0,+,2]])

B A = ([[-,1,-,3,6],[17,1,1,4],[10,5,2]])

Passo 2

Vamos usar o Teorema de Laplace e verificar se os determinantes são iguais:

det (A B) = 9 |[[1,17],[0,2]]| - ( - 1 ) |[[31,17],[10,2]]| + 8 |[[31,1],[10,0]]|

det (A B) = 9 ((1 xx 2) - (17 xx 0)) + ((31 xx 2) - (17 xx 10)) + 8 ((31 xx 0) - (1 xx 10))

det (A B) = 9 ((2) - (0)) + ((62) - (170)) + 8 ( (0) - (10) )

det (A B) = 18 - 108 - 80

det (A B) = - 170

det (B A) = ( - 1 ) |[[11,4],[5,2]]| - ( - 3 ) |[[17,4],[10,2]]| + 6 |[[17,11],[1,0,5]]|

det (B A) = (- 1) ((11 xx 2) - (4 xx 5)) + 3 ((17 xx 2) - (4 xx 10)) + 6 ((17 xx 5) - (11 xx 10))

det (B A) = (- 1) ((22) - (2 0)) + 3 ((34) - (4 0)) + 6 ((85) - (110))

det (B A) = - 2 - 18 - 150

det (B A) = - 170

Passo 3

Vamos achar agora a matriz A + B

A + B = ([[2,1,0],[3,4,0],[0,0,2]]) + ([[1,-,1,3],[7,1,2],[5,0,1]])

A + B = ([[2,+,1,1,+,(,-,1,),0,+,3],[3,+,7,4,+,1,0,+,2],[0,+,5,0,+,0,2,+,1]])

A + B = ([[3,0,3],[10,5,2],[5,0,3]])

Passo 4

Vamos agora comparar d e t ( A + B ) e det (A) + det (B)

det (A + B) = 3 |[[5,2],[0,3]]| - 0 |[[10,2],[5,3]]| + 3 |[[10,5],[5,0]]|

det (A + B) = 3 ((5 xx 3) - (2 xx 0)) + 3 ((10 xx 0) - (5 xx 5))

det (A + B) = 3 ((15) - (0)) + 3 ((0) - (25))

det (A + B) = 45 - 75

det (A B) = - 30

det (A) = 2 |[[4,0],[0,2]]| - 1 |[[3,0],[0,2]]| + 0 |[[3,4],[0,0]]|

det (A) = 2 ((4 xx 2) - (0 xx 0)) - 1 ((3 xx 2) - (0 xx 0))

det (A) = 16 - 6

det (A) = 10

det (B) = 1 |[[1,2],[0,1]]| - ( - 1 ) |[[7,2],[5,1]]| + 3 |[[7,1],[5,0]]|

det (B) = 1 ((1 xx 1) - (2 xx 0)) + ((7 xx 1) - (2 xx 5)) + 3 ((7 xx 0) - (1 xx 5))

det (B) = 1 - 3 - 15

det (B) = - 17

Logo, det (A + B) != det (A) + det (B)

Resposta

Logo, det (A B) = d e t ( B A ) e det (A + B) != det (A) + det (B)

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