Seja A(t)=(ccce^t & 2 e^-t & e^2 t 2 e^t & e^-t & -e^2 t -e^t & 3 e^-t & 2 e^2 t) e B(t)=(ccc2 e^t & e^-t & 3 e^2 t -e^t & 2 e^-t & e^2 t 3 e^t & -e^-t & -e^2 t), encontre
A B

Passo 1

Estamos multiplicando duas matrizes 3times3 que dependem de t:
A(t)=l(aijr), B(t)=l(bjkr),
onde
(AB)ik=sumj=1^3aij,bjk
Vamos usar exatamente essa definição para cada entrada do produto.

Passo 2

Extraímos as linhas de A e as colunas de B:
– Linha 1 de A: l(e^t,;2e^-t,;e^2tr)
– Linha 2 de A: l(2e^t,;e^-t,;-e^2tr)
– Linha 3 de A: l(-e^t,;3e^-t,;2e^2tr)
– Coluna 1 de B: l(2e^t,;-e^t,;3e^tr)^T
– Coluna 2 de B: l(e^-t,;2e^-t,;-e^-tr)^T
– Coluna 3 de B: l(3e^2t,;e^2t,;-e^2tr)^T

Passo 3

Calculamos a primeira linha de AB:
(AB)11 &=e^tcdot2e^t+2e^-tcdot(-e^t)+e^2tcdot3e^t =2e^2t-2+3e^3t, (AB)12 &=e^tcdot e^-t+2e^-tcdot2e^-t+e^2tcdot(-e^-t) =1+4e^-2t-e^t, (AB)13 &=e^tcdot3e^2t+2e^-tcdot e^2t+e^2tcdot(-e^2t) =3e^3t+2e^t-e^4t

Passo 4

Calculamos a segunda linha de AB:
(AB)21 &=2e^tcdot2e^t+e^-tcdot(-e^t)+(-e^2t)cdot3e^t =4e^2t-1-3e^3t, (AB)22 &=2e^tcdot e^-t+e^-tcdot2e^-t+(-e^2t)cdot(-e^-t) =2+2e^-2t+e^t, (AB)23 &=2e^tcdot3e^2t+e^-tcdot e^2t+(-e^2t)cdot(-e^2t) =6e^3t+e^t+e^4t

Passo 5

Calculamos a terceira linha de AB:
(AB)31 &=-e^tcdot2e^t+3e^-tcdot(-e^t)+2e^2tcdot3e^t =-2e^2t-3+6e^3t, (AB)32 &=-e^tcdot e^-t+3e^-tcdot2e^-t+2e^2tcdot(-e^-t) =-1+6e^-2t-2e^t, (AB)33 &=-e^tcdot3e^2t+3e^-tcdot e^2t+2e^2tcdot(-e^2t) =-3e^3t+3e^t-2e^4t

Resposta

A(t),B(t) = 3e^3t+2e^2t-2 & 1+4e^-2t-e^t & 3e^3t+2e^t-e^4t[6pt] -3e^3t+4e^2t-1 & 2+2e^-2t+e^t & 6e^3t+e^t+e^4t[6pt] 6e^3t-2e^2t-3 & -1+6e^-2t-2e^t & -3e^3t+3e^t-2e^4t ]

Exercícios relacionados