Sejam
E=\left[\begin{array}{ccc} 1 & 0 & 2 \\\\ -1 & 3 & 1 \end{array}\right], \quad F=\left[\begin{array}{ccc} 2 & 1 & 0 \\\\ 0 & -1 & 4 \\\\ 3 & 0 & 1 \end{array}\right], \quad G=\left[\begin{array}{ccc} 1 & 2 & -1 \\\\ 0 & 1 & 1 \\\\ 2 & -2 & 0 \end{array}\right]
Mostre que (E F) G=E(F G).

Passo 1

Resumindo o enunciado: temos as matrizes
E = \begin{pmatrix}1 & 0 & 2\\\\ -1 & 3 & 1\end{pmatrix},\quad F = \begin{pmatrix}2 & 1 & 0\\\\ 0 & -1 & 4\\\\ 3 & 0 & 1\end{pmatrix},\quad G = \begin{pmatrix}1 & 2 & -1\\\\ 0 & 1 & 1\\\\ 2 & -2 & 0\end{pmatrix}.
O objetivo é mostrar que a multiplicação de matrizes é associativa neste caso, ou seja,
(EF)\,G \;=\; E\,(F G).
Para isso, vamos calcular explicitamente EF, depois (EF)G, por um lado, e FG, depois E(FG), por outro, e comparar os resultados.

Passo 2

Cálculo de EF. Como E é 2\times3 e F é 3\times3, o produto EF é 2\times3.
Linha 1 de EF:
\bigl[1,0,2\bigr]\begin{pmatrix}2\\\ 0\\\ 3\end{pmatrix} =1\cdot2+0\cdot0+2\cdot3=8,\quad \bigl[1,0,2\bigr]\begin{pmatrix}1\\\ -1\\\ 0\end{pmatrix} =1\cdot1+0\cdot(-1)+2\cdot0=1,
\bigl[1,0,2\bigr]\begin{pmatrix}0\\\ 4\\\ 1\end{pmatrix} =1\cdot0+0\cdot4+2\cdot1=2.
Linha 2 de EF:
\bigl[-1,3,1\bigr]\begin{pmatrix}2\\\ 0\\\ 3\end{pmatrix} =-1\cdot2+3\cdot0+1\cdot3=1,\quad \bigl[-1,3,1\bigr]\begin{pmatrix}1\\\ -1\\\ 0\end{pmatrix} =-1\cdot1+3\cdot(-1)+1\cdot0=-4,
\bigl[-1,3,1\bigr]\begin{pmatrix}0\\\ 4\\\ 1\end{pmatrix} =-1\cdot0+3\cdot4+1\cdot1=13.
Logo,
EF = \begin{pmatrix}8 & 1 & 2\\[6pt]1 & -4 & 13\end{pmatrix}.]

Passo 3

Cálculo de (EF)G. Agora (EF) é 2\times3 e G é 3\times3, então (EF)G é 2\times3.
Linha 1 de (EF)G:
\bigl[8,1,2\bigr]\begin{pmatrix}1\\\ 0\\\ 2\end{pmatrix} =8\cdot1+1\cdot0+2\cdot2=12,\quad \bigl[8,1,2\bigr]\begin{pmatrix}2\\\ 1\\\ -2\end{pmatrix} =8\cdot2+1\cdot1+2\cdot(-2)=13,
\bigl[8,1,2\bigr]\begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix} =8\cdot(-1)+1\cdot1+2\cdot0=-7.
Linha 2 de (EF)G:
\bigl[1,-4,13\bigr]\begin{pmatrix}1\\\ 0\\\ 2\end{pmatrix} =1\cdot1+(-4)\cdot0+13\cdot2=27,\quad \bigl[1,-4,13\bigr]\begin{pmatrix}2\\\ 1\\\ -2\end{pmatrix} =1\cdot2+(-4)\cdot1+13\cdot(-2)=-28,
\bigl[1,-4,13\bigr]\begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix} =1\cdot(-1)+(-4)\cdot1+13\cdot0=-5.
Portanto,
(EF)\,G = \begin{pmatrix}12 & 13 & -7\\[6pt]27 & -28 & -5\end{pmatrix}.]

Passo 4

Cálculo de FG. Como F e G são 3\times3, o produto FG é 3\times3.
Linha 1 de FG:
\bigl[2,1,0\bigr]\begin{pmatrix}1\\\ 0\\\ 2\end{pmatrix}=2,\quad \bigl[2,1,0\bigr]\begin{pmatrix}2\\\ 1\\\ -2\end{pmatrix}=5,\quad \bigl[2,1,0\bigr]\begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix}=-1.
Linha 2 de FG:
\bigl[0,-1,4\bigr]\begin{pmatrix}1\\\ 0\\\ 2\end{pmatrix}=8,\quad \bigl[0,-1,4\bigr]\begin{pmatrix}2\\\ 1\\\ -2\end{pmatrix}=-9,\quad \bigl[0,-1,4\bigr]\begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix}=-1.
Linha 3 de FG:
\bigl[3,0,1\bigr]\begin{pmatrix}1\\\ 0\\\ 2\end{pmatrix}=5,\quad \bigl[3,0,1\bigr]\begin{pmatrix}2\\\ 1\\\ -2\end{pmatrix}=4,\quad \bigl[3,0,1\bigr]\begin{pmatrix}-1\\\ 1\\\ 0\end{pmatrix}=-3.
Assim,
F\,G = \begin{pmatrix}2 & 5 & -1\\[6pt]8 & -9 & -1\\[6pt]5 & 4 & -3\end{pmatrix}.]

Passo 5

Cálculo de E(FG). Agora E é 2\times3 e FG é 3\times3, então E(FG) é 2\times3.
Linha 1 de E(FG):
\bigl[1,0,2\bigr]\begin{pmatrix}2\\\ 8\\\ 5\end{pmatrix} =1\cdot2+0\cdot8+2\cdot5=12,\quad \bigl[1,0,2\bigr]\begin{pmatrix}5\\\ -9\\\ 4\end{pmatrix} =1\cdot5+0\cdot(-9)+2\cdot4=13,
\bigl[1,0,2\bigr]\begin{pmatrix}-1\\\ -1\\\ -3\end{pmatrix} =1\cdot(-1)+0\cdot(-1)+2\cdot(-3)=-7.
Linha 2 de E(FG):
\bigl[-1,3,1\bigr]\begin{pmatrix}2\\\ 8\\\ 5\end{pmatrix} =-1\cdot2+3\cdot8+1\cdot5=27,\quad \bigl[-1,3,1\bigr]\begin{pmatrix}5\\\ -9\\\ 4\end{pmatrix} =-1\cdot5+3\cdot(-9)+1\cdot4=-28,
\bigl[-1,3,1\bigr]\begin{pmatrix}-1\\\ -1\\\ -3\end{pmatrix} =-1\cdot(-1)+3\cdot(-1)+1\cdot(-3)=-5.
Logo,
E\,(F G) = \begin{pmatrix}12 & 13 & -7\\[6pt]27 & -28 & -5\end{pmatrix}.]

Passo 6

Comparando
(EF)\,G = \begin{pmatrix}12 & 13 & -7\\[3pt]27 & -28 & -5\end{pmatrix} \quad\text{com}\quad E\,(F G) = \begin{pmatrix}12 & 13 & -7\\[3pt]27 & -28 & -5\end{pmatrix},
vemos que são iguais. Isso prova que, para estas matrizes,
(EF)\,G = E\,(F G).]

Resposta

(EF)G = E(FG).

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