Considere as matrizes 3 x 3 \mathrm{~A} e B abaixo. Calcule AB e BA e verifique que A B \neq B A.
A=\left[\begin{array}{ccc} 1 & -1 & 1 \\\\ -3 & 2 & -1 \\\\ -2 & 1 & 0 \end{array}\right]_{3 x 3} \quad B=\left[\begin{array}{lll} 1 & 2 & 3 \\\\ 2 & 4 & 6 \\\\ 1 & 2 & 3 \end{array}\right]_{3 x 3}

Passo 1

O enunciado dá as matrizes
A=\begin{pmatrix} 1 & -1 & 1\\\\ -3 & 2 & -1\\\\ -2 & 1 & 0 \end{pmatrix}, \quad B=\begin{pmatrix} 1 & 2 & 3\\\\ 2 & 4 & 6\\\\ 1 & 2 & 3 \end{pmatrix},
e pede para calcular AB e BA e mostrar que AB\neq BA. O conteúdo principal é a definição de produto de matrizes, em que
(AB)_{ij}=\sum_{k=1}^{3}A_{ik}\,B_{kj}, \quad (BA)_{ij}=\sum_{k=1}^{3}B_{ik}\,A_{kj}.

Passo 2

Calcular AB pelas regras de linha de A vezes coluna de B:
AB =\begin{pmatrix} 1\cdot1 +(-1)\cdot2 +1\cdot1 & 1\cdot2 +(-1)\cdot4 +1\cdot2 & 1\cdot3 +(-1)\cdot6 +1\cdot3\\\\ -3\cdot1 +2\cdot2 +(-1)\cdot1 & -3\cdot2 +2\cdot4 +(-1)\cdot2 & -3\cdot3 +2\cdot6 +(-1)\cdot3\\\\ -2\cdot1 +1\cdot2 +0\cdot1 & -2\cdot2 +1\cdot4 +0\cdot2 & -2\cdot3 +1\cdot6 +0\cdot3 \end{pmatrix} =\begin{pmatrix} 0 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix}.

Passo 3

Calcular BA pelas regras de linha de B vezes coluna de A:
BA =\begin{pmatrix} 1\cdot1 +2\cdot(-3)+3\cdot(-2) & 1\cdot(-1)+2\cdot2 +3\cdot1 & 1\cdot1 +2\cdot(-1)+3\cdot0\\\\ 2\cdot1 +4\cdot(-3)+6\cdot(-2) & 2\cdot(-1)+4\cdot2 +6\cdot1 & 2\cdot1 +4\cdot(-1)+6\cdot0\\\\ 1\cdot1 +2\cdot(-3)+3\cdot(-2) & 1\cdot(-1)+2\cdot2 +3\cdot1 & 1\cdot1 +2\cdot(-1)+3\cdot0 \end{pmatrix} =\begin{pmatrix} -11 & 6 & -1\\\\ -22 & 12 & -2\\\\ -11 & 6 & -1 \end{pmatrix}.

Passo 4

Observamos que
AB = \begin{pmatrix} 0 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix} \neq \begin{pmatrix} -11 & 6 & -1\\\\ -22 & 12 & -2\\\\ -11 & 6 & -1 \end{pmatrix} = BA,
logo AB\neq BA.

Resposta

AB = \begin{pmatrix} 0 & 0 & 0\\\\ 0 & 0 & 0\\\\ 0 & 0 & 0 \end{pmatrix}, \quad BA = \begin{pmatrix} -11 & 6 & -1\\\\ -22 & 12 & -2\\\\ -11 & 6 & -1 \end{pmatrix},
e, portanto, AB\neq BA.

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