Question

PROVA!!
Dadas as matrizes A= $\left[\begin{array}{ccc}1&4&0\\1&-3&1\end{array}\right]$ e B = $\left[\begin{array}{ccc}1&-1\\-1&1\\5&0\end{array}\right]$ , Calcule :


a) A.B


b)B.A


c) compare os resultados


d) justifique a resposta

Answer 1

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$\boxed{\begin{array}{l}\sf o~produto~de~matrizes~existe~se~e~somente~se\\\sf o~n\acute umero~de~colunas~da~1^a~matriz~for~igual~ao\\\sf n\acute umero~de~linhas~da~2^a~matriz.\\\sf a~matriz~produto~ter\acute a~o~n\acute umero~de~linhas~da~1^a\\\sf e~o~n\acute umero~de~colunas~da~2^a.\end{array}}$

$\boxed{\begin{array}{l}\tt a)~\rm A\cdot B=\begin{bmatrix}\sf1&\sf4&\sf0\\\sf1&\sf-3&\sf1\end{bmatrix}\cdot\begin{bmatrix}\sf1&\sf-1\\\sf-1&\sf1\\\sf5&\sf0\end{bmatrix}\\\rm A\cdot B=\begin{bmatrix}\sf1\cdot1+4\cdot-1+0\cdot5&\sf1\cdot(-1)+4\cdot1+0\cdot0\\\sf1\cdot1-3\cdot(-1)+1\cdot5&\sf1\cdot(-1)-3\cdot1+1\cdot0\end{bmatrix}\\\rm A\cdot B=\begin{bmatrix}\sf-3&\sf3\\\sf9&-4\end{bmatrix}\end{array}}$

$\boxed{\begin{array}{l}\tt b)~\rm B\cdot A=\begin{bmatrix}\sf1&\sf-1\\\sf-1&\sf1\\\sf5&\sf0\end{bmatrix}\cdot\begin{bmatrix}\sf1&\sf4&\sf0\\\sf1&\sf-3&\sf1\end{bmatrix}\\\rm B\cdot A=\begin{bmatrix}\sf 1\cdot1-1\cdot1&\sf1\cdot4-1\cdot(-3)&\sf1\cdot0-1\cdot1\\\sf-1\cdot1+1\cdot1&\sf-1\cdot4+1\cdot(-3)&\sf-1\cdot0+1\cdot1\\\sf5\cdot1+0\cdot1&\sf5\cdot4+0\cdot(-3)&\sf5\cdot0+0\cdot1\end{bmatrix}\\\rm B\cdot A=\begin{bmatrix}\sf0&\sf7&\sf-1\\\sf0&\sf-7&\sf1\\\sf5&\sf20&\sf0\end{bmatrix}\end{array}}$

$\large\boxed{\begin{array}{l}\tt c)~A\cdot B\ne B\cdot A\\\tt d)~\sf o~produto~de~matrizes~n\tilde ao~\acute e~comutativo.\end{array}}$