Matemática, perguntado por siquaras, 1 ano atrás

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Respondido por Usuário anônimo
1
Sei que devia começar calculando a letra A, mas para encurtar, já vamos calcular parte da letra C, que pede pela matriz transposta das matrizes A e B.

A^t = \left[\begin{array}{ccc}2&1&0\\4&5&7\end{array}\right] \\\\\\\\ B^t = \left[\begin{array}{ccc}3&-1&9\\-2&6&8\end{array}\right]

Agora que já arrumamos isso, vamos começar.

a) \\\\ 3 \cdot   \left[\begin{array}{ccc}2&4\\1&5\\0&7\end{array}\right]  +   \left[\begin{array}{ccc}3&-2\\-1&6\\9&8\end{array}\right] \\\\\\\\   \left[\begin{array}{ccc}6&12\\3&15\\0&21\end{array}\right]  +   \left[\begin{array}{ccc}3&-2\\-1&6\\9&8\end{array}\right] \\\\\\\\ \left[\begin{array}{ccc}6+3&12-2\\3-1&15+6\\0+9&21+8\end{array}\right]  \\\\\\\\ {\left[\begin{array}{ccc}9&10\\2&21\\9&29\end{array}\right]

b) \\\\ \left[\begin{array}{ccc}2&4\\1&5\\0&7\end{array}\right] - 3 \cdot \left[\begin{array}{ccc}3&-2\\-1&6\\9&8\end{array}\right] \\\\\\\\ \left[\begin{array}{ccc}2&4\\1&5\\0&7\end{array}\right] - \left[\begin{array}{ccc}9&-6\\-3&18\\27&24\end{array}\right] \\\\\\\\ \left[\begin{array}{ccc}2-9&4+6\\1+3&5-18\\0-27&7-24\end{array}\right] \\\\\\\\ {\left[\begin{array}{ccc}-7&10\\4&-13\\-27&-17\end{array}\right]

c) \\\\ 2 \cdot \left[\begin{array}{ccc}2&1&0\\4&5&7\end{array}\right] + 3 \cdot \left[\begin{array}{ccc}3&-1&9\\-2&6&8\end{array}\right] \\\\\\\\ \left[\begin{array}{ccc}4&2&0\\8&10&14\end{array}\right] + \left[\begin{array}{ccc}9&-3&27\\-6&18&24\end{array}\right] \\\\\\\\ \left[\begin{array}{ccc}4+9&2-3&0+27\\8-6&10+18&14+24\end{array}\right] \\\\\\\\  \left[\begin{array}{ccc}13&-1&27\\2&28&38\end{array}\right]

Espero ter ajudado!
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