• Matéria: Matemática
  • Autor: hahaha2068
  • Perguntado 7 anos atrás

ME AJUDEM POR FAVOR!!

Anexos:

Respostas

respondido por: antoniosbarroso2011
1

Resposta:

Explicação passo-a-passo:

Temos que:

a)

A + 2D = \left[\begin{array}{cc}2&-1\\3&4\\2&0\end{array}\right]+2\left[\begin{array}{cc}-3&12\\6&-6\\0&9\end{array}\right]=\left[\begin{array}{cc}2&-1\\3&4\\2&0\end{array}\right]+\left[\begin{array}{cc}-6&24\\12&-12\\0&18\end{array}\right]=\left[\begin{array}{cc}-4&23\\15&-8\\2&18\end{array}\right]

b)

3B - 4C = 3B + (-4)C

3\left[\begin{array}{ccc}-2&7&0\\8&-1&5\\\end{array}\right]+(-4)\left[\begin{array}{ccc}0&4&-2\\6&2&8\\\end{array}\right]=\left[\begin{array}{ccc}-6&21&0\\24&-3&15\\\end{array}\right]+\left[\begin{array}{ccc}0&-16&8\\-24&-8&-32\\\end{array}\right]=\left[\begin{array}{ccc}-6&5&8\\0&-11&-17\\\end{array}\right]

c)

A+\frac{C^{t} }{2}-\frac{4D}{3}

C^{t}=\left[\begin{array}{ccc}0&4&-2\\6&2&8\\\end{array}\right]^{t}=>C^{t}=\left[\begin{array}{cc}0&6\\4&2\\-2&8\end{array}\right]÷2=\left[\begin{array}{cc}0&3\\2&1\\-1&4\end{array}\right]

4\left[\begin{array}{cc}-3&12\\6&-6\\0&9\end{array}\right]=\left[\begin{array}{cc}-12&48\\24&-24\\0&36\end{array}\right]÷3=\left[\begin{array}{cc}-4&16&\\8&-8\\0&12\end{array}\right]

\left[\begin{array}{cc}2&-1\\3&4\\2&0\end{array}\right]+\left[\begin{array}{cc}0&3\\2&1\\1&4\end{array}\right]-\left[\begin{array}{cc}-4&16&\\8&-8\\0&12\end{array}\right]=\left[\begin{array}{cc}2&2\\5&4\\3&4\end{array}\right]-\left[\begin{array}{cc}-4&16&\\8&-8\\0&12\end{array}\right]=\left[\begin{array}{cc}-6&8\\-3&12\\3&-8\end{array}\right].

Perguntas similares