Respostas

2014-07-01T00:58:47-03:00
E aí Gabriel,

dadas as matrizes A e B

  A=\left(\begin{array}{ccc}2&3&4\\5&-1&6\\\end{array}\right)~~~e~~~B=  \left(\begin{array}{ccc}-2&1&0\\3&-5&7\\\end{array}\right)

A soma das duas, A e B, teremos:

 A+B= \left(\begin{array}{ccc}2+(-2)&3+1&4+0\\5+3&-1+(-5)&6+7\\\end{array}\right)\\\\\\
A+B=  \left(\begin{array}{ccc}0&4&4\\8&-6&13\\\end{array}\right)

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Para subtração entre duas matrizes, use a mesma regra para soma algébrica, troque os sinais da 2ª matriz:

B-A=  \left(\begin{array}{ccc}-2&1&0\\3&-5&7\\\end{array}\right)-  \left(\begin{array}{ccc}2&3&4\\5&-1&6\\\end{array}\right)\\\\\\
B-A=  \left(\begin{array}{ccc}-2-2&1-3&0-4\\3-5&-5+1&7-6\\\end{array}\right)\\\\\\
B-A=  \left(\begin{array}{ccc}-4&-2&-4\\-2&-4&1\\\end{array}\right)

_______________________

Segue o mesmo processo, o que muda é que ao final da diferença entre as matrizes, teremos que obter a sua transposta:

(A-B)^t=\left\{  \left(\begin{array}{ccc}2-(-2)&3-1&4-0\\5-3&-1-(-5)&6-7\\\end{array}\right)\right\}\\\\\\
(A-B)^t=\left(\begin{array}{ccc}4&2&4\\2&4&-1\\\end{array}\right)\\\\\\
\boxed{(A-B)^t=  \left(\begin{array}{ccc}4&2\\2&4\\4&-1\end{array}\right)}

Espero ter ajudado e tenha ótimos estudos =))
1 5 1
2014-07-01T01:12:34-03:00
10)
  A + B => \left[\begin{array}{ccc}2&3&4\\5&-1&6\\\end{array}\right] -   \left[\begin{array}{ccc}-2&1&0\\3&-5&7\\\end{array}\right]  =   \left[\begin{array}{ccc}0&4&4\\8&-6&13\\\end{array}\right]  \\  \\  \\  \\ A - B => \left[\begin{array}{ccc}2&3&4\\5&-1&6\\\end{array}\right] -   \left[\begin{array}{ccc}-2&1&0\\3&-5&7\\\end{array}\right] =   \left[\begin{array}{ccc}-4&-2&-4\\-2&-4&1\\\end{array}\right]  \\  \\

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

 \left[\begin{array}{cc}4&2&\\2&1&\\1&1\end{array}\right]


1 5 1
de nada.
Helvio, da uma olhada na sua pq tem algo errado =/
ok, obrigado.