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2014-08-26T16:06:35-03:00

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Boa tarrde Thayss2465!

a) x+  \left[\begin{array}{ccc}4&3\\1&1\\2&0\end{array}\right] = \left[\begin{array}{ccc}5&0\\2&3\\7&8\end{array}\right]  \\  \\ \\  x=\left[\begin{array}{ccc}5&0\\2&3\\7&8\end{array}\right]  -  \left[\begin{array}{ccc}4&3\\1&1\\2&0\end{array}\right]  \\  \\  \\ x= \left[\begin{array}{ccc}5-4&0-3\\2-1&3-1\\7-2&8-0\end{array}\right] = \left[\begin{array}{ccc}1&-3\\1&2\\5&8\end{array}\right]
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b) x-    \left[\begin{array}{ccc}1&4&7\\-2&5&3\\\end{array}\right] =\left[\begin{array}{ccc}-1&2&11\\-3&4&1\\\end{array}\right]  \\  \\  \\ x=\left[\begin{array}{ccc}-1&2&11\\-3&4&1\\\end{array}\right] + \left[\begin{array}{ccc}1&4&7\\-2&5&3\\\end{array}\right] \\  \\  \\ x=\left[\begin{array}{ccc}-1+1&2+4&11+7\\-3-2&4+5&1+3\\\end{array}\right]=\left[\begin{array}{ccc}0&6&18\\-5&9&4\\\end{array}\right]
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c) \left[\begin{array}{ccc} \frac{1}{2} &1\\0&2\\\end{array}\right] + \left[\begin{array}{ccc} \frac{3}{2} &4\\3&7\\\end{array}\right]=x-\left[\begin{array}{ccc} -1 &-3\\-2&4\\\end{array}\right] \\ \\ \\ \left[\begin{array}{ccc} \frac{1}{2} &1\\0&2\\\end{array}\right]+ \left[\begin{array}{ccc} \frac{3}{2} &4\\3&7\\\end{array}\right] +\left[\begin{array}{ccc} -1 &-3\\-2&4\\\end{array}\right]=x \\ \\ \\ \left[\begin{array}{ccc} \frac{1}{2}+ \frac{3}{2}-1&1+4-3\\0+3-2&2+7+4\\\end{array}\right]=x\\ \\ \\

x=  \left[\begin{array}{ccc} \frac{3}{2}&2\\1&13\\\end{array}\right]

Tenha uma excelente tarde e bons estudos! ^.^ 


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