Respostas

2013-11-23T01:55:27-02:00
  \left[\begin{array}{ccccc}x&5&1&x&5\\2&3&1&2&3\\4&1&1&4&1\end{array}\right] = 0




3x + 20 + 2 - 12 -x -10 = 0
2x + 22 -22 = 0
2x = 0
x = 0/2
x = 0
2013-11-23T12:19:56-02:00
Para ser colinear deve ser igual a zero 

(3x + 20 - 2) - (12 + x - 10) = 0 
3x+18-2-x = 0 
2x+16 = 0 
x = -8 
Letra D.