Is it possible for a system of equations to have a non-zero determinant and no solution at the same time? I am quite confused by the solution I was given for the following problems: 
a) Solve the following system of equations using Gauss elimination only:
$2x - y = 5$
$-x + 2y = -4$
$3x - y = -1$
b) Based on part (a), what can you state about the relationship of vectors
$u = (2, -1, 3)$, $v = (-1, 2, -1)$, $w = (5, -4, -1)$?
c) What should the last co-ordinate of w be (the one which is currently -1) so that the behaviour of
the system of equations in (a) changes (i.e. if you found that it has no solutions, so that is now
has a unique solution; if you found it has a unique or infinite solutions, so that it now has no
solutions). Explain your answer.
The official solution I was given is this: http://screencast.com/t/FD3ZSPbUaL
Among the many problems I found with it, the one that confuses me the most is how it is possible for the determinant of the matrix to be non-zero (therefore the vectors to be independent) and the system to have no solution at the same time.
 A: $$   
 \left(  \begin{array}{rr}
  2  &  -1   \\
  -1   &  2   \\
  3  &  -1     
\end{array} 
  \right)  \; \;
 \left(  \begin{array}{r}
  x   \\
  y    
\end{array} 
  \right)  \; = \;
 \left(  \begin{array}{r}
  5   \\
  -4 \\
-1    
\end{array} 
  \right) 
  $$
I see, in the final question you could combine into a 3 by 3 matrix to most quickly alter that last entry. Sigh. 
Not how i would do the final part; it is easy enough to invert and solve a 2 by 2 system, finding out what $x,y$ must be, leading to
$$   
 \left(  \begin{array}{rr}
  2  &  -1   \\
  -1   &  2   \\
  3  &  -1     
\end{array} 
  \right)  \; \;
 \left(  \begin{array}{r}
  2   \\
  -1    
\end{array} 
  \right)  \; = \;
 \left(  \begin{array}{r}
  5   \\
  -4 \\
?    
\end{array} 
  \right) 
  $$
after which we calculate
$$   
 \left(  \begin{array}{rr}
  2  &  -1   \\
  -1   &  2   \\
  3  &  -1     
\end{array} 
  \right)  \; \;
 \left(  \begin{array}{r}
  2   \\
  -1    
\end{array} 
  \right)  \; = \;
 \left(  \begin{array}{r}
  5   \\
  -4 \\
7   
\end{array} 
  \right) 
  $$
