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.
$$ \left( \begin{array}{rr} 2 & -1 \\ -1 & 2 \\ 3 & -1 \end{array} \right) \; \; \left( \begin{array}{r} x \\ y \end{array} \right) \; = \; \left( \begin{array}{r} 5 \\ -4 \\ -1 \end{array} \right) $$
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 $$ \left( \begin{array}{rr} 2 & -1 \\ -1 & 2 \\ 3 & -1 \end{array} \right) \; \; \left( \begin{array}{r} 2 \\ -1 \end{array} \right) \; = \; \left( \begin{array}{r} 5 \\ -4 \\ 7 \end{array} \right) $$