Matrice, solving the system, need to put it in right form

[mr_coffee]

Hello everyone I had this system:
-2x1 + x2 = 5
6x1 - 3x2 = -15
\left( {\begin{array}{*{20}c}<br /> {-2} &amp; 1 &amp; 5 \\<br /> 6 &amp; -3 &amp; {-15}\\<br /> \end{array} } \right)

I then solved it down too:
\left( {\begin{array}{*{20}c}<br /> {-2} &amp; 1 &amp; 5 \\<br /> 0 &amp; 0 &amp; 0\\<br /> \end{array} } \right)

It wants me to put it parametric form:
so i did the following:
-2x1 + x2 = 5
2x1 = -5 + x2
x1 = (-5+x2)/2
let x2 = s;
[-5/2] + [1/2] S
[0 ] [0 ]

but its wrong! any ideas? Thanks

 

[TD]

You have to be careful, when you let x_2 = s, it doesn't just disappear/become zero. You still have then:

\left\{ \begin{gathered}<br /> x_1 = \frac{{ - 5 + x_2 }}<br /> {2} \hfill \\<br /> x_2 = x_2 \hfill \\ <br /> \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}<br /> x_1 = \frac{1}<br /> {2}s - \frac{5}<br /> {2} \hfill \\<br /> x_2 = s \hfill \\ <br /> \end{gathered} \right

Can you get the correct vector-notation now?

 

[mr_coffee]

awsome, thanks again TD!

 

[TD]

No problem