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

  • Thread starter mr_coffee
  • Start date
  • #1
mr_coffee
1,629
1
Hello everyone I had this system:
-2x1 + x2 = 5
6x1 - 3x2 = -15
[tex]\left( {\begin{array}{*{20}c}
{-2} & 1 & 5 \\
6 & -3 & {-15}\\
\end{array} } \right)[/tex]

I then solved it down too:
[tex]\left( {\begin{array}{*{20}c}
{-2} & 1 & 5 \\
0 & 0 & 0\\
\end{array} } \right)[/tex]

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
 

Answers and Replies

  • #2
TD
Homework Helper
1,022
0
You have to be careful, when you let [itex]x_2 = s[/itex], it doesn't just disappear/become zero. You still have then:

[tex]\left\{ \begin{gathered}
x_1 = \frac{{ - 5 + x_2 }}
{2} \hfill \\
x_2 = x_2 \hfill \\
\end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}
x_1 = \frac{1}
{2}s - \frac{5}
{2} \hfill \\
x_2 = s \hfill \\
\end{gathered} \right[/tex]

Can you get the correct vector-notation now?
 
  • #3
mr_coffee
1,629
1
awsome, thanks again TD!
 
  • #4
TD
Homework Helper
1,022
0
No problem :smile:
 

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