Reduced a big matrix, now the parametric form is not right, :\

[mr_coffee]

Hello everyone I did the following problem:
Click http://img220.imageshack.us/img220/8486/untitled1copy4oq.jpg to view the problem and my answer.

The row reduced form is:
1 5 0 0 -7 6 -7
0 0 1 0 -1 1 -1
0 0 0 1 - 2 -4 8

Any help would be great

 

[TD]

Your matrix seems to be correct, and is:

$$\left( {\begin{array}{*{20}c} 1 & 5 & 0 & 0 & { - 7} & 6 &\vline & { - 7} \\ 0 & 0 & 1 & 0 & { - 1} & 1 &\vline & { - 1} \\ 0 & 0 & 0 & 1 & { - 2} & { - 4} &\vline & 8 \\ \end{array} } \right)$$

Now we have 3 lineair indepedant equations with 6 variables which means we can 'choose' 3 variables. We keep the variables $$x_1$$, $$x_3$$ and $$x_4$$. Let $$x_2 = s$$, $$x_5 = t$$ and $$x_6 = u$$.

This gives the solution of the system:

$$\left\{ \begin{gathered} x_1 + 5s - 7t + 6u = - 7 \hfill \\ x_3 - t + u = - 1 \hfill \\ x_4 - 2t - 4u = 8 \hfill \\ \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered} x_1 = - 7 - 5s + 7t - 6u \hfill \\ x_3 = - 1 + t - u \hfill \\ x_4 = 8 + 2t + 4u \hfill \\ \end{gathered} \right$$

Can you put it in vectorial notation now?

 

[mr_coffee]

hm...i don't see how 'im still missing this:
i entered in:
-7 -5 7 6
0 1 0 0
-1 0 1 -1
8 0 0 4
0 0 1 0
0 0 0 1

still wrong, thanks for the reply!

 

[TD]

I think you for got the 2 in column 3 for $x_4$ since that has "2t) in its solution.

 

[mr_coffee]

I'm stilling screwing somthing up...I fixed what you said, but still not the right answer, and I know its row reduced correctly because I checked it with my calculator! here is the picture of what I wrote and submitted:
http://img214.imageshack.us/img214/8780/gfsdgfd6wq.jpg Thanks for the help!

 

[TD]

I see you entered "6" in the last column for $x_1$ and I think that has to be "-6" (check my last system).

 

[mr_coffee]

ah ha! my bad! thanks for the help!

 

[TD]

It was correct now? Glad I could help