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

AI Thread Summary
The discussion revolves around solving a system of equations derived from a row-reduced matrix. The user initially presents their row-reduced form and seeks assistance in expressing the solution in vector notation. After some back-and-forth, corrections are made regarding coefficients in the equations, particularly for variable x_4. The user confirms that the matrix was row-reduced correctly and ultimately resolves the issue with the help of others. The interaction highlights the importance of attention to detail in matrix solutions and collaborative problem-solving.
mr_coffee
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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
 
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Your matrix seems to be correct, and is:

\left( {\begin{array}{*{20}c}<br /> 1 &amp; 5 &amp; 0 &amp; 0 &amp; { - 7} &amp; 6 &amp;\vline &amp; { - 7} \\<br /> 0 &amp; 0 &amp; 1 &amp; 0 &amp; { - 1} &amp; 1 &amp;\vline &amp; { - 1} \\<br /> 0 &amp; 0 &amp; 0 &amp; 1 &amp; { - 2} &amp; { - 4} &amp;\vline &amp; 8 \\<br /> <br /> \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}<br /> x_1 + 5s - 7t + 6u = - 7 \hfill \\<br /> x_3 - t + u = - 1 \hfill \\<br /> x_4 - 2t - 4u = 8 \hfill \\ <br /> \end{gathered} \right. \Leftrightarrow \left\{ \begin{gathered}<br /> x_1 = - 7 - 5s + 7t - 6u \hfill \\<br /> x_3 = - 1 + t - u \hfill \\<br /> x_4 = 8 + 2t + 4u \hfill \\ <br /> \end{gathered} \right

Can you put it in vectorial notation now?
 
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!
 
I think you for got the 2 in column 3 for x_4 since that has "2t) in its solution.
 
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!
 
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I see you entered "6" in the last column for x_1 and I think that has to be "-6" (check my last system).
 
ah ha! my bad! thanks for the help! :biggrin:
 
It was correct now? Glad I could help :smile:
 
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