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jordo6889
Nov 18, 2009, 10:53 AM
Solve the system of equations below by the Gaussian Elimination method.

2x-6y-5z=11

4x-2y-6z=1

2x+4y-z=11

justcurious55
Nov 18, 2009, 11:37 AM
We will not do your homework on this site. If you show us that you have at least attempted the problems yourself, we are here to help. Or if you have a specific question, there is probably someone here who can help you. But we're not going to just do your homework. Do as much as you can and post it back here for help

Chris-infj
Nov 18, 2009, 12:02 PM
Start by writing the augmented matrix like so:

\begin{pmatrix} 2 & -6 & -5 & 11 \\ 4 & -2 & -6 & 1 \\ 2 & 4 & -1 & 11 \end{pmatrix}

Then by adding and subtracting rows, try to get three zeroes in the bottom left corner of the matrix. Bleh, I'll stop here and let someone more competent with the latex add more details.

Chris-infj
Nov 18, 2009, 12:24 PM
What I meant about the three zeroes in the bottom left corner of the matrix was this:

Obtain this form by subtracting each rows, for example, row 1 minus row 3 or say, row 2 minus twice row 1 - in as many steps as it takes to get the three zeroes that look like a staircase (math purists, pardon my lack of rigour in the terminology!)

\begin{pmatrix} a_{1,1} & a_{1,2} & a_{1,3} \\ 0 & a_{2,2} & a_{2,3} \\ 0 & 0 & a_{3,3} \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} b_{1,1} \\ b_{2,1} \\ b_{3,1}\end{pmatrix}

Then you equate a_{3,3} z = b_{3,1} and obtain z and then substitute again to get y and then x

But can you check that you correctly posted the right equations? They don't look right...