Found this info on a site.
* Solve the following system of equations using Gaussian elimination.
–3x + 2y – 6z = 6
5x + 7y – 5z = 6
x + 4y – 2z = 8
No equation is solved for a variable, so I'll have to do the multiplication-and-addition thing to simplify this system. In order to keep track of my work, I'll write down each step as I go. But I'll do my computations on scratch paper. Here is how I did it:
The first thing to do is to get rid of the leading x-terms in two of the rows. For now, I'll just look at which rows will be easy to clear out; I can switch rows later to get the system into upper triangular form. There is no rule that says I have to use the x-term from the first row, and, in this case, I think it will be simpler to use the x-term from the third row. So I'll multiply the third row by 3, and add it to the first row. I do the computations on scratch paper:
[-3x + 2y - 6z = 6] + [3x + 12y - 6z = 24] = [14y - 12z = 30]
...and then I write down the results:
system of equations with updated first row
(When we were solving two-variable systems, we could multiply a row, rewriting the system off to the side, and then add down. There is no space for this in a three-variable system, which is why we need the scratch paper.)
Since I didn't actually do anything to the third row, I copied it down, unchanged, into the new matrix. I used the third row, but I didn't actually change it. Don't let this confuse you.
To get smaller numbers, I'll multiply the first row by one-half:
first row: 7y - 6z = 15
Now I'll multiply the third row by –5 and add this to the second row. I do my work on scratch paper:
[5x + 7y - 5z = 6] + [-5x - 20y + 10z = -40] = [-13y + 5z = -34]
...and then I write down the results: Copyright © Elizabeth Stapel 2006-2008 All Rights Reserved
system of equations with updated second row
I didn't do anything with the first row, so I copied it down unchanged. I worked with the third row, but I only worked on the second row, so the second row is updated and the third row is copied over unchanged.
Okay, now the x-column is cleared out except for the leading term in the third row. So now I have to work on the y-column. If I add twice the first row to the second this will give me a leading 1 in the second row. I won't have gotten rid of the leading y-term in the second row, but I will have converted it (without getting involved in fractions) to a form that is simpler to deal with. (You should keep an eye out for this sort of simplification.) First I do the scratch work:
[-13y + 5z = -34] + [14y - 12z = 30] = [y - 7z = -4]
...and then I write down the results:
system of equations with updated second row
Now I can use the second row to clear out the y-term in the first row. I'll multiply the second row by –7 and add. First I do the scratch work:
[7y - 6z = 15] + [-7y + 49z = 28] = [43z = 43]
...and then I write down the results:
system of equations with updated first row
I can tell what z is now, but, just to be thorough, I'll divide the first row by 43. Then I'll rearrange the rows to put them in upper-triangular form:
x + 4y - 2z = 8; y - 7z = -4; z = 1
Now I can start the process of back-solving:
y – 7(1) = –4
y – 7 = –4
y = 3
x + 4(3) – 2(1) = 8
x + 12 – 2 = 8
x + 10 = 8
x = –2
Then the solution is (x, y, z) = (–2, 3, 1).
and the above has me confused!!!
