The
Purplemath Forums 
Systems
of Linear Equations: Sections: Definitions, Solving by graphing, Substitition, Elimination/addition, Gaussian elimination. Solving threevariable, threeequation linear systems is more difficult, at least initially, than solving the twovariable systems, because the computations involved are more messy. You will need to be very neat in your working, and you should plan to use lots of scratch paper. The method for solving these systems is an extension of the twovariable solvingbyaddition method, so make sure you know this method well and can use it consistently correctly. Though the method of solution is based on addition/elimination, trying to do actual addition tends to get very messy, so there is a systematized method for solving the threeormorevariables systems. This method is called "Gaussian elimination" (with the equations ending up in what is called "rowechelon form"). Let's start simple, and work our way up to messier examples.
5x
+ 4y – z =
0
It's fairly easy to see how to proceed in this case. I'll just backsubstitute the zvalue from the third equation into the second equation, solve the result for y, and then plug z and y into the first equation and solve the result for x. 10y
– 3(3) = 11 5x +
4(2) – (3) = 0 Then the solution is (x, y, z) = (–1, 2, 3). The reason this system was easy to solve is that the system was "triangular"; this refers to the equations having the form of a triangle, because of the lower equations containing only the later variables. The point is that, in this format, the system is simple to solve. And Gaussian elimination is the method we'll use to convert systems to this upper triangular form, using the row operations we learned when we did the addition method.
–3x
+ 2y – 6z = 6 No equation is solved for a variable, so I'll have to do the multiplicationandaddition 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 xterms 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 xterm from the first row, and, in this case, I think it will be simpler to use the xterm from the third row, since its coefficient is simply "1". So I'll multiply the third row by 3, and add it to the first row. I do the computations on scratch paper: ...and then I write down the results: (When we were solving twovariable 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 threevariable system, which is why we need the scratch paper.) Warning: Since I didn't actually do anything to the third row, I copied it down, unchanged, into the new matrix of equations. I used the third row, but I didn't actually change it. Don't confuse "using" with "changing". To get smaller numbers for coefficients, I'll multiply the first row by onehalf: Now I'll multiply the third row by –5 and add this to the second row. I do my work on scratch paper: ...and then I write down the results: Copyright © Elizabeth Stapel 20032011 All Rights Reserved 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 xcolumn is cleared out except for the leading term in the third row. So next I have to work on the ycolumn. Warning: Since the third equation has an xterm, I cannot use it on either of the other two equations any more (or I'll undo my progress). I can work on the equation, but not with it. If I add twice the first row to the second row, this will give me a leading 1 in the second row. I won't have gotten rid of the leading yterm 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: ...and then I write down the results: Now I can use the second row to clear out the yterm in the first row. I'll multiply the second row by –7 and add. First I do the scratch work: ...and then I write down the results: 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 uppertriangular form: Now I can start the process of backsolving: y –
7(1) = –4 x +
4(3) – 2(1) = 8 Then the solution is (x, y, z) = (–2, 3, 1). Note: There is nothing sacred about the steps I used in solving the above system; there was nothing special about how I solved this system. You could work in a different order or simplify different rows, and still come up with the correct answer. These systems are sufficiently complicated that there is unlikely to be one right way of computing the answer. So don't stress over "how did she know to do that next?", because there is no rule. I just did whatever struck my fancy; I did whatever seemed simplest or whatever came to mind first. Don't worry if you would have used completely different steps. As long as each step along the way is correct, you'll come up with the same answer. In the above example, I could have gone further in my computations and been more thoroughgoing in my row operations, clearing out all the yterms other than that in the second row and all the zterms other than that in the first row. This is what the process would then have looked like: This way, I can just read off the values of x, y, and z, and I don't have to bother with the backsubstitution. This morecomplete method of solving is called "GaussJordan elimination" (with the equations ending up in what is called "reducedrowechelon form"). Many texts only go as far as Gaussian elimination, but I've always found it easier to continue on and do GaussJordan. Note that I did two row operations at once in that last step before switching the rows. As long as I'm not working with and working on the same row in the same step, this is okay. In this case, I was working with the first row and working on the second and third rows. << Previous Top  1  2  3  4  5  6  7  Return to Index Next >>



Copyright © 20032012 Elizabeth Stapel  About  Terms of Use 




