Systems of Linear Equations: Gaussian Elimination

It's fairly easy to see how to get started in solving this system. The third equation in the system, the one at the bottom of the stack above, is a simple one-variable equality; namely, that z = 3. The second equation, the middle one in the stack above, has only the variables y and z. The simplest next step for me, then, it to substitute the z-value from the third equation back into the second equation. Then I'll solve the resulting one-variable equation for the value of y; in other words, I'll use that second equation to back-solve for y.

This gives me the second value in this three-variable system. The only variable left is x. The first equation in the system, the one at the top of the stack above, contains this variable. I already have the values of y and z, so I'll plug their value in for their variables in the first equation, and back-solve for x.

This is the third value for the solution to this system of equations. Keeping in mind that the variables will be listed in this three-coordinate point alphabetically, my answer is:

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 the rows' order later, if needed, to put 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, since its coefficient is simply "1". So I'll multiply the third row by 3, and add it to the first row. I first do my actual computations on scratch paper:

...and then I write down the results:

(When we were solving two-variable systems on the previous page, 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.)

(To make my steps clear for the grader (and for me, when I review later for the test), I have annotated the arrow between the two steps to show that I'd multiplied the third row by three, and then had added the result to the first row, in order to get a new first row.)

I generally prefer to work with smaller numbers, and I notice that all the coefficients in the (new) first row are even. So, to get smaller coefficients, I'll multiply the first row by one-half (or, which is the same thing, divide through by 2):

Now that's I've removed the x-term from R₁, I turn my attention to R₂. I could divide this row by −5 and then add this to the third row (leaving the in-use second row unchanged), but this will give me loads of fractions in the third row. I'd like to avoid fractions as long as possible. So, instead, I'll multiply the third row by −5 and add the result to the second row (leaving the in-use third row unchanged). I do this work on scratch paper:

I haven't done anything with the first row, so I'll copy it to the new system unchanged. I worked with the third row (in my scratch-work), but I only worked on the second row, so the second row will be updated in the new system, but the third row will be copied over unchanged.

Okay, now thex-column is cleared out except for the leading term in the third row. So, to work further toward triangular form, I next have to work on the y-column.

Note: Since the third equation has an x-term, I cannot use it on either of the other two equations any more — or I'll undo my progress in clearing those rows of x-terms. I can work on the equation (using other rows to clear the y-term), but not with it (that is, I cannot apply it to the other two rows).

Looking at the first and second rows, I see that I could divide the first row by 7 and the second row by 13, and then the y-terms will cancel. But this will give me loads of fractions. I could multiply to the least common multiple, which is 91, but then the numbers will start getting kinda big. Neither of these options is "wrong"; it's just that I'm lazy, so I'm going to see if there might be a better option.

If I multiply the first row by 2 and add the result to the second row, 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 bogging down 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:

I update the second row, leaving the first (that I'd worked with but not on) and third (with which I'd done nothing) unchanged.

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 the result to the first row. First I do my scratch work:

I worked with R₂, but only on R₁; I did nothing with R₃. So I leave R₂ and R₃ unchanged, updating the system with the new R₁.

I can now see what the value of z is but, just to be thorough, I'll divide through on the first row by 43. Then I'll rearrange the rows to put them in upper-triangular form:

Now I can start the process of back-solving:

That's a second variable done. Now I'll do the final back-solving:

I now have values for each of the three variables. I found these values in, as it happened to turn out, reverse-alphabetical order. I'll need to remember to write my solution with the variables in the correct, alphabetical order.