Systems of Linear Equations: More Examples

I think I'll use the first row to clear out the x-terms from the second and third rows:

Technically, I should now divide the first row by 2 to get a leading 1, but that will give me fractions, and I'd like to avoid that for as long as possible. Instead, I'll move on to using the second row to clear out the y-term from the third row:

I can divide the third row by 4:

To be technically correct, I'll now divide the second row by 5 and the first row by 2:

Now that I've got leading 1s in each row (that is, now that the coefficient of the first variable term in each row is 1), I can find the system's solution. I already have the solution's value for z. Back-solving from the third row's equation into the second equation, I get:

That's the second variable done. Now for the last one:

Remembering that the variables must be listed in alphabetical order (and, in particular, not in the order in which I happened to solve for them), my answer is:

I think I'll use the second row to work on the x-terms in the first and third rows. I'll be able to clear out the third row, and I'll be able to produce a 1x as the leading term in the first row. This will let me finish the job of clearing out the x-column, and I'll be able to do it without having to deal with fractions:

(Many instructors would teach you always to divide through on one of the rows to get a leading coefficient of 1, but I would rather take an extra step or two and use addition to get a leading 1. That's just a personal preference, but I'm sure you can see the advantage of avoiding fractions for as long as possible. It can really cut down on computational errors.)

Now I'll use that nice leading x in the first row to clear out the leading term in the second row. While I'm at it, I'll also divide the third row by 6, to get smaller numbers:

Hm... The second and third rows are the same. I guess I can use the second row to clear out the third row entirely:

Thinking back to the two-variable case, getting a line like "0 = 0" (which is true, but unhelpful) means that this is a dependent system, and the solution is going to have variables in it. If you get into linear algebra much, you will learn that the answer above means that the solution is a line in three-dimensional space rather than a single point. For now, all you need to know is how to write the solution.

To find the solution, I have to solve the two remaining equations for x and y in terms of z. From the first row, I get:

And from the second row, I get:

Now, in my classes, it would be perfectly okay to say that the answer is (z − 2, 3z − 4, z). But your textbook will almost certainly pull out a new and different variable, such as a, s, or t. Let's say that my textbook is using t, and I'm expected to follow the book's lead in my own work. This means that I'll replace every z in (z − 2, 3z − 4, z) with t:

You should be getting the hang of things by now, so I'll just show the steps that I used:

As soon as I get a nonsense row (like "0 = 1"), I know that this is an inconsistent system, and I can quit. In starting the solving process, I had assumed that this system had a solution. By arriving at garbage, I now know that this assumption was false; this system has no solution.