CHAPTER 04.06: GAUSSIAN ELIMINATION: Round-off Error Issues: Example: Part 3 of 3

In this segment, we're going to talk about Gauss elimination with partial pivoting, which is a way to find solution to simultaneous linear equations, and we are, right now, talking about an example. And, of course, in this example we are using something special, we are using five significant digits with chopping for all our intermediate as well as final calculations.

So, having said that, this is what we obtained at the end of the forward elimination steps. So the example is requiring us to now conduct the back substitution steps. So this is what we obtained at the end of the forward elimination steps for this example. We had 20, 15, 10, 0, -2.75, 0.5, 0, 0, 8.5001, x1, x2, x3, and we had 45, -2.25, and 8.5001.  So what we want to be able to do is we want to be able to use our back substitution steps.  So we have to show the back substitution steps for this example.  But again, keep in mind that all the back substitution steps which we have to show is that they have to follow this rule of five significant digits with chopping, that all intermediate as well as final calculations have to be done with five significant digits with chopping.  So let's go ahead and do the back substitution.  What that means is that we'll have to take the last equation, then the second-last equation, the third-last equation, and so on and so forth, to be able to calculate our constants . . . to be able to calculate our solution vector.  So the last equation is 8.5001 x3 is equal to 8.5001, so I get x3 is equal to 8.5001 divided by 8.5001, and that turns out to be 1. Now, surely I can show all the significant digits which I was supposed to show, but I'm just not showing them for purposes of keeping things a little bit clean, but you can understand that if I need to show all the significant digits, then I do show them. Now x2, in order to calculate x2, we have -2.75 x2, plus 0.5 x3 is equal to -2.25, right?  So we have that, so what that is meaning x2 is equal to -2.25, minus 0.5 x3, divided by -2.75, and that value turns out to be -2.75, minus 0.5 times 1, divided by 2.75. So keep in mind that I have to show each intermediate calculation has to be done with five significant digits, that's why I'm not just simply putting these numbers in a calculator and telling you what the answer is.  So I'll do -2.25 minus 0.5, so I need to do this multiplication first, divided by 2.75 . . . -2.75, and then I will do this addition first, so I get -2.75 divided by -2.75, and now I'll do the division, and I get 1 for x2. So that's what I get as the value of x2.  So now let's go ahead and see what x3 is, that will be obtained from the third-last equation, which in this case is the last . . . first equation, so I get 20 x1, plus 15 x2, plus 10 x3 is equal to 45.  So x1, the way the equation will be written is, 45, minus 15 x2, minus 10 x3, divided by 20.  So I get 45, minus 15 times 1, minus 10 times 1, divided by 20, because those are the values which I obtained for x2 and x3. So what I'll have to do is I'll do the multiplications first, I get 45, minus 15, minus 10, divided by 20, and then I'll subtract 45 . . . subtract 15 from 45 first, I get 30 minus 10, divided by 20, then I do the 30 minus 10, I get 20 divided by 20, and I get equal to 1. 

So the solution which I get by using the back substitution for this problem with five significant digits with chopping, I get x1, x2, x3 is equal to 1, 1, and 1, and that, in fact, is the exact solution.  So keep in mind that we are illustrating this example to show that how we can reduce round off error.  In this case, the round off error got reduced totally, but that's just a coincidence, so please don't think that when you're going to use Gauss elimination with partial pivoting that you will be able to reduce the round off error to 0. That's not the case.  In this case it turns out to be, but that's just a coincidence.  And that's the end of this segment.