CHAPTER 04.06: GAUSSIAN ELIMINATION: Gaussian Elimination With Partial Pivoting: Example: Part 3 of 3 (Back Substitution)


In this segment, we'll continue with the example which we have, we have Gaussian elimination . . . what we're trying to do is we are developing algorithms called Gauss elimination with partial pivoting to solve simultaneous linear equations, and we're taking an example here, and we'll look at the back substitution part of that example. So at the end of the forward elimination steps, we got the equations in the matrix form, where . . . for the set of equations to be as follows, we got 144, 12, 1, a1, a2, a3, and we've got this as an upper triangular matrix. The right-hand side is 279.2, 58.33, and -0.23.  So this is what we obtain at the end of the forward elimination steps, and you can see that it is in that form, where the coefficient matrix is an upper triangular matrix, everything below the diagonal is 0.


So in the back substitution, the reason why it's called back substitution because we're going backwards, we're going to start from the last equation, find a3, then go to the second-last equation, find a2, go to the third-last equation, which in this case is the first equation, and find a1, because in this last equation a3 is the only unknown, in the second equation, there are two unknowns, a2 and a3, but a3 we just found out, so we have only one unknown, which is a2.  In the first equation, or the third-last equation, we have three unknowns, a1, a2, a3, but we just found a2 and a3, so a1 can be found from the first equation, or the third-last equation. So let's go ahead and go through this process of back substitution, so what I will get a3, I will get -0.2 a3, that's the last equation, is equal to -0.23, and a3 is -0.23 divided by -0.2, which is 1. . . . 1.15, so that's the value which I get for a3. Now let's calculate a2, which will be from the second equation here, so let's see what I get for a2.  I get 2.917 a2, plus 0.8264 a3 is equal to 58.33. And so a2 will be equal to 58.33, minus, we'll take this to the right-hand side, because we know a3, and then I'm going to divide by this 2.917. So I get 58.33, minus 0.8264, a3 is 1.15, divided by 2.917, so that's what I'm doing, and I'm getting 19.67.  Again, keep in mind, as mentioned before, I'm using only four significant digits in showing the calculations.  If you use more, you're going to get a slightly different answer from than what I have.  I'm only doing this for the purpose of keeping things simple.  Now let's look at the last equation, we have 144 a1, plus 12 a2, plus a3 is equal to 279.2. So we've got to find a1 from here, so, again, a1 will be 279.2, I'm going to take 12 a2 to the right-hand side, because a2 is known, minus a3, because a3 is known, divided by 144. So that gives me 279.2, minus 12 times a2, a2 we just found is 19.67, a3 we found out previously is 1.15, divided by 144, and this value here turns out to be 0.2917. So we have all the values which we need . . . which we need to find, a1, a2, and a3.  So let me go ahead and write down these values. So the resulting unknown vector which we had, or the solution vector which we had, a1, a2, a3, has been just found by using the back substitution algorithm, a1 turning out to be 0.2917, a2 turning out to be 19.67, and 1.15. 


So that . . . that is the solution there, so as homework, what I want to do, I've been harping about this throughout this segment and a previous segment that I'm only showing you . . . showing you only four significant digits in the calculations. I'm not using only four significant digits, but I'm only showing you four significant digits in your calculations.  So what I would like you to do is to redo the problem . . . redo the problem showing five significant digits.  I'm not talking about that you need to do individual calculations using five significant digits with chopping or rounding, but as you are showing the final results . . . intermediate results, that you're going to use only five significant digits, so redo the problem showing five significant digits, and see whether you get a slightly different answer, which you should.  That is the end of this segment.