.
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.
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