.
In this segment, I'm going to continue with the
example we had for Naive Gaussian elimination.|We're looking
at an example to solve a set of simultaneous linear equations,
and we're going to, in this part, do back substitution
part.|There are two parts to Naive Gaussian
elimination, or even the Gauss elimination with partial pivoting, which we will do later, is
there's a forward elimination part to it, and there is a back substitution part of it.
The forward elimination part for the example which we just did resulted
in the coefficient matrix being turned into an upper triangular matrix
so that we can apply the algorithm of back substitution to do that, and this is
what we obtained, 25, 5, 1, 0,
-4.8, -1.56, 0, 0,
0.7, and we had three unknowns, a1, a2, and a3,
and we had 106.8, -96.208,
and 0.76.
So the reason why we convert the coefficient matrix into an upper triangular matrix,
because you're seeing that everything below the diagonal is 0, is so that we can solve one equation at a time,
is that when we look at the last equation, it only has one unknown in it, when we
look at the second . . . second-last equation, it has two unknowns, but a3 is already known, and
the last, or the first equation, we have three unknowns, a1, a2, and a3,
but these two have already been calculated, hence we can calculate a1.|So that's the
whole point why we convert our coefficient matrix, or do the
matrix operations, so as to convert the coefficient matrix into an upper triangular matrix.
So we'll look at the last equation, so from the third equation, so that's how you start in the
back substitution, that's why it's called back substitution, because you're going backwards, so the
third equation is simply 0.7 a3 is
0.76, which gives you a3 is 0.76 divided by
0.7, and that gives you 1.08571,
so that's the value of a3.|Now we want to calculate a2, what we'll have to
do is we'll have to now write down this equation here, which is the second-last equation.
So that turns out to be -4.8
a2, minus 1.56 a3, and
that's equal to -96.208.
So if I'm going to write down this equation in terms of a2, I get -96.208,
plus 1.56 a3,
divided by -4.8, that's what I get.|So I'm going to substitute the value
of a3, which I just obtained from the previous equation, or
the previous step, a3 is 1.08571,
divided by 4.8, and this value here turns out
to be 19.6905, so we have already calculated
now a3 and a2.|Now let's go ahead and substitute
now the values of a2 and a3 to get a1,
so for that we'll have to look at the first equation.|The first equation is, when we expand it will
be 25 a1, plus 5 a2, plus a3
is 106.8, and since we have already calculated
a2 and a3, calculating a1 is pretty straightforward, so what we're going to do is we're going to write
down the equation for a1, which is 106.8, minus 5
a2, minus a3, divided by 25.|The reason why I'm writing the equation
like this is because that's following the algorithm, that what you do is
in order to calculate a particular unknown, you take the element, the coefficient corresponding to it goes
in the denominator, the right-hand side goes there, and then there's a summation of all the contributions
coming from the previous values which have already been calculated.|So that is . . .
for that you should look at the algorithm part of this . . . of the Naive Gauss
elimination to be able to see that how that would work in a program.|So here this is my
106.8 minus 5, a2 we just calculated as
19.6905, minus a3,
which is 1.08571, divided by
25, and this value here turns out to be
0.290472.|So we have been able to calculate
all the unknowns which we have.|So if I write down the
vector of the unknowns, a1, a2, a3,
we get this to be equal to 0.290472,
because that's just what we calculated here, that's a2, which is
19.6905, that's a3,
that's 1.08571.
And that's the solution which we obtain for the three equations, three
unknowns by using the back substitution part of the Naive
Gauss elimination algorithm.|And that is the end of this segment.
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