In this segment, we will continue to talk about how the norm of a matrix is related to the conditioning
of a system of equations. We'll take now another example ,and we say that hey if we have a
equations like this [1 2, 2 3] [x, y] is equal to [4, 7]. So let's suppose we have this system
of equations, then we know that for this particular system of equations the solution is [2, 1]
So that if that is the solution then we want to find the norm of the [A] matrix, so
let's say that hey if we considered the above set of equations to be of the form [A][X]=[C], where [A] is
the coefficient matrix, [X] is the solution matrix and [C] is the right hand side, then
we are able to see that the norm of [X] in this case would
be 2 and the norm of [C] in this case will be 7.
Now let's go and make a small change in the right hand side vector and see what happens
there. So if we make a small change in the right hand side
vector [1 2, 2 3] [x, y] and, instead of [4, 7], we say [4.001, 7.001].
In this case, we find the solution by any means we have, and we get the solution
to be 1.999 and 1.001. It seems to be close to
2 and 1 but lets see what it means in terms of the norm of
the matrices. [A][X']=[C']. So what we are basically saying is that hey
now since we are changing the [C] vector the right hand
side vector a little bit, then we'll get a different
solution which is [X']. [A] staying the same, we're keeping the coefficient matrix the same.
So now in this case, what we'll do is calculate our
delta x so we'll say delta [X]. So we'll say delta [X] vector will be equal to
[X']-[X]. And in this case, that will be equal to -0.001 and 0.001. And we'll
get norm of delta [X] to be equal to 0.001. Now again
the same way, we can find delta [C] equal to [C']
minus [C]. So this is the new right hand side minus the previous right hand side. And in this
case, we will get delta [C] is equal to 0.001 and 0.001. So
in this case, we will find out that the norm of delta [C], which
is the difference between this right hand side and this right hand side right here, is
turning out to be equal to 0.001. So let's look at the relative
change that's taking place in our [X]. So the relative change in the norm of the difference
of the solution vector divided by the norm of the
solution vector of the original set of equations turns out to be this which is 5 times 10 to
the power of -4. And now if you look at
hey what is the relative change in the right hand
side vector which we had so we get this relative change in the norm of the right hand side
vector. We get 0.001 divided by 7 which turns out
to be 1.429 times to ten to the power of -4. So
what you're finding out is that a relative change in the norm of the right hand side vector
which is this much is resulting in a relative change
in the solution vector norms to be of this much.
So they seem to be compatible to each other. If I was going to divide the relative changes
I'd get delta [X] infinity or the norm of delta [X] divided by norm of [X] divided by the norm
of the delta [C] divided by the norm of [C] will turn out to be
5 times 10 to the power of -4 divided by 1.429 times ten to the power
of -4. And that turns out to be 3.5.
So what that basically means is that a relative change in the right hand side vector, resulted
in a compatible change in the solution vector. So this solution of equations we started with
seems to be well-conditioned. And how much well-conditioned
it is will be quantified in future segments, but
here you got to see the difference between part one and part two of this particular video
to see that in the previous case we got a very large
ratio between the two relative norms and here we are getting
a small one. So that should tell you which one is well-conditioned and which one is ill-conditioned.
And that's the end of this segment.