CHAPTER 09.05: ADEQUACY OF SOLUTIONS: How is the norm related to the conditioning of a system of equations Part 2 of 2

 

In this segment we will continue to talk about how the norm is related to the conditioning of a system of equations. We'll take now another example and we say that A if we have an equation like this 1 2 2 3, xy is equal to 4 and 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 comma 1. So that if that is the solution then we want to find the norm of the A matrix so letís say that A if we considered the above set of equations to [A] [X]=[C] what a is the caution 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 letís go and make a small change in the right hand side vector and see what happens there. Take a small change in the right hand side vector 1 2 2 3 x y and instead of 4 and 7 we say 4.001 and 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 closer to 11 but letís see what it means in terms of the norm of the matrices. [A][X] prime equal to c prime.

 

So what we are basically saying is that the A 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 prime. A staying the same, we're keeping the caution matrix the same. So now in this case what we'll do is calculate our delta x so we'll say delta x vector will be equal to x prime minus x. And in this case that will be equal to minus 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 prime minus c so this is the new right hand side minus the previous right hand side so 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 .001.

 

So letís look at the relative change thatís 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 minus four. 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 minus four. So what you're finding out is that a relative change in the norm of the right hand side which is this much is resulting in a relative change in the solution vector norms to be of this much. So they are simply 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 five times to the power of minus four divided by 1. 429 times ten to the power of minus four. And that turns out to be three point five. So what that basically means is that the relative change in the right hand side vector, resulted in a really 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 norm. So that should tell you which one is well-conditioned and which one is ill-conditioned. And that is the end of this segment.