11 – 12 Omega And Xi QUIZ RENDER V1

The last thing I can tell you before we go into programming is, why this makes any sense. Suppose you fill the matrix, which I call omega, and the vector, which will give the Greek name of xi. Then I can find the best solution for all the landmark position or the work position, by a very simple mathematical trick and that is completely counter intuitive. I invert the omega, I right multiply with xi, and out comes a vector mu, which gives me the best estimates for all the locations and the landmark locations. Now, that is quite amazing, because all it means in graphs lemmas, that we keep adding numbers to these matrices every time sequence straight and when you’re done with it, you run a very simple procedure, and out comes the best places for your robot. So, let’s go and try it and I’m going to ask you to program this and I’m giving you my matrix class, so you can do this easily. What I’m asking you to do is to build a three by three matrix, and of course a three by one vector, by which you shall state that our initial location is minus three X 1 and exploitation is obtained by adding five to 0 and X 2 is obtained by adding three to X1. So, our expectation, what we should get out, we run the mu equals omega minus one times xi trick. That X 0 becomes minus three X 1 becomes 2, and X 2 becomes five. So, diving straight into our programming environment, I’m giving you a matrix class, you might want to take a moment to look over it. It’s really augmented to what I’ve given you before and I fixed a bug with the inversion code, which is quite essential and if I run it, I construct an omega matrix. Piece-by-piece that’s the one that you should come out with xi vector and then I run and print out using the show command the result of omega to the minus one times xi. You can see -3, 2 and 5 are the correct result. The result from the omega matrix and xi vector. So, what I want you to do is write code that incrementally step-by-step constructs the omega vector and the xi function, and then returns to me those results over here. There’s an empty function in the code that accepts as parameter the initial position minus three and the demotion values five and three.

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