jblsmith Posted July 5, 2011 Share Posted July 5, 2011 Hi guys, Working on the math in my Master's Thesis and I'm stuck on one part. I was clicking around here and thought I'd offer it up to you guys for some fun: Solve the system, b_{1}=(y+z)/(1-x-y-z) b_{2}=(x+z)/(1-x-y-z) b_{3}=(x+y)/(1-x-y-z) for x, y, and z in terms of b_{1}, b_{2}, and b_{3}. I'm sure there's some easy, self-evident step that makes everything reduce nicely but after staring at it for a couple hours I can't come up with it. Let me know if you guys out there figure it out. Enjoy! Link to comment Share on other sites More sharing options...

Matt W Posted July 5, 2011 Share Posted July 5, 2011 Hi guys, Working on the math in my Master's Thesis and I'm stuck on one part. I was clicking around here and thought I'd offer it up to you guys for some fun: Solve the system, b_{1}=(y+z)/(1-x-y-z) b_{2}=(x+z)/(1-x-y-z) b_{3}=(x+y)/(1-x-y-z) for x, y, and z in terms of b_{1}, b_{2}, and b_{3}. I'm sure there's some easy, self-evident step that makes everything reduce nicely but after staring at it for a couple hours I can't come up with it. Let me know if you guys out there figure it out. Enjoy! Basic arithmetic gives: b_{1}=b_{1}x+(b_{1}+1)y+(b_{1}+1)z b_{2}=(b_{2}+1)x+b_{2}y+(b_{2}+1)z b_{3}=(b_{3}+1)x+(b_{3}+1)y+b_{3}z (when x+y+z!=1) This is a simple system of linear equations to solve. coffeeintotheorems 1 Link to comment Share on other sites More sharing options...

jblsmith Posted July 6, 2011 Author Share Posted July 6, 2011 Basic arithmetic gives: b_{1}=b_{1}x+(b_{1}+1)y+(b_{1}+1)z b_{2}=(b_{2}+1)x+b_{2}y+(b_{2}+1)z b_{3}=(b_{3}+1)x+(b_{3}+1)y+b_{3}z (when x+y+z!=1) This is a simple system of linear equations to solve. Yeah I got that but matrix arithmatic keeps giving me shit like 1=1 and using Cramer's Rule gets quite messy when calculating the determinants. How would you solve it from here? There's gotta be something I'm missing. Link to comment Share on other sites More sharing options...

coffeeintotheorems Posted July 7, 2011 Share Posted July 7, 2011 Yeah I got that but matrix arithmatic keeps giving me shit like 1=1 That's interesting. The first thing I would do would be to substitute some values for b1, b2, and b3 -- say 2, 3, and 5 -- and then check two things: 1) Make sure the determinant isn't zero for those values. If it is, then funny things might happen... 2) If the determinant is nonzero (or even if it is zero), solve the system for those values and see what happens. 3) If 1 and 2 check out, try other values for the b's, such as all zeros, all 1's, etc. This is just a little tinkering to make sure something weird isn't going on. After that, just solve algebraically with Gaussian elimination (or any ad-hoc technique that might be faster), and see what you get. If you're getting "1=1" then I'm guessing your scratch work isn't very organized and maybe your substitutions are sending you in circles. (Hey, we've all been there.) Out of curiosity, what's the application in which this equation has shown up? Link to comment Share on other sites More sharing options...

jblsmith Posted July 7, 2011 Author Share Posted July 7, 2011 (edited) Thanks for the reply! 1) Make sure the determinant isn't zero for those values. If it is, then funny things might happen... 2) If the determinant is nonzero (or even if it is zero), solve the system for those values and see what happens. I'm fairly certain the determinants are non-zero and that there is a unique solution. Earlier in the paper I solve a similar 2 equation system by simply reducing and substituting. After that, just solve algebraically with Gaussian elimination (or any ad-hoc technique that might be faster), and see what you get. If you're getting "1=1" then I'm guessing your scratch work isn't very organized and maybe your substitutions are sending you in circles. (Hey, we've all been there.) I've been using (or at least trying) Gaussian elimination but I cant get it into reduced escelon form. Instead of 2 of the variables reducing they all reduce and I end up with shit like b3/(b1-b2)=b3/(b1-b2). That's what I mean by "1=1" Out of curiosity, what's the application in which this equation has shown up? My masters thesis is a study of how physical capital, education capital, and healthcare capital effect economic growth. A certain portion of growth can be attributed to each of these three forms of capital, these proportions are x, y, and z in the above system. These proportions are unobserved an must be estimated statistically. b1, b2, and b3 are the estimated coefficients from an econometric model and are know, If I can put x, y, and z in terms of b1, b2, and b3 I can then calculate x, y, and z as a non-linear combination of the estimated coefficients. Edited July 7, 2011 by jblsmith Link to comment Share on other sites More sharing options...

coffeeintotheorems Posted July 7, 2011 Share Posted July 7, 2011 I've been using (or at least trying) Gaussian elimination but I cant get it into reduced escelon form. Instead of 2 of the variables reducing they all reduce and I end up with shit like b3/(b1-b2)=b3/(b1-b2). That's what I mean by "1=1" With all due respect, I think you're making a fundamental error somewhere in the GE (Gaussian elimination) process. For example, maybe you're forgetting about the variables x, y, and z implicitly being "behind the scenes," and then you're, say, subtracting b1 from all sides of an equation. When all else fails, you can always solve the system the old-fashioned way--by writing out all variables (x, y, and z), eliminating them one at a time, then substituting. In fact, I started doing this because I was curious. I've got to say, the algebra is pretty ugly; I stopped after I solved for z (and I didn't even solve it completely; just got close.) Here are a few ideas: 1) Instead of b1, b2, and b3, I would use three variables without subscripts, to avoid confusion. Maybe just a, b, c? 2) One practical problem with GE is that when you're writing everything within the confines of a matrix, you limit your space and end up having to cram stuff in. Unless you have really big paper or turn it sideways. 3) Another option you might try that would require you to write a little less of the b variables would be to actually find the inverse of the matrix and express your final solution in terms of that. Recall: if you're starting from the matrix-vector equation Ax=b and the inverse A' exists, then your solution is x = A'b -- in other words, express your final solutions as dot products, if that would be acceptable to your audience. Remember, to find the inverse you juxtapose the non-augmented matrix A with I3 (the 3x3 identity) and row-reduce (A | I3 ). Another important question: do you absolutely need an explicit final solution, or do you just need to demonstrate that it has certain properties? If your thesis audience does not need to see the final solution, then maybe you could just declare that "given b1, b2, and b3, x in terms of these coefficients is the first entry in the vector given by performing GE on the augmented matrix." Link to comment Share on other sites More sharing options...

jblsmith Posted July 7, 2011 Author Share Posted July 7, 2011 (edited) With all due respect, I think you're making a fundamental error somewhere in the GE (Gaussian elimination) process. For example, maybe you're forgetting about the variables x, y, and z implicitly being "behind the scenes," and then you're, say, subtracting b1 from all sides of an equation. Haha, I know what you're saying but I wasn't making that mistake When all else fails, you can always solve the system the old-fashioned way--by writing out all variables (x, y, and z), eliminating them one at a time, then substituting. In fact, I started doing this because I was curious. I've got to say, the algebra is pretty ugly; I stopped after I solved for z (and I didn't even solve it completely; just got close.) When I started working on this, that's what I did as well. I solved for x in terms of b1, b2, b3, and y but gave up before substituting again. Another important question: do you absolutely need an explicit final solution, or do you just need to demonstrate that it has certain properties? Unfortunately I need the explicit solutions, they are a pretty significant part of the paper. Where's Matt W? He said this was supposed to be easy. Edited July 7, 2011 by jblsmith Link to comment Share on other sites More sharing options...

coffeeintotheorems Posted July 7, 2011 Share Posted July 7, 2011 Ok, I found the inverse using row-reduction of the non-augmented juxtaposed with I3. It took me three pages, and the only conditions needed to avoid dividing by zero are c != 0 , b+c != -1, and a+b+c != -2. I promise you this will work and is probably the easiest way. (Crack open your linear algebra text if you've forgotten the process.) I recommend doing the following three things to keep your work as organized as possible: 1) Turn your paper sideways to get maximum lateral space (i.e. so you can keep long equations on a single line without going off the page) 2) In between row-equivalent representations of your matrix, make sure you explicitly write down every row operation you're performing 3) Simplify your algebra at every step whenever possible. Good luck. Let me know if you get it to work. Link to comment Share on other sites More sharing options...

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