You see nonlinear equations, they see linear algebra! (Harvard-MIT math tournament)

so I just know a very nice way to use linear  algebra to solve this system of equations but   I'm not talking about the gaussian elimination  because this right here is not linear because we   have the x * Y and also the x * Z this question is  from the harvard MIT math tournament in the year   of 2023 I will have the link in the description  so you can go check it out and I learned the   following approach from their solution but for  now you know the deal please pause the video   and try this first done now let's talk about it  of course you can kind of combine the equations   do substitutions and all that but I will leave  that to you how can we use linear algebra for   a nonlinear system well earlier we mentioned it  because we have the x * Y and the x * Z that's   not linear but what if you cannot just preset the  X to be a number let say 177 then in that case we   just have Y and Z to be the variables right so  here's the deal I think this super cool right   we are going to first by treating X as a constant  so we can narrow this down to a 3X two system of   equations and let me rewrite it here we will just  have the x * y + z = 40 x * Z so I'll put it here   line up right and then the Y here and then equals  51 and because X is a constant now I'll just put   x to the other side so y + z = 19 - x and now we  have a 3x2 syst of equations H okay but what good   does this do for us though well not let's think  about it we have more equations than the unnown   it's kind of dangerous sometimes because for  example let's say I give you five y + 2 Z = 10   and I give you another one I say y + z = 3 we can  solve this and then we do have a unique solution   for that because these two lines they do intercept  at a point very nice but if you just give me   another equation let say 2 y - Z = 1 unfortunately  this system no longer have a solution because this   three lines they do not intersect one single  point anymore H so how can we have a solution   in this kind of situation though well there's  one condition or one situation that can happen   it's like the three equations that redundant what  I mean by that is just imagine you just go ahead   and add this and that you get six you add this and  that you gu three and you add this and 13 now what   this system of equations they will of course have  solution because you would really just have the   same as the original one yeah in fact this is  not the only way in general for such system to   have a solution you just have to make sure that  one of the equations is a linear combination of   the other two it means the following look at  that the first equation may be multiply by any   constant you want but not zero let's say multiply  by two so I'm looking at 10 y + 4 Z = 20 and then   look here another equation and multiply by any  constant you want but not zero let say multiply   by -1 so - y - Z = -3 then you combine them this  is what we mean by a linear combination so this   and that was nine this and that is + three this  and that is 17 good this right here will still   have a solution okay but how can we be sure that  one of this right here is a linear combination of   the other two though well this is where the linear  algebra comes in let's look at this right here as   a matrix let's write down the coefficients and  I'll just put on the original 519 and then 2 1   3 and then 10 3 17 if one of the rows is a linear  combination of the other two then if you compute   the determinant of it you will for sure get zero  but I'm not going to compete this right here just   take a look at won Alpha right now and now how  can we use this to our Vantage well let's just   go ahead and do what we did let's put this into a  3X3 Matrix here we have X 1 1 and then this is 1   X1 and this right here is 40 51 and 19 - x now we  have a 3X3 Matrix and let's just go ahead look at   its determinant and then hope that we can get  zero out of it but be careful though I want to   mention this before we actually compute it when  we do this sometimes even though the determinant   of such Matrix gives a zero but such system might  not still have a solution for example let me show   you guys an example real quick a redundant one  let's say we have y + z equal 1 and then y + z   equals 2 of course this system will not have a  solution right because they contradict already   and if you have a third one I say I will just do  2 y + 2 Z = 3 guess what this right here has no   solution but if you compute such determinant  I'm looking at one one2 1 one2 and then 1 2   3 this determinant will still be equal to zero  so we have to be careful after we compute the X   from here we still have to go back and make sure  that the X does work so now how do we compute the   determinant of a 3X3 Matrix tricks well we can do  this trick let's write down the first two columns   so we have X11 and also 1 X1 and then I'm just  going to multiply them diagonally like this so   we'll get x square * that so I will just put this  down right here and then next we do this times   this times that and we add 1 * 51 * 1 is + 51 and  then we keep adding this * this * that 40 * 1 * 1   is 40 then we go the other way 1 * 1 * X well  19 - x when we go the other way we subtract so   minus we have just 19 - x and then we do x * 51  * 1 so that's - 51x lastly 40 * x * 1 so that's   Min - 40x then as we mentioned earlier we want to  set this to be equal to zero and hope for the best   now let's solve this real quick multiply this out  we will get 19 x² - X3 power this and that + 91   distribute we get minus 19 + x and then this and  is minus 50 - 91x and that's equal to Zer and then   this right here ISX to 3 power + 19 x^2 and then  this and that is minus 90x and lastly we have this   and that that will be plus 72 equal Z and that's  solve this cubic and to do so perhaps let multiply   Everybody by negative 1 or divide it by negative  1 x to thir power - 19 x^ 2 power + 90 x - 72 = 0   this is actually not so bad because notice 1 - 19  is -8 + 90 is precisely 72 and then - 72 is equal   to Z that means x = 1 is a root so we can do the  following take this divide it by x - one and then   we'll do the synth deficient real quick write  down the coefficients we have 1 9 and then + 90   and then - 72 because we are trying to divide  it by x - one we put a one right here and then   go ahead and just write down the one right here  and then put a 1 * 1 which is one and we add so   we get 18 continue we get 18 here that will give  us 72 lastly positive 72 the remainder is equal to   zero it has to be because we know this right  here has a factor of xus one then this right   here will tell us the coefficient of the next  factor which will be a quadratic because cubic   divided by a linear we get quadratic 1 x 2 - 18  x + 72 Factor this we get x - 12 and then x - 6   so as we can see we have three possibilities for  X to consider 1 12 and six now let's go ahead and   plug in these three values into X solve them so  plugging one first when we have X is equal to 1   well we will just get 1 y + + z = 40 and then when  we plug in one in here this is the same as y + 1 Z   equals 51 and do we have to continue no because  y + z this one says it's 40 and this one says   it's 51 so in fact X actually cannot be equal to  one otherwise we get we don't get a solution for   that now try xal 6 so put six into here we are  looking at 6 y + z = 40 and then put 6 in here   let's write down y first so y + 6 Z = 51 and then  put 6 into here subtract six on both sides so we   have y + z = 13 now I want to show you something  real quick this one right here is technically a   linear combination of the first two why because  if you add them up well we'll just get 7 y + 7   Z = 91 and we can divide everything by seven if  we divide everything by seven we do get the y + c   equal 13 so as you can see we have a chance all  right now let's just go ahead and solve this the   regular way I'm going to pick these two equations  I'm just going to multiply this one by negative   and then solve it this and that is zero this minus  that is 5 Z equals this minus that is 38 and then   let's divide both sides by five C is equal to  let's use decimal we have 7.6 put this back so   we get y + 7.6 = 13 and then y = this minus that  is 5 4 yes so 6 5.4 and 7.6 now let's try when   X is equal 12 so when X is equal to 12 we'll get  the following we have 12 y + z = 40 and then let's   write down y + 12 C equals 51 and then put 12 in  here subtract 12 on both sides so we get y + z = 7 okay now we have 3x2 system of equations again  and I'm just going to do the same thing like there   over there so subtract this is zero this is 11  Z that will will be 44 so that means Z = 4 and   then put a four in here we will get y + 4 = 7  so Y is equal to 3 so now ladies and gentlemen   here are the answers the first one is six 5.4  and then 7.6 and the other one is 12 3 and 4 whoow how cool is this if this is the kind of  question that interest you and want to learn   more about how to solve difficult questions  or just want to enhance your problem solving   skill then I will highly recommend you guys to  check out today's sponsor brillian thatw work   brillian is online learning platform that helps  you to Exel in math and science their goal is to   provide the most effective way for you to learn  I really like brilliant not only because they   got thousands of interactive lessons ranging  from basic algebra to advanced math but also   each lesson is filled with Hands-On problem  solving that let you test out the concepts   along the way as a maap teacher for over 10  years I always advise my students that you   need to be constantly learning and practicing  in order to reach your goal no matter if that's   to participate and win the math competition or  simply just to master a subject BR is the best   choice for you now you can try Bri for free  for 30 days use the link in the description   b/ blackpen red pan so that you can also get 20%  off discount for the annual premium subscription   and I want to thank B for sponsoring this video I  also want to thank you guys for checking them out

Channel: blackpenredpen

Views: 138,816

Keywords: harvard mit math tournament, hmmt, blackpenredpen, math for fun, math competition problem, hard algebra problem, linear algebra problem

Id: WsNnmB8NCRo

Length: 15min 16sec (916 seconds)

Published: Sat Mar 30 2024