the last time i spent solving asystem of equations dealing with the chilling of this hardboiledegg being put in an ice bath. we called t1 the temperature ofthe yoke and t2 the temperature of the white.what i am going to do is revisit that same system ofequations, but basically the topic for today is to learn tosolve that system of equations by a completely differentmethod. it is the method that isnormally used in practice. elimination is used mostly bypeople who have forgotten how to
do it any other way.now, in order to make it a little more general,i am not going to use the dependent variables t1 and t2because they suggest temperature a little too closely.let's change them to neutral variables.i will use x equals t1, and for t2 i will just usey. i am not going to re-deriveanything. i am not going to resolveanything. i am not going to repeatanything of what i did last
time, except to write down toremind you what the system was in terms of these variables,the system we derived using the particular conductivityconstants, two and three, respectively.the system was this one, minus 2x plus 2y.and the y prime was 2x minus 5y.and so we solved this by elimination.we got a single second-order equation with constantcoefficients, which we solved in the usualway.
from that i derived what the xwas, from that we derived what the y was, and then i put themall together. i will just remind you what thefinal solution was when written out in terms of arbitraryconstants. it was c1 times e to thenegative t plus c2 e to the negative 6t, and y was c1 over 2 e to the negative t minus 2c2 e tothe negative 6t. that was the solution we got. and then i went on to put ininitial conditions,
but we are not going to explorethat aspect of it today. we will in a week or so.this was the general solution because it had two arbitraryconstants in it. what i want to do now isrevisit this and do it by a different method,which makes heavy use of matrices.that is a prerequisite for this course, so i am assuming thatyou reviewed a little bit about matrices.and it is in your book. your book puts in a nice littlereview section.
two-by-two and three-by-threewill be good enough for 18.03 mostly because i don't want youto calculate all night on bigger matrices, bigger systems.so nothing serious, matrix multiplication,solving systems of linear equations, end-by-end systems.i will remind you at the appropriate places today of whatit is you need to remember. the very first thing we aregoing to do is, let's see.i haven't figured out the color coding for this lecture yet,but let's make this system in
green and the solution can be inpurple. invisible purple,but i have a lot of it. let's abbreviate,first of all, the system using matrices.i am going to make a column vector out of (x,y). then you differentiate a columnvector by differentiating each component.i can write the left-hand side of the system as (x,y) prime. how about the right-hand side?
well, i say i can just writethe matrix of coefficients to negative 2, 2,2, negative 5 times x,y. and i say that this matrixequation says exactly the same thing as that green equationand, therefore, it is legitimate to put it upin green, too. the top here is x prime.what is the top here? after i multiply these two iget a column vector. and what is its top entry?it is negative 2x plus 2y. there it is.
and the bottom entry the sameway is 2x minus 5y, just as it is down there.now, what i want to do is, well, maybe i should translatethe solution. what does the solution looklike? we got that,too. how am i going to write this asa matrix equation? actually, if i told you to usematrices, use vectors, the point at which you might bemost hesitant is this one right here, the very next step.because how you should write it
is extremely well-concealed inthis notation. but the point is,this is a column vector and i am adding together two columnvectors. and what is in each one of thecolumn vectors? think of these two things as acolumn vector. pull out all the scalars fromthem that you can. well, you see that c1 is acommon factor of both entries and so is e to the negative t,that function. now, if i pull both of thoseout of the vector,
what is left of the vector?well, you cannot even see it. what is left is a 1 up here anda one-half there. so i am going to write that inthe following form. i will put out the c1,it's the common factor in both, and put that out front.then i will put in the guts of the vector, even though youcannot see it, the column vector 1,one-half. and then i will put the other scalar function in back.the only reason for putting one
of these in front and one inback is visual so to make it easy to read.there is no other reason. you could put the c1 here,you could put it here, you could put the e negative tin front if you want to, but people will fire you.don't do that. write it the standard waybecause that is the way that it is easiest to read.the constants out front, the functions behind,and the column vector of numbers in the middle.and so the other one will be
written how?well, here, that one is a little more transparent.c2, 1, 2 and the other thing is e to the negative 6t. there is our solution.that is going to need a lot of purple, but i have it. and now i want to talk abouthow the new method of solving the equation.it is based just on the same idea as the way we solvesecond-order equations. yes, question.
oh, here.sorry. this should be negative two.thanks very much. what i am going to use is atrial solution. remember when we had asecond-order equation with constant coefficients the veryfirst thing i did was i said we are going to try a solution ofthe form e to the rt. why that?well, because oiler thought of it and it has been known for 200or 300 years that that is the thing you should do.well, this has not been known
nearly as long because matriceswere only invented around 1880 or so, and people did not reallyuse them to solve systems of differential equations until themiddle of the last century, 1950-1960.if you look at books written in 1950, they won't even talk aboutsystems of differential equations, or talk very littleanyway and they won't solve them using matrices.this is only 50 years old. i mean, my god,in mathematics that is very up to date, particularly elementarymathematics.
anyway, the method of solvingis going to use as a trial solution.now, if you were left to your own devices you might say,well, let's try x equals some constant times e to the lambda1t and y equals some other constant timese to the lambda2 t. now, if you try that, it is a sensible thing to try,but it will turn out not to work.and that is the reason i have written out this particularsolution, so we can see what
solutions look like.the essential point is here is the basic solution i am tryingto find. here is another one.their form is a column vector of constants.but they both use the same exponential factor,which is the point. in other words,i should not use here, in my trial solution,two different lambdas, i should use the same lambda.and so the way to write the trial solution is (x,y) equals two unknown numbers,
that or that or whatever,times e to a single unknown exponent factor.let's call it lambda t. it is called lambda.it is called r. it is called m.i have never seen it called anything but one of those threethings. i am using lambda.your book uses lambda. it is a common choice.let's stick with it. now what is the next step?well, we plug into the system. substitute into the system.what are we going to get?
well, let's do it.first of all, i have to differentiate.the left-hand side asks me to differentiate this.how do i differentiate this? column vector times a function.well, the column vector acts as a constant.and i differentiate that. that is lambda e to the lambdat. so the (x, y) prime is (a1,a2) times e to the lambda t times lambda. now, it is ugly to put the
lambda afterwards because it isa number so you should put it in front, again,to make things easier to read. but this lambda comes fromdifferentiating e to the lambda t and using the chain rule. this much is the left-handside. that is the derivative (x,y) prime. i differentiate the x and idifferentiated the y. how about the right-hand side.well, the right-hand side is negative 2, 2,2, negative 5 times what?
well, times (x, y), which is (a1,a2) e to the lambda t. now, the same thing that happened a month or a month anda half ago happens now. the whole point of making thatsubstitution is that the e to the lambda t,the function part of it drops out completely.and one is left with what? an algebraic equation to besolved for lambda a1 and a2. in other words,by means of that substitution,
and it basically uses the factthat the coefficients are constant, what you have done isreduced the problem of calculus, of solving differentialequations, to solving algebraic equations.in some sense that is the only method there is,unless you do numerical stuff. you reduce the calculus toalgebra. the laplace transform isexactly the same thing. all the work is algebra.you turn the original differential equation into analgebraic equation for y of s,
you solve it,and then you use more algebra to find out what the originallittle y of t was. it is not different here.so let's solve this system of equations.now, the whole problem with solving this system,first of all, what is the system?let's write it out explicitly. well, it is really twoequations, isn't it? the first one says lambda a1 isequal to negative 2 a1 plus 2 a2.that is the first one.
the other one says lambda a2 isequal to 2 a1 minus 5 a2. now, purely,if you want to classify that, that is two equations and threevariables, three unknowns. the a1, a2, and lambda are allunknown. and, unfortunately,if you want to classify them correctly, they are nonlinearequations because they are made nonlinear by the fact that youhave multiplied two of the variables.well, if you sit down and try to hack away at solving thosewithout a plan,
you are not going to getanywhere. it is going to be a mess.also, two equations and three unknowns is indeterminate.you can solve three equations and three unknowns and get adefinite answer, but two equations and threeunknowns usually have an infinity of solutions.well, at this point it is the only idea that is required.well, this was a little idea, but i assume one would think ofthat. and the idea that is requiredhere is, i think,
not so unnatural,it is not to view these a1, a2, and lambda as equal.not all variables are created equal.some are more equal than others.a1 and a2 are definitely equal to each other,and let's relegate lambda to the background.in other words, i am going to think of lambdaas just a parameter. i am going to demote it fromthe status of variable to parameter.if i demoted it further it
would just be an unknownconstant. that is as bad as you can be.i am going to focus my attention on the a1,a2 and sort of view the lambda as a nuisance.now, as soon as i do that, i see that these equations arelinear if i just look at them as equations in a1 and a2.and moreover, they are not just linear,they are homogenous. because if i think of lambdajust as a parameter, i should rewrite the equationsthis way.
i am going to subtract this andmove the left-hand side to the right side, and it is going tolook like (minus 2 minus lambda) times a1 plus 2 a2 is equal tozero. and the same way for the other one.it is going to be 2a1 plus, what is the coefficient,(minus 5 minus lambda) a2 equals zero. that is a pair of simultaneouslinear equations for determining a1 and a2, and the coefficientsinvolved are parameter lambda.
now, what is the point of doingthat? well, now the point is whateveryou learned about linear equations, you should havelearned the most fundamental theorem of linear equations.the main theorem is that you have a square system ofhomogeneous equations, this is a two-by-two system soit is square, it always has the trivialsolution, of course, a1, a2 equals zero.now, we don't want that trivial solution because if a1 and a2are zero, then so are x and y
zero.now that is a solution. unfortunately,it is of no interest. if the solution were x,y zero, it corresponds to the fact that this is an ice bath.the yoke is at zero, the white is at zero and itstays that way for all time until the ice melts.so that is the solution we don't want.we don't want the trivial solution.well, when does it have a nontrivial solution?nontrivial means non-zero,
in other words.if and only if this determinant is zero. in other words,by using that theorem on linear equations, what we find is thereis a condition that lambda must satisfy, an equation in lambdain order that we would be able to find non-zero values for a1and a2. let's write it out.i will recopy it over here. what was it?negative 2 minus lambda, two, here it was 2 and minus 5minus lambda.
all right.you have to expand the determinant.in other words, we are trying to find out forwhat values of lambda is this determinant zero.those will be the good values which lead to nontrivialsolutions for the a's. this is the equation lambdaplus 2. see, this is minus that andminus that, the product of the two minus ones is plus one.so it is lambda plus 2 times lambda plus 5,
which is the product of the twodiagonal elements, minus the product of the twoanti-diagonal elements, which is 4, is equal to zero.and if i write that out, what is that,that is the equation lambda squared plus 7 lambda, 5 lambda plus 2 lambda,and then the constant term is 10 minus 4 which is 6.how many of you have long enough memories,two-day memories that you remember that equation?when i did the method of
elimination, it led to exactlythe same equation except it had r's in it instead of lambda.and this equation, therefore, is given the samename and another color. let's make it salmon. and it is called thecharacteristic equation for this method.all right. now i am going to use now theword from last time. you factor this.from the factorization we get its root easily enough.the roots are lambda equals
negative 1 andlambda equals negative 6 by factoring theequation. now what i am supposed to do?you have to keep the different parts of the method together.now i have found the only values of lambda for which iwill be able to find nonzero values for the a1 and a2.for each of those values of lambda, i now have to find thecorresponding a1 and a2. let's do them one at a time.let's take first lambda equals negative one.my problem is now to find a1
and a2.where am i going to find them from?well, from that system of equations over there.i will recopy it over here. what is the system?the hardest part of this is dealing with multiple minussigns, but you had experience with that in determinants so youknow all about that. in other words,there is the system of equations over there.let's recopy them here. minus 2, minus minus 1 makesminus 1.
what's the other coefficient?it is just plain old 2. good.there is my first equation. and when i substitute lambdaequals negative one for the second equation,what do you get? 2 a1 plus negative 5 minusnegative 1 makes negative 4. there is my system that willfind me a1 and a2. what is the first thing younotice about it? you immediately notice thatthis system is fake because this second equation is twice thefirst one.
something is wrong.no, something is right. if that did not happen,if the second equation were not a constant multiple of the firstone then the only solution of the system would be a1 equalszero, a2 equals zero because the determinant of the coefficientswould not be zero. the whole function of thisexercise was to find the value of lambda, negative 1,for which the system would be redundant and,therefore, would have a nontrivial solution.do you get that?
in other words,calculate the system out, just as i have done here,you have an automatic check on the method.if one equation is not a constant multiple of the otheryou made a mistake. you don't have the right valueof lambda or you substituted into the system wrong,which is frankly a more common error.go back, recheck first the substitution,and if convinced that is right then recheck where you gotlambda from.
but here everything is goingfine so we can now find out what the value of a1 and a2 are.you don't have to go through a big song and dance for thissince most of the time you will have two-by-two equations andnow and then three-by-three. for two-by-two all you do is,since we really have the same equation twice,to get a solution i can assign one of the variables any valueand then simply solve for the other.the natural thing to do is to make a2 equal one,then i won't need fractions and
then a1 will be a2.so the solution is (2, 1).i am only trying to find one solution.any constant multiple of this would also be a solution,as long as it wasn't zero, zero which is the trivial one.and, therefore, this is a solution to thissystem of algebraic equations. and the solution to the wholesystem of differential equations is, this is only the (a1,a2) part. i have to add to it,as a factor,
lambda is negative,therefore, e to the minus t. there is our purple thing. see how i got it?starting with the trial solution, i first found outthrough this procedure what the lambda's have to be.then i took the lambda and found what the corresponding a1and a2 that went with it and then made up my solution out ofthat. now, quickly i will do the samething for lambda equals negative 6.each one of these must be
treated separately.they are separate problems and you are looking for separatesolutions. lambda equals negative 6.what do i do? how do my equations look now?well, the first one is minus 2 minus negative 6 makes plus 4.it is 4a1 plus 2a2 equals zero. then i hold my breath while i calculate the second one to seeif it comes out to be a constant multiple.i get 2a1 plus negative 5 minus negative 6, which makes plus 1.and, indeed,
one is a constant multiple ofthe other. i really only have on equationthere. i will just write downimmediately now what the solution is to the system.well, the (a1, a2) will be what?now, it is more natural to make a1 equal 1 and then solve to getan integer for a2. if a1 is 1, then a2 is negative2. and i should multiply that by eto the negative 6t because negative 6 is thecorresponding value.
there is my other one.and now there is a superposition principle,which if i get a chance will prove for you at the end of thehour. if not, you will have to do ityourself for homework. since this is a linear systemof equations, once you have two separatesolutions, neither a constant multiple of the other,you can multiply each one of these by a constant and it willstill be a solution. you can add them together andthat will still be a solution,
and that gives the generalsolution. the general solution is the sumof these two, an arbitrary constant.i am going to change the name since i don't want to confuse itwith the c1 i used before, times the first solution whichis (2, 1) e to the negative t plus c2, another arbitraryconstant, times 1 negative 2 e to the minus 6t. now you notice that is exactly the same solution i got before.the only difference is that i
have renamed the arbitraryconstants. the relationship between them,c1 over 2, i am now calling c1 tilda,and c2 i am calling c2 tilda. if you have an arbitraryconstant, it doesn't matter whether you divide it by two.it is still just an arbitrary a constant.it covers all values, in other words.well, i think you will agree that is a different procedure,yet it has only one coincidence.it is like elimination goes
this way and comes to theanswer. and this method goes acompletely different route and comes to the answer,except it is not quite like that.they walk like this and then they come within viewingdistance of each other to check that both are using the samecharacteristic equation, and then they again go theirseparate ways and end up with the same answer. there is something special ofthese values.
you cannot get away from thosetwo values of lambda. somehow they are reallyintrinsically connected. occurs the exponentialcoefficient, and they are intrinsically connected with theproblem of the egg that we started with.now what i would like to do is very quickly sketch how thismethod looks when i remove all the numbers from it.in some sense, it becomes a little clearerwhat is going on. and that will give me a chanceto introduce the terminology
that you need when you talkabout it. well, you have notes.let me try to write it down in general. i will first write it outtwo-by-two. i am just going to sketch.the system looks like (x, y) equals, i will still put itup in colors. except now, instead of usingtwos and fives, i will use (a,b; c, d). the trial solution will lookhow?
the trial is going to be (a1,a2). that i don't have to change thename of. i am going to substitute in,and what the result of substitution is going to belambda (a1, a2). i am going to skip a step andpretend that the e to the lambda t's have alreadybeen canceled out. is equal to (a,b; c, d) times (a1, a2).what does that correspond to? that corresponds to the systemas i wrote it here.
and then we wrote it out interms of two equations. and what was the resultingthing that we ended up with? well, you write it out,you move the lambda to the other side.and then the homogeneous system is we will look in general how?well, we could write it out. it is going to look like aminus lambda, b, c, d minus lambda. that is just how it looks there and the general calculation isthe same.
times (a1, a2) is equal tozero. this is solvable nontrivially.in other words, it has a nontrivial solution ifan only if the determinant of coefficients is zero. let's now write that out,calculate out once and for all what that determinant is.i will write it out here. it is a minus lambda times dminus lambda, the product of the diagonalelements, minus the anti-diagonal minus bc is equalto zero.
and let's calculate that out. it is lambda squared minus alambda minus d lambda plus ad, the constant term from here,negative bc from there, plus ad minus bc, where have i seen that before?this equation is the general form using letters of what wecalculated using the specific numbers before.again, i will code it the same way with that color salmon.now, most of the calculations will be for two-by-two systems.i advise you,
in the strongest possibleterms, to remember this equation.you could write down this equation immediately for thematrix. you don't have to go throughall this stuff. for god's sakes,don't say let the trial solution be blah,blah, blah. you don't want to do that.i don't want you to repeat the derivation of this every timeyou go through a particular problem.it is just like in solving
second order equations.you have a second order equation.you immediately write down its characteristic equation,then you factor it, you find its roots and youconstruct the solution. it takes a minute.the same thing, this takes a minute,too. what is the constant term?ad minus bc, what is that?matrix is (a, b; c, d).ad minus bc is its determinant.
this is the determinant of thatmatrix. i didn't give the matrix aname, did i? i will now give the matrix aname a. what is this?well, you are not supposed to know that until now.i will tell you. this is called the trace of a.put that down in your little books.the abbreviation is trace a, and the word is trace.the trace of a square matrix is the sum of the d elements downits main diagonal.
if it were a three-by-threethere would be three terms in whatever you are up to.here it is a plus b, the sum of the diagonalelements. you can immediately write downthis characteristic equation. let's give it a name.this is a characteristic equation of what?of the matrix, now.not of the system, of the matrix.you have a two-by-two matrix. you could immediately writedown its characteristic
equation.watch out for this sign, minus.that is a very common error to leave out the minus sign becausethat is the way the formula comes out.its roots. if it is a quadratic equationit will have roots; lambda1, lambda2 for the momentlet's assume are real and distinct. for the enrichment of yourvocabulary, those are called the eigenvalues.
they are something whichbelonged to the matrix a. they are two secret numbers.you can calculate from the coefficients a,b, and c, and d, but they are not in thecoefficients. you cannot look at a matrix andsee what its eigenvalues are. you have to calculatesomething. but they are the most importantnumbers in the matrix. they are hidden,but they are the things that control how this system behaves.those are called the
eigenvalues.now, there are various purists, there are a fair number of themin the world who do not like this word because it beginsgerman and ends english. eigenvalues were firstintroduced by a german mathematician,you know, around the time matrices came into being in 1880or so. a little while aftereigenvalues came into being, too.and since all this happened in germany they were namedeigenvalues in german,
which begins eigen and endsvalue. but people who do not like thatcall them the characteristic values.unfortunately, it is two words and takes a lotmore space to write out. an older generation even callsthem something different, which you are not so likely tosee nowadays, but you will in slightly olderbooks. you can also call them theproper values. characteristic is not atranslation of eigen,
but proper is,but it means it in a funny sense which has almostdisappeared nowadays. it means proper in the sense ofbelong to. the only example i can think ofis the word property. property is something thatbelongs to you. that is the use of the wordproper. it is something that belongs tothe matrix. the matrix has its propervalues. it does not mean proper in thesense of fitting and proper or i
hope you will behave properlywhen we go to aunt agatha's or something like that.but, as i say, by far the most popular thing,slowly the word eigenvalue is pretty much taking over theliterature. just because it's just oneword, that is a tremendous advantage.okay. what now is still to be done?well, there are those vectors to be found.so the very last step would be to solve the system to find thevectors a1 and a2.
for each (lambda)i,find the associated vector. the vector, we will call it(alpha)i. that is the a1 and a2.of course it's going to be indexed.you have to put another subscript on it because thereare two of them. and a1 and a2 is stretched alittle too far. by solving the system,and the system will be the system which i will write thisway, (a minus lambda, b, c, d minus lambda).
it is just that system that was over there, but i will recopyit, (a1, a2) equals zero, zero.and these are called the eigenvectors.each of these is called the eigenvector associated with orbelonging to, again, in that sense ofproperty. eigenvector,let's say belonging to, i see that a little morefrequently, belonging to lambda i.so we have the eigenvalues,
the eigenvectors and,of course, the people who call them characteristic values alsocall these guys characteristic vectors.i don't think i have ever seen proper vectors,but that is because i am not old enough.i think that is what they used to be called a long time ago,but not anymore. and then, finally,the general solution will be, by the superposition principle,(x, y) equals the arbitrary constant times the firsteigenvector times the eigenvalue
times the e to the correspondingeigenvalue. and then the same thing for thesecond one, (a1, a2), but now the second indexwill be 2 to indicate that it goes with the eigenvalue e tothe lambda 2t. i have done that twice.and now in the remaining five minutes i will do it a thirdtime because it is possible to write this in still a morecondensed form. and the advantage of the morecondensed form is a, it takes only that much spaceto write, and b,
it applies to systems,not just the two-by-two systems, but to end-by-endsystems. the method is exactly the same.let's write it out as it would apply to end-by-end systems.the vector i started with is (x, y) and so on,but i will simply abbreviate this, as is done in 18.02,by x with an arrow over it. the matrix a i will abbreviatewith a, as i did before with capital a.and then the system looks like x prime is equal to --
x prime is what?ax. that is all there is to it. there is our green system.now notice in this form i did not even tell you whether this atwo-by-two matrix or an end-by-end.and in this condensed form it will look the same no matter howmany equations you have. your book deals from thebeginning with end-by-end systems.that is, in my view, one of its weaknesses because idon't think most students start
with two-by-two.fortunately, the book double-talks.the theory is end-by-end, but all the examples aretwo-by-two. so just read the examples.read the notes instead, which just do two-by-two tostart out with. the trial solution is x equalswhat? an unknown vector alpha times eto the lambda t. alpha is what we called a1 and a2 before.plug this into there and cancel
the e to the lambda t's. what do you get?well, this is lambda alpha e to the lambda t equals a alpha e tothe lambda t. these two cancel.and the system to be solved, a alpha equals lambda alpha. and now the question is how doyou solve that system? well, you can tell if a book iswritten by a scoundrel or not by how they go --a book, which is in my opinion completely scoundrel,simply says you subtract one
from the other,and without further ado writes a minus lambda,and they tuck a little i in there and write alpha equalszero. why is the i put in there?well, this is what you would like to write.what is wrong with this equation?this is not a valid matrix equation because that is asquare end-by-end matrix, a square two-by-two matrix ifyou like. this is a scalar.you cannot subtract the scalar
from a matrix.it is not an operation. to subtract matrices they haveto be the same size, the same shape.what is done is you make this a two-by-two matrix.this is a two-by-two matrix with lambdas down the maindiagonal and i elsewhere. and the justification is thatlambda alpha is the same thing as the lambda i times alphabecause i is an identity matrix. now, in fact,jumping from here to here is not something that would occurto anybody.
the way it should occur to youto do this is you do this, you write that,you realize it doesn't work, and then you say to yourself idon't understand what these matrices are all about.i think i'd better write it all out.and then you would write it all out and you would write thatequation on the left-hand board there.oh, now i see what it should look like.i should subtract lambda from the main diagonal.that is the way it will come
out.and then say, hey, the way to save lambdafrom the main diagonal is put it in an identity matrix.that will do it for me. in other words,there is a little detour that goes from here to here.and one of the ways i judge books is by how well theyexplain the passage from this to that.if they don't explain it at all and just write it down,they have never talked to students.they have just written books.
where did we get finally here?the characteristic equation from that, i had forgotten whatcolor. that is in salmon.the characteristic equation, then, is going to be the thingwhich says that the determinant of that is zero.that is the circumstances under which it is solvable.in general, this is the way the characteristic equation looks.and its roots, once again, are theeigenvalues. and from then you calculate thecorresponding eigenvectors.
okay.go home and practice. in recitation you will practiceon both two-by-two and three-by-three cases,and we will talk more next time.
The Quadratic Formula Coloring Activity Egg Answers