 So, in the previous class I had given you a bunch of problems to try out many of them I had already told you the solutions to I hope you have tried to complete at least this one particular problem that I had given which was to find rational solutions of this have you completed the solution of this because if you do you will find out there are some issues with it even after we have parameterized. If you work this through you will see that you get that radical and when we are asking for rational solutions right things inside the radicals must be perfect squares right. So, if you just substitute all possible rational values for that parameter which is the ratio of x to y yeah I am referring to the problem that we discussed the previous day right then not every value would work only those values corresponding to which the entry inside that radical inside that square root is a perfect square of a rational number only those would work right. So, please try this it is very important particularly because subsequently we shall be talking about what kind of solutions are we looking for in general are we looking for solutions that are real that are natural numbers that are complex numbers and so on and so forth, but all that in good time before that let us just do some usual routine stuff that you are familiar with as I said it is linear algebra. So, we will be solving linear equations right. So, of the form let us say 2x plus 5y is equal to 12 and 3x plus 2y is equal to let us say 7 something like this right you know how to solve this you have done this there are so many methods one of the methods is just you know what you learnt in middle school just multiply this by a certain term this by a certain term subtract one from the other get rid of one of these unknowns right. As it turns out in this course we shall give it a name a special name which we call taking linear combinations of equations right. So, that cancellation that you did what you essentially did was a linear combination there are of course, other methods you might have used the determinant particularly when you had more unknowns like a 3 by 3 system you might not have preferred the elimination of one variable at a time you might have used the determinant and then you know the inverse of a matrix you could have used for example, if a matrix is invertible you know that it is given by the adjugate of the matrix A divided by its determinant yeah and all these things. But you might have lost much of the intuition behind what is going on when you actually cancelled these equations which is what we call now linear combinations. So, let us investigate what initially worked for us right when we first encountered these objects in our middle school and let us try to see where that takes us right try to get an algorithmic kind of a structure if I may call it that towards solving these equations and then we can have some hope that we will be able to mimic it for a host of other problems. Because once we have a structure and algorithm that works we can apply it to other cases too right it should be scalable. So, that is the goal. So, what is it that we are essentially doing let us write it in the most generic form the equation looks something like this let us just deal with real numbers for the time being and we are looking for solutions that are also real. So, this let us say B is this let us say the unknowns that is this is this and of course, A has to then be of size m cross n. So, this is a standard way of representing m equations in n unknowns right. So, each equation is like a constraint once again I will provoke you to think a little geometrically of course, that geometric thought does not go beyond three dimensions, but let us think about it in three dimensions itself. So, if I give you the equation of say 3 x plus 4 y plus 7 z is equal to 8 what is this it is a plane right. So, if I ask you to solve for this equation what does it geometrically mean it means that I can sketch this yeah x y z whichever way you like some right handed system. So, I can just sketch this plane in some manner and basically every point on this plane is a solution equation. Now, if I add another plane's equation say 7 x plus 5 y plus 10 z is equal to 14 which is not necessarily parallel to this just think geometrically it means we are talking about a second plane which probably intersects this plane I do not know how, but in some manner. Now, when I am asking for the solution the set of solutions has been cut down you realize that right earlier every point on this first plane was a feasible solution to the linear equation, but remember that is also a system of linear equations you might call it just one equation it is still fitting in with this structure does not it right because you can just say that m is equal to 1 n is equal to 3 this is also of the form A x is equal to B. Now, we are adding one more equation where n becomes 2 n still remains 3, but what is happening is that now your solutions are confined to a line which is fewer points as you might expect right. Now, if I choose carefully a third equation and I am hoping I am choosing good enough equations that are not you know dependent or whatever we will come to that term dependent later then what we expect is that we will have a third plane and now we will talk about a single point hopefully right. So, we are cutting down the possibility of solutions it is as if with the imposition of every additional equation you are imposing more and more constraints initially you are free to choose solutions on this whole plane then I cut it down by one by adding one more constraint and now you could choose solutions only within a line and if I have imposed a hard constraint which is more restrictive then now I am asking you to choose only one point right. So, it is as if I first before someone enters this room you are free to loiter around this room right and then when probably the instructor comes in you are expected to take your seat somewhere yeah and if the instructor is like a very strict one and says no you cannot chitchat you cannot discuss with anybody then you are sort of like you have to look forward. So, your degrees of freedom are being curtailed with the imposition of every single constraint is a very fundamental principle. So, the number of constraints plus the number of degrees of freedom yeah this turns out to be a constant for those of you who have done a preliminary course on linear algebra you might recognize this as some form of the so called rank nullity theorem ok we will revisit it later the implications of this, but this is true alright we will see how this can be manifested through the rank nullity theorem, but at least at this level through this geometric construction of this whatever pathetic drawing this might look like I hope you appreciate that you are imposing more and more constraints and therefore, your degrees of freedom are being reduced right. So, that is what the addition of every equation does when you are keeping the number of unknowns fixed, but you are adding more and more constraints it might even so happen that your existence might be you know something undesirable in the sense that I give you some infeasible set of equations which have no solutions which means someone does not want you in his or her home you are an undesirable guest right. So, that means there is no solution no freedom for you at all forget about freedom you even your existence is not desirable right. So, that is what I hope you get an intuitive idea about what this means, but we will go about it in a more formal fashion and we will of course try to look beyond three dimensions for which we need abstraction we need some sort of an algorithm that works each and every time. So, what is it that we would want to do these equations with increase in the number of variables or unknowns become more and more challenging to eliminate one by one the number of variables and the number of operations and it seems like it is something based on observation you look at these numbers you look at 2 there you look at 5 there you look at 7 there and you think multiply this by 7 there has to be a systematic way right we want some structure so that someday a computer can do the same thing that you are doing through observation we do not want that vagueness in the picture. So, what is it that allows us to do so if this A matrix has a nice structure perhaps it will be easier for us to solve for this the simplest thing I can think of is when it is a square matrix and the A matrix is just identity the most desirable case the easiest case the trivial case right, but of course when you have m and n different you cannot have an identity. So, how do we go about this business of simplifying the structure of this A matrix while ensuring that we are looking for exactly the same solutions mind you. So, if you are tweaking this system of equations you should not be looking for a restricted subset of the solutions neither you should you be looking for spurious solutions that are not solutions of the original system. So, what do you have to preserve when you are trying to tweak this system to look nice you have to always bear in mind that you are not admitting some superfluous solutions neither are you ignoring some solutions that were there in the actual system, but now in your modified system that is no longer a solution. So, what we are trying to do now the goal before us that we are going to set for ourselves is this we are going to say we are going to get some nice looking structure for A we are going to try to do that at least whilst ensuring that this new modified system of equations has exactly the same solution set as the original system of equations. So, what is a way of ensuring that you generate a new system of equations which has the same set of solutions well let us do a simple trial. So, let me write it down in a little more elaborate fashion now. So, let us say this is a 1 1 x 1 plus a 1 2 x 2 plus so on a 1 n x n is equal to b 1 similarly a 2 1 x 1 plus a 2 2 x 2 plus a 2 n x n is equal to b 2 so on until the last one is a m 1 x 1 plus a m 2 x 2 plus plus a m n x n is equal to b m right. So, that is the way it is. Now suppose I want to cook up the m plus first equation. So, let us try to cook up the m plus first equation from these m equations by taking a linear combination. What do I mean by that? What I mean is quite simply this multiply this equation by some alpha 1 this equation by some alpha 2 so on until you have this equation some alpha m on both sides precisely. So, then what do we get? This is equation number 1 is equation number 2 is equation number m. So, what is equation number m plus 1? Equation number m plus 1 will look something like this I will write in a short hand summation notation. So, you can see it is going to be alpha 1 a 1 1 alpha 2 a 2 1 right. So, the sum is over alpha 1 and over the first index right. Please ask if there is any question about this notation alpha i a i 1 x 1 plus summation alpha i a i 2 x 2 plus dot dot dot till I have the last term which is summation alpha i a i n x n is equal to summation alpha i b i where of course all these i's are being summed from 1 through m right. So, i running from 1 through m 1 through m. So, that is my most general construction of a linear combination where I can choose these alpha 1 through alpha m to be any real number that I please yeah so far so good nothing wrong with this. Now, what is so interesting or what is so special about this m plus first equation that I have cooked up. So, consider any solution of this original system of equations. So, let so let zeta of course in R n be a solution of A x is equal to B right. I am going to claim claim that zeta satisfies the m plus first equation is that obvious how do you check well the conventional method left hand side must be equal to right hand side. So, you just open it up left hand side and right hand side and you equate that. So, on the left hand side also all these terms they equate to the same as this is just substitution really. If there is any doubt please ask I can just do a bit of laborious writing down of those expansions, but maybe I will leave it to you as an exercise to check it is really not that hard right yes. What I am saying is that I had m equations originally I have cooked up the m plus first equation this is something that was not originally given to me this is something that I have cooked up through a linear combination of the first m equations the original m equations. Now, I am saying that if I know that this zeta satisfies the original system of equations then this zeta must also satisfy this new equation that I have now cooked up. In order to prove that all you have to do is to substitute that zeta in place of x. So, zeta has n tuples and every one of those n tuples gets substituted in place of x 1 through x n yeah and because it satisfies the original system of equations you will see that on the left hand side also you will have the same term as on the right hand side here right. So, I am meaning that proof I mean it is not really much of a proof really, but this is just the assertion, but you will immediately realize that not every solution of this is a solution of this because this has a lot of degrees of freedom right just the discussion we had about freedom and constraints. This has presumably m constraints at most whereas, this has at most one constraints. So, there are obviously situations where this admits a way more number of solutions than the original system of equations. So, just by solving this you cannot say oh ok I have found a solution to the original problem. So, you might need to cook up more of these linear combinations, but what is the systematic way what is the guarantee that you will be able to do this and still retain and zero in on the original system of equations and their solutions right. So, this is where we are going to make this claim now or rather we should first define something. So, this observation although I have not proved it please complete this proof even if it is trivial it is apparently trivial, but you should still complete this and convince yourself that it is true because every subsequent result that we prove will build on the previous the preceding results. So, now we are going to define something we are going to define equivalent systems ok. So, what are equivalent systems suppose you have system 1 given by A x is equal to B and system 2 as A bar x is equal to B bar ok right. You know the sizes let us just say A is A comma A bar ok let us let us not talk about the sizes yet let us just keep them different in general right now ok. Now, what is it that makes them equivalent the definition here ok. The definition is as follows if every equation in system 1 is a linear combination of those in system 2 and the other way round that is every system every equation in system 2 is also a linear combination of every system in equation every I mean every equation in system 2 is a linear combination of the equations in system 1. So, it works both ways then system 1 and system 2 are equivalent. So, it is all very well may be another definition for you to gobble up and stuff you know what is the use of all this ok. Why are we introducing this definition because the next result that comes up is going to be incredibly interesting ok. So, what is the result which we are shooting for for which we have actually set up this definition because we do not want to have to say all this you see I told you the other day the reason we set up definitions is because we want to use here after when we use the term equivalent systems you will immediately know what it means I do not have to always say every equation in system 1 is a linear combination of those in system 2 every equation in system 2 is a linear combination of those in you see the number of words I am spending right instead of that I will just say that equivalent systems. So, now the first important result is going to be this if 2 systems of linear equations. So, let us call this a proposition ok equations are equivalent then they have the same or they admit the same solution set that is crucial because remember our goal is to get from one system of linear equations to another system of linear equations that are much nicer ok while preserving the fact that we are not admitting any extra solutions neither are we cutting down on any existing solutions. So, if it turns out that there is a way if this result is to be believed that is then it turns out our goal is now reformulated what is our goal a way of cooking up equivalent systems you see. So, if this result is true we should now aim to cook up equivalent systems that look nicer. So, once we have a recipe for cooking up equivalent systems yeah then we should try to achieve our goal of cooking them up in a in a way that makes it look nicer, but of course for that we need to prove this ok. So, there is the proposition maybe I will just try to give you a very sketchy idea of the proof, but an important element is here. So, how do you prove to sets are equal what is the most standard way of showing the 2 sets are equal I mean if it is 2 numbers you know there are so many different ways how do you generally approach this problem of proving the 2 sets are equal yes thank you. So, you try to show if there are 2 sets S 1 and S 2 first show that S 1 must be contained in S 2 then you show that S 2 must be contained in S 1 and if both of them have to be true then they cannot help, but be equal in fact the same holds for numbers if you show that one is less than or equal to the other and the other is also less than or equal to the first then they must be equal right. So, that is the standard approach. So, let us say suppose S 1 and S 2 are solution sets of A x is equal to b and A bar x is equal to b bar respectively then what do we need to do first we have to show that S 1 is contained in S 2 this is to show that S 1 is contained in S 2. So, we pick any arbitrary element from S 1 and if we can show that it must also belong to S 2 then it must be true that the entire entirety of S 1 must be contained in S 2 right. So, pick any element of S 1 say zeta 1 that means A zeta 1 is equal to b right, but by our preceding result what does it mean? What do we know about every equation in the second system of equations? Every equation in this is a linear combination of the equations in this right and by our preceding result therefore, if zeta 1 satisfies system 1 which it indeed does because it is a member of S 1 the set of solutions of system 1 therefore, it must also satisfy let us say this is m equations here then the m plus first equation is the first equation here the m plus second equation is the second equation here until the m plus m bar is equation here is the last equation here whether m bar equations in the second system let us say. So, that is true then can I not say that it is obvious that then so, always be where when people use the term obviously yeah be very suspicious even if I am using it it is a sure shot way of probably cheating people yeah obviously then obviously A bar zeta 1 is equal to b bar, but I hope at least this is obvious. So, I am allowed to use this obviously I have given you the explanation right. So, then any element that belongs to S 1 must belong to S 2 because this implies that zeta 1 must belong to S 2 yeah. So, anything that comes from S 1 must also belong to S 2. So, therefore, what we have essentially proved I may now erase this definition here what we have essentially proved now is the fact that S 1 is contained in S 2 that is therefore, S 1 is contained in S 2, but because the definition of equivalent systems entails that the same thing goes further when the story is flipped. So, now if you want to show that S 2 is contained in S 1 take any solution of the second system of equations that is A bar x is equal to b bar right and that must also satisfy each and every equation of the first system of equations yeah because every equation here is also linear combination of these fellows right. So, by similar reasoning of course, by flipping the indices S 2 must also be contained in S 1 ergo S 1 is equal to S 2. So, that means we now need to understand how to get to an equivalent system of equations given a system of equations what is the recipe for cooking up equivalent systems because it turns out that once we are living within the realm of equivalent systems it suffices if you solve for any one of them. So, let us solve for the nicest looking one the easiest one to solve. If you have done it you solve for the entire system of equivalent equivalent system of equations right all of them at one go. So, which is the nicest one we have to identify you might argue it is a matter of taste, but there is something that is universally agreed upon as to which is a nicer way of representing which is the nicest form yeah, but the question still remains what is the way to get there how to get to an equivalent system of equations for that we will make a few observations especially pertaining to this thing that we have been using so loosely up until this point which is taking linear combinations of equations.