CHAPTER 05.16: SYSTEM OF EQUATIONS: Can a system of equations have more then one but not infinite number of solutions?
In this segment we will talk about a real life problem of setting up simultaneously linear equations. Letís suppose somebody gives you a velocity of a rocket as a function of time. The rocket is going in an upward motion-a straight line, letís suppose- And we are given the velocity of the rocket at different times: 5 8 12 seconds. The values given are 106.8, 177.2 m/s, 279.2 m/s. And one of the questions which† people might ask you about when somebody gives you time data at (47-49)number of points is-Hey can you find the velocity at some point-Letís suppose 6 seconds or so or 9 seconds or so, somewhere in between 5 and 12. And one of the ways to do it is to simply draw a straight line between-letís suppose 2 consecutive points like if somebody is asking you to find the velocity at 6 seconds. You might say ok hey, I am going to take these two points and draw a straight line and find out what the velocity of 6 seconds. Or letís suppose somebody asked me to find the velocity of 10seconds I can always draw a line between these two data points and be able to figure that out.
But letís suppose somebody says hey, no you have to do better than a straight line. So if we plot these points on a piece of graph paper. We got velocity here and time here. So velocity at 5 is given as 5, 106.8-thatís the coordinate there- Then it is 8, 177.2. Then at 12 it is 279.2. And what somebody is asking you is, hey, I want you to draw a second order polynomial. I want to have a velocity profile or a velocity curve, which is a 2nd order polynomial which is going through these 3 points. So in that case, if we are going to call that velocity profile or velocity curve to be A T squared plus BT plus C. That is our 2nd order polynomial which is going to approximate the velocity from 5 to 12 seconds. So our velocity profile or what we call a interpolant also will be AT squared plus BT plus C 5 less than or equal to T less than or equal to 12 because you are going to write an interpolant for discrete data, it has to be also, what has to be given is the domain in which that particular interpolant is valid.
So the question is that: If I know, if I can find out what this velocity profile is, then Iíll be able to find the velocity at any point between time equal to 5 and time equal to 12 seconds. So the question rises, hey, how do I find out A B and C? Well the thing we are seeing here is that this velocity profile the 2nd order of the velocity profile, the 2nd polynomial of the velocity profile is going through 3 points. So that is going to help me set up the 3 equations. So what I mean by that is hey, if I say the velocity at 5-velocity at 5 is given by A times 5 squared plus B times 5 plus C. But what is the velocity at 5? It is 106.8. And what is the velocity at 8? It will be A times 8 squares plus B times 8 plus C is equal to 177.2. And what is the velocity at 12? It is A times 12 square plus B times 12 plus C equals 279.2. Now if I make the substitute and make the simplification here, the 1st equation, which I have here, will become 25A plus 5B plus C is equal to 106.8. Then the 2nd equation, which is right here, will become 64A plus 8B plus C is equal to 177.2. The 3rd equation, which is right here, will be 144A plus 12B plus C is equal to 279.2.
So what I have is 3 equations and 3 unknowns. So this is my first equation, this my 2nd equation and this is my 3rd equation. So I have 3 equations and the 3 unknowns are A B and C. SO this is an example of a real life problem of a simultaneous linear equation. So we got 3 equations and we got 3 unknowns. If we are able to solve these 3 simultaneous linear equations we will be able to find the values of A B and C. Once we know the values of A B and C is, we can have the velocity profile or the velocity cure with time equal to 5 to time equal to 12 hence being able to find the approximate value of the velocity at any other time other than 5 8 and 12 which are already given to us. And that is the end of this segment. In this segment weíll talk about if a system of equations can have more than one but not infinite number of solutions. We only talked about that if for a system of equations like AX equal to C itís consistent. Then the only two possibilities are that it has a unique solution or an infinite number of solutions. We cannot say that hey it has only five solutions or six solutions or seven solutions. It cannot choose a number which is finite other than one. So, letís see for consistent system of equations. Letís see how we prove that, that we cannot have just the finite number of solutions greater than one. So let Y, so letís suppose somebody says hey you know, I know that Y and Z these are the vectors. So letís suppose this is a n by n matrix, n by one and this is n by one. And lets these be column vectors. So letís suppose somebody says hey let Y and Z be the solution to this set of equations AX equal to C. So somebody is giving you an equation of unknowns and you say hey Iím finding two vectors to be the solution Y and Z. And what we want to be able to show is that hey it is not possible to just have two solutions or three solutions or four solutions, it has to be infinite if we have more than one solution. So if Y and Z are solutions to AX equal C, that means that A times Y has to be equal to C. Because we just said that Y is the solution to the AX equal to C. And A times Z will also will have to be equal to C because Z is also a solution. Now what that means is that if I look at this particular equation set of equations, Iíll have R times A times Y will be equal to R times C. So if R is a scalar I can multiply both sides by R, R is a scalar. And what I do is I multiply this one by one minus r because r is a scalar, one minus r is also scalar and it is a perfectly good scalar to use so I say A times Z is equal to one minus r times C like this. So once that is happened letís see what we get by now we have these two sets of equations one and two and we can add one and two. Letís see if we add one and two what we get. Weíre going to get R times A times Y plus one minus r times A times Z is equal to r times c plus one minus r times c. And what I can do is, I can take A common here and Iíll get r times† plus one minus r times z here will be equal to r times c plus c minus r times c. This and this cancels so that is equal to c. So what youíre basically finding out is that for that A multiplied by this is equal to the right hand side vector here. That means that this is a solution. If Y and Z are solutions to AX equal to C then this is also a solution because it satisfies A times some vector equal to the right hand side vector. But we can very well see that here if this is a solution r can take any value. R is a scalar. R can take a value of 2, 3, 5, 6.5, one billion, two point two billion; whatever value which you might want, whatever real number which you might want to choose for R. So that means that Iíll have infinite combinations of Y and Z which are going to give me the solutions to AX equal to C. So thatís why itís not possible to have finite numbers of solutions which is a number greater than one. You can have no solution, unique solution, and infinite solutions. You cannot have two solutions, three solutions, four solutions or any finite number other than zero or one when somebody is telling you to solve an equation or unknowns. And thatís the end of this proof and thatís also the end of this segment.