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. Let’s 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, let’s 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-Let’s
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-let’s
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 let’s 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
let’s 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-that’s 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 I’ll 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 we’ll 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 it’s 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, let’s see for consistent system
of equations. Let’s see how we prove that, that we cannot have just the
finite number of solutions greater than one. So let Y, so let’s suppose
somebody says hey you know, I know that Y and Z these are the vectors. So
let’s suppose this is a n by n matrix, n by one and
this is n by one. And lets these be column vectors. So let’s 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 I’m 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, I’ll 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 let’s see what we get by now we have these two sets of equations one
and two and we can add one and two. Let’s see if we add one and two what we get.
We’re 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 I’ll 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 you’re
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 I’ll have infinite
combinations of Y and Z which are going to give me the solutions to AX equal
to C. So that’s why it’s 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 that’s the end
of this proof and that’s also the end of this segment. |