WEBVTT
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The last time I spent solving a
system of equations dealing with
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the chilling of this hardboiled
egg being put in an ice bath.
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We called T1 the temperature of
the yoke and T2 the temperature
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of the white.
What I am going to do is
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revisit that same system of
equations, but basically the
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topic for today is to learn to
solve that system of equations
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by a completely different
method.
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It is the method that is
normally used in practice.
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Elimination is used mostly by
people who have forgotten how to
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do it any other way.
Now, in order to make it a
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little more general,
I am not going to use the
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dependent variables T1 and T2
because they suggest temperature
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a little too closely.
Let's change them to neutral
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variables.
I will use x equals T1,
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and for T2 I will just use
y.
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I am not going to re-derive
anything.
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I am not going to resolve
anything.
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I am not going to repeat
anything of what I did last
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time, except to write down to
remind you what the system was
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in terms of these variables,
the system we derived using the
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particular conductivity
constants, two and three,
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respectively.
The system was this one,
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minus 2x plus 2y.
And the y prime was
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2x minus 5y.
And so we solved this by
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elimination.
We got a single second-order
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equation with constant
coefficients,
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which we solved in the usual
way.
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From that I derived what the x
was, from that we derived what
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the y was, and then I put them
all together.
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I will just remind you what the
final solution was when written
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out in terms of arbitrary
constants.
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It was c1 times e to the
negative t plus c2 e to the
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negative 6t,
and y was c1 over 2 e
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to the negative t minus 2c2 e to
the negative 6t.
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That was the solution we got.
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And then I went on to put in
initial conditions,
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but we are not going to explore
that aspect of it today.
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We will in a week or so.
This was the general solution
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because it had two arbitrary
constants in it.
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What I want to do now is
revisit this and do it by a
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different method,
which makes heavy use of
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matrices.
That is a prerequisite for this
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course, so I am assuming that
you reviewed a little bit about
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matrices.
And it is in your book.
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Your book puts in a nice little
review section.
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Two-by-two and three-by-three
will be good enough for 18.03
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mostly because I don't want you
to calculate all night on bigger
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matrices, bigger systems.
So nothing serious,
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matrix multiplication,
solving systems of linear
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equations, end-by-end systems.
I will remind you at the
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appropriate places today of what
it is you need to remember.
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The very first thing we are
going to do is,
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let's see.
I haven't figured out the color
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coding for this lecture yet,
but let's make this system in
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green and the solution can be in
purple.
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Invisible purple,
but I have a lot of it.
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Let's abbreviate,
first of all,
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the system using matrices.
I am going to make a column
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vector out of (x,
y).
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Then you differentiate a column
vector by differentiating each
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component.
I can write the left-hand side
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of the system as (x,
y) prime.
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How about the right-hand side?
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Well, I say I can just write
the matrix of coefficients to
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negative 2, 2,
2, negative 5 times x,y.
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And I say that this matrix
equation says exactly the same
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thing as that green equation
and, therefore,
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it is legitimate to put it up
in green, too.
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The top here is x prime.
What is the top here?
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After I multiply these two I
get a column vector.
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And what is its top entry?
It is negative 2x plus 2y.
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There it is.
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And the bottom entry the same
way is 2x minus 5y,
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just as it is down there.
Now, what I want to do is,
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well, maybe I should translate
the solution.
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What does the solution look
like?
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We got that,
too.
How am I going to write this as
a matrix equation?
Actually, if I told you to use
matrices, use vectors,
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the point at which you might be
most hesitant is this one right
here, the very next step.
Because how you should write it
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is extremely well-concealed in
this notation.
But the point is,
this is a column vector and I
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am adding together two column
vectors.
And what is in each one of the
column vectors?
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Think of these two things as a
column vector.
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Pull out all the scalars from
them that you can.
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Well, you see that c1 is a
common factor of both entries
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and so is e to the negative t,
that function.
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Now, if I pull both of those
out of the vector,
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what is left of the vector?
Well, you cannot even see it.
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What is left is a 1 up here and
a one-half there.
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So I am going to write that in
the following form.
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I will put out the c1,
it's the common factor in both,
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and put that out front.
Then I will put in the guts of
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the vector, even though you
cannot see it,
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the column vector 1,
one-half.
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And then I will put the other
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scalar function in back.
The only reason for putting one
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of these in front and one in
back is visual so to make it
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easy to read.
There is no other reason.
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You could put the c1 here,
you could put it here,
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you could put the e negative t
in front if you want
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to, but people will fire you.
Don't do that.
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Write it the standard way
because that is the way that it
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is easiest to read.
The constants out front,
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the functions behind,
and the column vector of
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numbers in the middle.
And so the other one will be
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written how?
Well, here, that one is a
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little more transparent.
c2, 1, 2 and the other thing is
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e to the negative 6t.
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There is our solution.
That is going to need a lot of
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purple, but I have it.
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And now I want to talk about
how the new method of solving
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the equation.
It is based just on the same
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idea as the way we solve
second-order equations.
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Yes, question.
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Oh, here.
Sorry.
This should be negative two.
Thanks very much.
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What I am going to use is a
trial solution.
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Remember when we had a
second-order equation with
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constant coefficients the very
first thing I did was I said we
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are going to try a solution of
the form e to the rt.
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Why that?
Well, because Oiler thought of
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it and it has been known for
or 300 years that that is the
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thing you should do.
Well, this has not been known
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nearly as long because matrices
were only invented around
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or so, and people did not really
use them to solve systems of
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differential equations until the
middle of the last century,
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1950-1960.
If you look at books written in
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1950, they won't even talk about
systems of differential
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equations, or talk very little
anyway and they won't solve them
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using matrices.
This is only 50 years old.
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I mean, my God,
in mathematics that is very up
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to date, particularly elementary
mathematics.
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Anyway, the method of solving
is going to use as a trial
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solution.
Now, if you were left to your
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own devices you might say,
well, let's try x equals some
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constant times e to the lambda1
t and y
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equals some other constant times
e to the lambda2 t.
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Now, if you try that,
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it is a sensible thing to try,
but it will turn out not to
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work.
And that is the reason I have
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written out this particular
solution, so we can see what
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solutions look like.
The essential point is here is
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the basic solution I am trying
to find.
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Here is another one.
Their form is a column vector
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of constants.
But they both use the same
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exponential factor,
which is the point.
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In other words,
I should not use here,
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in my trial solution,
two different lambdas,
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I should use the same lambda.
And so the way to write the
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trial solution is (x,
y) equals two unknown numbers,
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that or that or whatever,
times e to a single unknown
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exponent factor.
Let's call it lambda t.
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It is called lambda.
It is called r.
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It is called m.
I have never seen it called
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anything but one of those three
things.
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I am using lambda.
Your book uses lambda.
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It is a common choice.
Let's stick with it.
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Now what is the next step?
Well, we plug into the system.
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Substitute into the system.
What are we going to get?
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Well, let's do it.
First of all,
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I have to differentiate.
The left-hand side asks me to
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differentiate this.
How do I differentiate this?
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Column vector times a function.
Well, the column vector acts as
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a constant.
And I differentiate that.
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That is lambda e to the lambda
t.
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So the (x, y) prime is (a1,
a2) times e to the lambda t
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times lambda.
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Now, it is ugly to put the
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lambda afterwards because it is
a number so you should put it in
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front, again,
to make things easier to read.
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But this lambda comes from
differentiating e to the lambda
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t and using the chain rule.
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This much is the left-hand
side.
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That is the derivative (x,
y) prime.
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I differentiate the x and I
differentiated the y.
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How about the right-hand side.
Well, the right-hand side is
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negative 2, 2,
2, negative 5 times what?
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Well, times (x,
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y), which is (a1,
a2) e to the lambda t.
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Now, the same thing that
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happened a month or a month and
a half ago happens now.
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The whole point of making that
substitution is that the e to
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the lambda t,
the function part of it drops
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out completely.
And one is left with what?
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An algebraic equation to be
solved for lambda a1 and a2.
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In other words,
by means of that substitution,
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and it basically uses the fact
that the coefficients are
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constant, what you have done is
reduced the problem of calculus,
00:12:53.000 --> 00:12:59.000
of solving differential
equations, to solving algebraic
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equations.
In some sense that is the only
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method there is,
unless you do numerical stuff.
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You reduce the calculus to
algebra.
00:13:08.000 --> 00:13:14.000
The Laplace transform is
exactly the same thing.
00:13:11.000 --> 00:13:17.000
All the work is algebra.
You turn the original
00:13:14.000 --> 00:13:20.000
differential equation into an
algebraic equation for Y of s,
00:13:19.000 --> 00:13:25.000
you solve it,
and then you use more algebra
00:13:22.000 --> 00:13:28.000
to find out what the original
little y of t was.
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It is not different here.
So let's solve this system of
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equations.
Now, the whole problem with
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solving this system,
first of all,
00:13:37.000 --> 00:13:43.000
what is the system?
Let's write it out explicitly.
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Well, it is really two
equations, isn't it?
00:13:44.000 --> 00:13:50.000
The first one says lambda a1 is
equal to negative 2 a1
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plus 2 a2.
That is the first one.
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The other one says lambda a2 is
equal to 2 a1 minus 5 a2.
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Now, purely,
if you want to classify that,
00:14:09.000 --> 00:14:15.000
that is two equations and three
variables, three unknowns.
00:14:13.000 --> 00:14:19.000
The a1, a2, and lambda are all
unknown.
00:14:16.000 --> 00:14:22.000
And, unfortunately,
if you want to classify them
00:14:19.000 --> 00:14:25.000
correctly, they are nonlinear
equations because they are made
00:14:24.000 --> 00:14:30.000
nonlinear by the fact that you
have multiplied two of the
00:14:28.000 --> 00:14:34.000
variables.
Well, if you sit down and try
00:14:32.000 --> 00:14:38.000
to hack away at solving those
without a plan,
00:14:35.000 --> 00:14:41.000
you are not going to get
anywhere.
00:14:37.000 --> 00:14:43.000
It is going to be a mess.
Also, two equations and three
00:14:41.000 --> 00:14:47.000
unknowns is indeterminate.
You can solve three equations
00:14:46.000 --> 00:14:52.000
and three unknowns and get a
definite answer,
00:14:49.000 --> 00:14:55.000
but two equations and three
unknowns usually have an
00:14:53.000 --> 00:14:59.000
infinity of solutions.
Well, at this point it is the
00:14:56.000 --> 00:15:02.000
only idea that is required.
Well, this was a little idea,
00:15:02.000 --> 00:15:08.000
but I assume one would think of
that.
00:15:05.000 --> 00:15:11.000
And the idea that is required
here is, I think,
00:15:09.000 --> 00:15:15.000
not so unnatural,
it is not to view these a1,
00:15:13.000 --> 00:15:19.000
a2, and lambda as equal.
Not all variables are created
00:15:17.000 --> 00:15:23.000
equal.
Some are more equal than
00:15:20.000 --> 00:15:26.000
others.
a1 and a2 are definitely equal
00:15:23.000 --> 00:15:29.000
to each other,
and let's relegate lambda to
00:15:27.000 --> 00:15:33.000
the background.
In other words,
00:15:31.000 --> 00:15:37.000
I am going to think of lambda
as just a parameter.
00:15:34.000 --> 00:15:40.000
I am going to demote it from
the status of variable to
00:15:39.000 --> 00:15:45.000
parameter.
If I demoted it further it
00:15:41.000 --> 00:15:47.000
would just be an unknown
constant.
00:15:44.000 --> 00:15:50.000
That is as bad as you can be.
I am going to focus my
00:15:48.000 --> 00:15:54.000
attention on the a1,
a2 and sort of view the lambda
00:15:52.000 --> 00:15:58.000
as a nuisance.
Now, as soon as I do that,
00:15:55.000 --> 00:16:01.000
I see that these equations are
linear if I just look at them as
00:15:59.000 --> 00:16:05.000
equations in a1 and a2.
And moreover,
00:16:04.000 --> 00:16:10.000
they are not just linear,
they are homogenous.
00:16:07.000 --> 00:16:13.000
Because if I think of lambda
just as a parameter,
00:16:11.000 --> 00:16:17.000
I should rewrite the equations
this way.
00:16:14.000 --> 00:16:20.000
I am going to subtract this and
move the left-hand side to the
00:16:19.000 --> 00:16:25.000
right side, and it is going to
look like (minus 2 minus lambda)
00:16:23.000 --> 00:16:29.000
times a1 plus 2 a2 is equal to
zero.
00:16:28.000 --> 00:16:34.000
And the same way for the other
00:16:32.000 --> 00:16:38.000
one.
It is going to be 2a1 plus,
00:16:35.000 --> 00:16:41.000
what is the coefficient,
(minus 5 minus lambda) a2
00:16:39.000 --> 00:16:45.000
equals zero.
00:16:43.000 --> 00:16:49.000
That is a pair of simultaneous
linear equations for determining
00:16:48.000 --> 00:16:54.000
a1 and a2, and the coefficients
involved are parameter lambda.
00:16:53.000 --> 00:16:59.000
Now, what is the point of doing
that?
00:16:58.000 --> 00:17:04.000
Well, now the point is whatever
you learned about linear
00:17:02.000 --> 00:17:08.000
equations, you should have
learned the most fundamental
00:17:06.000 --> 00:17:12.000
theorem of linear equations.
The main theorem is that you
00:17:10.000 --> 00:17:16.000
have a square system of
homogeneous equations,
00:17:13.000 --> 00:17:19.000
this is a two-by-two system so
it is square,
00:17:16.000 --> 00:17:22.000
it always has the trivial
solution, of course,
00:17:20.000 --> 00:17:26.000
a1, a2 equals zero.
Now, we don't want that trivial
00:17:24.000 --> 00:17:30.000
solution because if a1 and a2
are zero, then so are x and y
00:17:28.000 --> 00:17:34.000
zero.
Now that is a solution.
00:17:32.000 --> 00:17:38.000
Unfortunately,
it is of no interest.
00:17:35.000 --> 00:17:41.000
If the solution were x,
y zero, it corresponds to the
00:17:39.000 --> 00:17:45.000
fact that this is an ice bath.
The yoke is at zero,
00:17:44.000 --> 00:17:50.000
the white is at zero and it
stays that way for all time
00:17:48.000 --> 00:17:54.000
until the ice melts.
So that is the solution we
00:17:52.000 --> 00:17:58.000
don't want.
We don't want the trivial
00:17:56.000 --> 00:18:02.000
solution.
Well, when does it have a
00:17:59.000 --> 00:18:05.000
nontrivial solution?
Nontrivial means non-zero,
00:18:03.000 --> 00:18:09.000
in other words.
If and only if this determinant
00:18:09.000 --> 00:18:15.000
is zero.
00:18:18.000 --> 00:18:24.000
In other words,
by using that theorem on linear
00:18:22.000 --> 00:18:28.000
equations, what we find is there
is a condition that lambda must
00:18:28.000 --> 00:18:34.000
satisfy, an equation in lambda
in order that we would be able
00:18:33.000 --> 00:18:39.000
to find non-zero values for a1
and a2.
00:18:38.000 --> 00:18:44.000
Let's write it out.
I will recopy it over here.
00:18:41.000 --> 00:18:47.000
What was it?
Negative 2 minus lambda,
00:18:44.000 --> 00:18:50.000
two, here it was 2 and minus 5
minus lambda.
00:18:49.000 --> 00:18:55.000
All right.
You have to expand the
00:18:51.000 --> 00:18:57.000
determinant.
In other words,
00:18:54.000 --> 00:19:00.000
we are trying to find out for
what values of lambda is this
00:18:58.000 --> 00:19:04.000
determinant zero.
Those will be the good values
00:19:03.000 --> 00:19:09.000
which lead to nontrivial
solutions for the a's.
00:19:06.000 --> 00:19:12.000
This is the equation lambda
plus 2.
00:19:09.000 --> 00:19:15.000
See, this is minus that and
minus that, the product of the
00:19:14.000 --> 00:19:20.000
two minus ones is plus one.
So it is lambda plus 2 times
00:19:18.000 --> 00:19:24.000
lambda plus 5,
00:19:21.000 --> 00:19:27.000
which is the product of the two
diagonal elements,
00:19:25.000 --> 00:19:31.000
minus the product of the two
anti-diagonal elements,
00:19:29.000 --> 00:19:35.000
which is 4, is equal to zero.
And if I write that out,
00:19:35.000 --> 00:19:41.000
what is that,
that is the equation lambda
00:19:39.000 --> 00:19:45.000
squared plus 7 lambda,
00:19:43.000 --> 00:19:49.000
5 lambda plus 2 lambda,
and then the constant term is
00:19:48.000 --> 00:19:54.000
10 minus 4 which is 6.
How many of you have long
00:19:52.000 --> 00:19:58.000
enough memories,
two-day memories that you
00:19:56.000 --> 00:20:02.000
remember that equation?
When I did the method of
00:20:02.000 --> 00:20:08.000
elimination, it led to exactly
the same equation except it had
00:20:10.000 --> 00:20:16.000
r's in it instead of lambda.
And this equation,
00:20:15.000 --> 00:20:21.000
therefore, is given the same
name and another color.
00:20:21.000 --> 00:20:27.000
Let's make it salmon.
00:20:30.000 --> 00:20:36.000
And it is called the
characteristic equation for this
00:20:34.000 --> 00:20:40.000
method.
All right.
00:20:36.000 --> 00:20:42.000
Now I am going to use now the
word from last time.
00:20:40.000 --> 00:20:46.000
You factor this.
From the factorization we get
00:20:44.000 --> 00:20:50.000
its root easily enough.
The roots are lambda equals
00:20:48.000 --> 00:20:54.000
negative 1 and
lambda equals negative 6
00:20:54.000 --> 00:21:00.000
by factoring the
equation.
00:20:57.000 --> 00:21:03.000
Now what I am supposed to do?
You have to keep the different
00:21:03.000 --> 00:21:09.000
parts of the method together.
Now I have found the only
00:21:08.000 --> 00:21:14.000
values of lambda for which I
will be able to find nonzero
00:21:12.000 --> 00:21:18.000
values for the a1 and a2.
For each of those values of
00:21:17.000 --> 00:21:23.000
lambda, I now have to find the
corresponding a1 and a2.
00:21:21.000 --> 00:21:27.000
Let's do them one at a time.
Let's take first lambda equals
00:21:26.000 --> 00:21:32.000
negative one.
My problem is now to find a1
00:21:32.000 --> 00:21:38.000
and a2.
Where am I going to find them
00:21:35.000 --> 00:21:41.000
from?
Well, from that system of
00:21:38.000 --> 00:21:44.000
equations over there.
I will recopy it over here.
00:21:42.000 --> 00:21:48.000
What is the system?
The hardest part of this is
00:21:47.000 --> 00:21:53.000
dealing with multiple minus
signs, but you had experience
00:21:52.000 --> 00:21:58.000
with that in determinants so you
know all about that.
00:21:58.000 --> 00:22:04.000
In other words,
there is the system of
00:22:01.000 --> 00:22:07.000
equations over there.
Let's recopy them here.
00:22:06.000 --> 00:22:12.000
Minus 2, minus minus 1 makes
minus 1.
00:22:09.000 --> 00:22:15.000
What's the other coefficient?
It is just plain old 2.
00:22:15.000 --> 00:22:21.000
Good.
There is my first equation.
00:22:18.000 --> 00:22:24.000
And when I substitute lambda
equals negative one
00:22:24.000 --> 00:22:30.000
for the second equation,
what do you get?
00:22:30.000 --> 00:22:36.000
2 a1 plus negative 5 minus
negative 1 makes negative 4.
00:22:37.000 --> 00:22:43.000
There is my system that will
find me a1 and a2.
00:22:43.000 --> 00:22:49.000
What is the first thing you
notice about it?
00:22:48.000 --> 00:22:54.000
You immediately notice that
this system is fake because this
00:22:56.000 --> 00:23:02.000
second equation is twice the
first one.
00:23:03.000 --> 00:23:09.000
Something is wrong.
No, something is right.
00:23:06.000 --> 00:23:12.000
If that did not happen,
if the second equation were not
00:23:12.000 --> 00:23:18.000
a constant multiple of the first
one then the only solution of
00:23:17.000 --> 00:23:23.000
the system would be a1 equals
zero, a2 equals zero because the
00:23:23.000 --> 00:23:29.000
determinant of the coefficients
would not be zero.
00:23:29.000 --> 00:23:35.000
The whole function of this
exercise was to find the value
00:23:33.000 --> 00:23:39.000
of lambda, negative 1,
for which the system would be
00:23:37.000 --> 00:23:43.000
redundant and,
therefore, would have a
00:23:40.000 --> 00:23:46.000
nontrivial solution.
Do you get that?
00:23:43.000 --> 00:23:49.000
In other words,
calculate the system out,
00:23:47.000 --> 00:23:53.000
just as I have done here,
you have an automatic check on
00:23:51.000 --> 00:23:57.000
the method.
If one equation is not a
00:23:54.000 --> 00:24:00.000
constant multiple of the other
you made a mistake.
00:24:00.000 --> 00:24:06.000
You don't have the right value
of lambda or you substituted
00:24:05.000 --> 00:24:11.000
into the system wrong,
which is frankly a more common
00:24:10.000 --> 00:24:16.000
error.
Go back, recheck first the
00:24:13.000 --> 00:24:19.000
substitution,
and if convinced that is right
00:24:17.000 --> 00:24:23.000
then recheck where you got
lambda from.
00:24:20.000 --> 00:24:26.000
But here everything is going
fine so we can now find out what
00:24:26.000 --> 00:24:32.000
the value of a1 and a2 are.
You don't have to go through a
00:24:32.000 --> 00:24:38.000
big song and dance for this
since most of the time you will
00:24:36.000 --> 00:24:42.000
have two-by-two equations and
now and then three-by-three.
00:24:40.000 --> 00:24:46.000
For two-by-two all you do is,
since we really have the same
00:24:44.000 --> 00:24:50.000
equation twice,
to get a solution I can assign
00:24:47.000 --> 00:24:53.000
one of the variables any value
and then simply solve for the
00:24:51.000 --> 00:24:57.000
other.
The natural thing to do is to
00:24:54.000 --> 00:25:00.000
make a2 equal one,
then I won't need fractions and
00:24:57.000 --> 00:25:03.000
then a1 will be a2.
So the solution is (2,
00:25:01.000 --> 00:25:07.000
1).
I am only trying to find one
00:25:04.000 --> 00:25:10.000
solution.
Any constant multiple of this
00:25:07.000 --> 00:25:13.000
would also be a solution,
as long as it wasn't zero,
00:25:12.000 --> 00:25:18.000
zero which is the trivial one.
And, therefore,
00:25:15.000 --> 00:25:21.000
this is a solution to this
system of algebraic equations.
00:25:20.000 --> 00:25:26.000
And the solution to the whole
system of differential equations
00:25:25.000 --> 00:25:31.000
is, this is only the (a1,
a2) part.
00:25:30.000 --> 00:25:36.000
I have to add to it,
as a factor,
00:25:32.000 --> 00:25:38.000
lambda is negative,
therefore, e to the minus t.
00:25:36.000 --> 00:25:42.000
There is our purple thing.
00:25:50.000 --> 00:25:56.000
See how I got it?
Starting with the trial
00:25:52.000 --> 00:25:58.000
solution, I first found out
through this procedure what the
00:25:57.000 --> 00:26:03.000
lambda's have to be.
Then I took the lambda and
00:26:00.000 --> 00:26:06.000
found what the corresponding a1
and a2 that went with it and
00:26:04.000 --> 00:26:10.000
then made up my solution out of
that.
00:26:06.000 --> 00:26:12.000
Now, quickly I will do the same
thing for lambda
00:26:11.000 --> 00:26:17.000
equals negative 6.
Each one of these must be
00:26:13.000 --> 00:26:19.000
treated separately.
They are separate problems and
00:26:17.000 --> 00:26:23.000
you are looking for separate
solutions.
00:26:19.000 --> 00:26:25.000
Lambda equals negative 6.
What do I do?
00:26:22.000 --> 00:26:28.000
How do my equations look now?
Well, the first one is minus 2
00:26:25.000 --> 00:26:31.000
minus negative 6 makes plus 4.
It is 4a1 plus 2a2 equals zero.
00:26:32.000 --> 00:26:38.000
Then I hold my breath while I
00:26:37.000 --> 00:26:43.000
calculate the second one to see
if it comes out to be a constant
00:26:44.000 --> 00:26:50.000
multiple.
I get 2a1 plus negative 5 minus
00:26:49.000 --> 00:26:55.000
negative 6, which makes plus 1.
And, indeed,
00:26:54.000 --> 00:27:00.000
one is a constant multiple of
the other.
00:27:00.000 --> 00:27:06.000
I really only have on equation
there.
00:27:04.000 --> 00:27:10.000
I will just write down
immediately now what the
00:27:09.000 --> 00:27:15.000
solution is to the system.
Well, the (a1,
00:27:13.000 --> 00:27:19.000
a2) will be what?
Now, it is more natural to make
00:27:19.000 --> 00:27:25.000
a1 equal 1 and then solve to get
an integer for a2.
00:27:25.000 --> 00:27:31.000
If a1 is 1, then a2 is negative
2.
00:27:30.000 --> 00:27:36.000
And I should multiply that by e
to the negative 6t
00:27:34.000 --> 00:27:40.000
because negative 6 is the
corresponding value.
00:27:38.000 --> 00:27:44.000
There is my other one.
And now there is a
00:27:41.000 --> 00:27:47.000
superposition principle,
which if I get a chance will
00:27:44.000 --> 00:27:50.000
prove for you at the end of the
hour.
00:27:47.000 --> 00:27:53.000
If not, you will have to do it
yourself for homework.
00:27:51.000 --> 00:27:57.000
Since this is a linear system
of equations,
00:27:54.000 --> 00:28:00.000
once you have two separate
solutions, neither a constant
00:27:58.000 --> 00:28:04.000
multiple of the other,
you can multiply each one of
00:28:02.000 --> 00:28:08.000
these by a constant and it will
still be a solution.
00:28:08.000 --> 00:28:14.000
You can add them together and
that will still be a solution,
00:28:11.000 --> 00:28:17.000
and that gives the general
solution.
00:28:13.000 --> 00:28:19.000
The general solution is the sum
of these two,
00:28:16.000 --> 00:28:22.000
an arbitrary constant.
I am going to change the name
00:28:19.000 --> 00:28:25.000
since I don't want to confuse it
with the c1 I used before,
00:28:22.000 --> 00:28:28.000
times the first solution which
is (2, 1) e to the negative t
00:28:26.000 --> 00:28:32.000
plus c2, another arbitrary
constant, times 1 negative 2 e
00:28:29.000 --> 00:28:35.000
to the minus 6t.
00:28:32.000 --> 00:28:38.000
Now you notice that is exactly
00:28:37.000 --> 00:28:43.000
the same solution I got before.
The only difference is that I
00:28:43.000 --> 00:28:49.000
have renamed the arbitrary
constants.
00:28:47.000 --> 00:28:53.000
The relationship between them,
c1 over 2,
00:28:53.000 --> 00:28:59.000
I am now calling c1 tilda,
and c2 I am calling c2 tilda.
00:29:00.000 --> 00:29:06.000
If you have an arbitrary
constant, it doesn't matter
00:29:04.000 --> 00:29:10.000
whether you divide it by two.
It is still just an arbitrary a
00:29:09.000 --> 00:29:15.000
constant.
It covers all values,
00:29:12.000 --> 00:29:18.000
in other words.
Well, I think you will agree
00:29:16.000 --> 00:29:22.000
that is a different procedure,
yet it has only one
00:29:20.000 --> 00:29:26.000
coincidence.
It is like elimination goes
00:29:24.000 --> 00:29:30.000
this way and comes to the
answer.
00:29:28.000 --> 00:29:34.000
And this method goes a
completely different route and
00:29:31.000 --> 00:29:37.000
comes to the answer,
except it is not quite like
00:29:34.000 --> 00:29:40.000
that.
They walk like this and then
00:29:36.000 --> 00:29:42.000
they come within viewing
distance of each other to check
00:29:40.000 --> 00:29:46.000
that both are using the same
characteristic equation,
00:29:44.000 --> 00:29:50.000
and then they again go their
separate ways and end up with
00:29:48.000 --> 00:29:54.000
the same answer.
00:29:58.000 --> 00:30:04.000
There is something special of
these values.
00:30:00.000 --> 00:30:06.000
You cannot get away from those
two values of lambda.
00:30:03.000 --> 00:30:09.000
Somehow they are really
intrinsically connected.
00:30:06.000 --> 00:30:12.000
Occurs the exponential
coefficient, and they are
00:30:09.000 --> 00:30:15.000
intrinsically connected with the
problem of the egg that we
00:30:12.000 --> 00:30:18.000
started with.
Now what I would like to do is
00:30:15.000 --> 00:30:21.000
very quickly sketch how this
method looks when I remove all
00:30:18.000 --> 00:30:24.000
the numbers from it.
In some sense,
00:30:20.000 --> 00:30:26.000
it becomes a little clearer
what is going on.
00:30:23.000 --> 00:30:29.000
And that will give me a chance
to introduce the terminology
00:30:26.000 --> 00:30:32.000
that you need when you talk
about it.
00:30:55.000 --> 00:31:01.000
Well, you have notes.
Let me try to write it down in
00:31:03.000 --> 00:31:09.000
general.
00:31:10.000 --> 00:31:16.000
I will first write it out
two-by-two.
00:31:13.000 --> 00:31:19.000
I am just going to sketch.
The system looks like (x,
00:31:18.000 --> 00:31:24.000
y) equals, I will still put it
up in colors.
00:31:30.000 --> 00:31:36.000
Except now, instead of using
twos and fives,
00:31:33.000 --> 00:31:39.000
I will use (a,
b; c, d).
00:31:40.000 --> 00:31:46.000
The trial solution will look
how?
00:31:44.000 --> 00:31:50.000
The trial is going to be (a1,
a2).
00:31:48.000 --> 00:31:54.000
That I don't have to change the
name of.
00:31:53.000 --> 00:31:59.000
I am going to substitute in,
and what the result of
00:31:59.000 --> 00:32:05.000
substitution is going to be
lambda (a1, a2).
00:32:06.000 --> 00:32:12.000
I am going to skip a step and
pretend that the e to the lambda
00:32:11.000 --> 00:32:17.000
t's have already
been canceled out.
00:32:16.000 --> 00:32:22.000
Is equal to (a,
b; c, d) times (a1,
00:32:19.000 --> 00:32:25.000
a2).
What does that correspond to?
00:32:22.000 --> 00:32:28.000
That corresponds to the system
as I wrote it here.
00:32:28.000 --> 00:32:34.000
And then we wrote it out in
terms of two equations.
00:32:31.000 --> 00:32:37.000
And what was the resulting
thing that we ended up with?
00:32:35.000 --> 00:32:41.000
Well, you write it out,
you move the lambda to the
00:32:38.000 --> 00:32:44.000
other side.
And then the homogeneous system
00:32:41.000 --> 00:32:47.000
is we will look in general how?
Well, we could write it out.
00:32:45.000 --> 00:32:51.000
It is going to look like a
minus lambda,
00:32:48.000 --> 00:32:54.000
b, c, d minus lambda.
00:32:50.000 --> 00:32:56.000
That is just how it looks there
00:32:53.000 --> 00:32:59.000
and the general calculation is
the same.
00:32:56.000 --> 00:33:02.000
Times (a1, a2) is equal to
zero.
00:33:05.000 --> 00:33:11.000
This is solvable nontrivially.
In other words,
00:33:13.000 --> 00:33:19.000
it has a nontrivial solution if
an only if the determinant of
00:33:25.000 --> 00:33:31.000
coefficients is zero.
00:33:35.000 --> 00:33:41.000
Let's now write that out,
calculate out once and for all
00:33:39.000 --> 00:33:45.000
what that determinant is.
I will write it out here.
00:33:43.000 --> 00:33:49.000
It is a minus lambda times d
minus lambda,
00:33:46.000 --> 00:33:52.000
the product of the diagonal
elements, minus the
00:33:50.000 --> 00:33:56.000
anti-diagonal minus bc is equal
to zero.
00:33:55.000 --> 00:34:01.000
And let's calculate that out.
00:34:00.000 --> 00:34:06.000
It is lambda squared minus a
lambda minus d lambda plus ad,
00:34:07.000 --> 00:34:13.000
the constant term from here,
negative bc from there,
00:34:14.000 --> 00:34:20.000
plus ad minus bc,
00:34:21.000 --> 00:34:27.000
where have I seen that before?
This equation is the general
00:34:29.000 --> 00:34:35.000
form using letters of what we
calculated using the specific
00:34:36.000 --> 00:34:42.000
numbers before.
Again, I will code it the same
00:34:45.000 --> 00:34:51.000
way with that color salmon.
Now, most of the calculations
00:34:54.000 --> 00:35:00.000
will be for two-by-two systems.
I advise you,
00:35:00.000 --> 00:35:06.000
in the strongest possible
terms, to remember this
00:35:03.000 --> 00:35:09.000
equation.
You could write down this
00:35:06.000 --> 00:35:12.000
equation immediately for the
matrix.
00:35:08.000 --> 00:35:14.000
You don't have to go through
all this stuff.
00:35:11.000 --> 00:35:17.000
For God's sakes,
don't say let the trial
00:35:13.000 --> 00:35:19.000
solution be blah,
blah, blah.
00:35:15.000 --> 00:35:21.000
You don't want to do that.
I don't want you to repeat the
00:35:19.000 --> 00:35:25.000
derivation of this every time
you go through a particular
00:35:23.000 --> 00:35:29.000
problem.
It is just like in solving
00:35:25.000 --> 00:35:31.000
second order equations.
You have a second order
00:35:29.000 --> 00:35:35.000
equation.
You immediately write down its
00:35:32.000 --> 00:35:38.000
characteristic equation,
then you factor it,
00:35:35.000 --> 00:35:41.000
you find its roots and you
construct the solution.
00:35:39.000 --> 00:35:45.000
It takes a minute.
The same thing,
00:35:42.000 --> 00:35:48.000
this takes a minute,
too.
00:35:43.000 --> 00:35:49.000
What is the constant term?
Ad minus bc,
00:35:47.000 --> 00:35:53.000
what is that?
Matrix is (a,
00:35:49.000 --> 00:35:55.000
b; c, d).
Ad minus bc is its determinant.
00:35:52.000 --> 00:35:58.000
This is the determinant of that
matrix.
00:35:55.000 --> 00:36:01.000
I didn't give the matrix a
name, did I?
00:36:00.000 --> 00:36:06.000
I will now give the matrix a
name A.
00:36:03.000 --> 00:36:09.000
What is this?
Well, you are not supposed to
00:36:07.000 --> 00:36:13.000
know that until now.
I will tell you.
00:36:10.000 --> 00:36:16.000
This is called the trace of A.
Put that down in your little
00:36:16.000 --> 00:36:22.000
books.
The abbreviation is trace A,
00:36:19.000 --> 00:36:25.000
and the word is trace.
The trace of a square matrix is
00:36:25.000 --> 00:36:31.000
the sum of the d elements down
its main diagonal.
00:36:31.000 --> 00:36:37.000
If it were a three-by-three
there would be three terms in
00:36:35.000 --> 00:36:41.000
whatever you are up to.
Here it is a plus b,
00:36:40.000 --> 00:36:46.000
the sum of the diagonal
elements.
00:36:43.000 --> 00:36:49.000
You can immediately write down
this characteristic equation.
00:36:48.000 --> 00:36:54.000
Let's give it a name.
This is a characteristic
00:36:52.000 --> 00:36:58.000
equation of what?
Of the matrix,
00:36:55.000 --> 00:37:01.000
now.
Not of the system,
00:36:57.000 --> 00:37:03.000
of the matrix.
You have a two-by-two matrix.
00:37:02.000 --> 00:37:08.000
You could immediately write
down its characteristic
00:37:06.000 --> 00:37:12.000
equation.
Watch out for this sign,
00:37:08.000 --> 00:37:14.000
minus.
That is a very common error to
00:37:11.000 --> 00:37:17.000
leave out the minus sign because
that is the way the formula
00:37:15.000 --> 00:37:21.000
comes out.
Its roots.
00:37:17.000 --> 00:37:23.000
If it is a quadratic equation
it will have roots;
00:37:20.000 --> 00:37:26.000
lambda1, lambda2 for the moment
let's assume are real and
00:37:25.000 --> 00:37:31.000
distinct.
00:37:37.000 --> 00:37:43.000
For the enrichment of your
vocabulary, those are called the
00:37:39.000 --> 00:37:45.000
eigenvalues.
00:37:50.000 --> 00:37:56.000
They are something which
belonged to the matrix A.
00:37:53.000 --> 00:37:59.000
They are two secret numbers.
You can calculate from the
00:37:56.000 --> 00:38:02.000
coefficients a,
b, and c, and d,
00:37:58.000 --> 00:38:04.000
but they are not in the
coefficients.
00:38:00.000 --> 00:38:06.000
You cannot look at a matrix and
see what its eigenvalues are.
00:38:05.000 --> 00:38:11.000
You have to calculate
something.
00:38:07.000 --> 00:38:13.000
But they are the most important
numbers in the matrix.
00:38:10.000 --> 00:38:16.000
They are hidden,
but they are the things that
00:38:13.000 --> 00:38:19.000
control how this system behaves.
Those are called the
00:38:17.000 --> 00:38:23.000
eigenvalues.
Now, there are various purists,
00:38:20.000 --> 00:38:26.000
there are a fair number of them
in the world who do not like
00:38:24.000 --> 00:38:30.000
this word because it begins
German and ends English.
00:38:29.000 --> 00:38:35.000
Eigenvalues were first
introduced by a German
00:38:32.000 --> 00:38:38.000
mathematician,
you know, around the time
00:38:35.000 --> 00:38:41.000
matrices came into being in
or so.
00:38:38.000 --> 00:38:44.000
A little while after
eigenvalues came into being,
00:38:41.000 --> 00:38:47.000
too.
And since all this happened in
00:38:44.000 --> 00:38:50.000
Germany they were named
eigenvalues in German,
00:38:47.000 --> 00:38:53.000
which begins eigen and ends
value.
00:38:50.000 --> 00:38:56.000
But people who do not like that
call them the characteristic
00:38:54.000 --> 00:39:00.000
values.
Unfortunately,
00:38:56.000 --> 00:39:02.000
it is two words and takes a lot
more space to write out.
00:39:02.000 --> 00:39:08.000
An older generation even calls
them something different,
00:39:06.000 --> 00:39:12.000
which you are not so likely to
see nowadays,
00:39:10.000 --> 00:39:16.000
but you will in slightly older
books.
00:39:13.000 --> 00:39:19.000
You can also call them the
proper values.
00:39:17.000 --> 00:39:23.000
Characteristic is not a
translation of eigen,
00:39:21.000 --> 00:39:27.000
but proper is,
but it means it in a funny
00:39:25.000 --> 00:39:31.000
sense which has almost
disappeared nowadays.
00:39:30.000 --> 00:39:36.000
It means proper in the sense of
belong to.
00:39:33.000 --> 00:39:39.000
The only example I can think of
is the word property.
00:39:37.000 --> 00:39:43.000
Property is something that
belongs to you.
00:39:40.000 --> 00:39:46.000
That is the use of the word
proper.
00:39:43.000 --> 00:39:49.000
It is something that belongs to
the matrix.
00:39:46.000 --> 00:39:52.000
The matrix has its proper
values.
00:39:49.000 --> 00:39:55.000
It does not mean proper in the
sense of fitting and proper or I
00:39:54.000 --> 00:40:00.000
hope you will behave properly
when we go to Aunt Agatha's or
00:39:59.000 --> 00:40:05.000
something like that.
But, as I say,
00:40:03.000 --> 00:40:09.000
by far the most popular thing,
slowly the word eigenvalue is
00:40:07.000 --> 00:40:13.000
pretty much taking over the
literature.
00:40:11.000 --> 00:40:17.000
Just because it's just one
word, that is a tremendous
00:40:15.000 --> 00:40:21.000
advantage.
Okay.
00:40:16.000 --> 00:40:22.000
What now is still to be done?
Well, there are those vectors
00:40:21.000 --> 00:40:27.000
to be found.
So the very last step would be
00:40:24.000 --> 00:40:30.000
to solve the system to find the
vectors a1 and a2.
00:40:35.000 --> 00:40:41.000
For each (lambda)i,
find the associated vector.
00:40:40.000 --> 00:40:46.000
The vector, we will call it
(alpha)i.
00:40:44.000 --> 00:40:50.000
That is the a1 and a2.
Of course it's going to be
00:40:50.000 --> 00:40:56.000
indexed.
You have to put another
00:40:53.000 --> 00:40:59.000
subscript on it because there
are two of them.
00:41:00.000 --> 00:41:06.000
And a1 and a2 is stretched a
little too far.
00:41:04.000 --> 00:41:10.000
By solving the system,
and the system will be the
00:41:09.000 --> 00:41:15.000
system which I will write this
way, (a minus lambda,
00:41:15.000 --> 00:41:21.000
b, c, d minus lambda).
00:41:19.000 --> 00:41:25.000
It is just that system that was
00:41:24.000 --> 00:41:30.000
over there, but I will recopy
it, (a1, a2) equals zero,
00:41:30.000 --> 00:41:36.000
zero.
And these are called the
00:41:34.000 --> 00:41:40.000
eigenvectors.
Each of these is called the
00:41:39.000 --> 00:41:45.000
eigenvector associated with or
belonging to,
00:41:44.000 --> 00:41:50.000
again, in that sense of
property.
00:41:48.000 --> 00:41:54.000
Eigenvector,
let's say belonging to,
00:41:52.000 --> 00:41:58.000
I see that a little more
frequently, belonging to lambda
00:41:58.000 --> 00:42:04.000
i.
So we have the eigenvalues,
00:42:01.000 --> 00:42:07.000
the eigenvectors and,
of course, the people who call
00:42:04.000 --> 00:42:10.000
them characteristic values also
call these guys characteristic
00:42:08.000 --> 00:42:14.000
vectors.
I don't think I have ever seen
00:42:11.000 --> 00:42:17.000
proper vectors,
but that is because I am not
00:42:13.000 --> 00:42:19.000
old enough.
I think that is what they used
00:42:16.000 --> 00:42:22.000
to be called a long time ago,
but not anymore.
00:42:20.000 --> 00:42:26.000
And then, finally,
the general solution will be,
00:42:24.000 --> 00:42:30.000
by the superposition principle,
(x, y) equals the arbitrary
00:42:30.000 --> 00:42:36.000
constant times the first
eigenvector times the eigenvalue
00:42:35.000 --> 00:42:41.000
times the e to the corresponding
eigenvalue.
00:42:40.000 --> 00:42:46.000
And then the same thing for the
second one, (a1,
00:42:44.000 --> 00:42:50.000
a2), but now the second index
will be 2 to indicate that it
00:42:50.000 --> 00:42:56.000
goes with the eigenvalue e to
the lambda 2t.
00:42:56.000 --> 00:43:02.000
I have done that twice.
And now in the remaining five
00:43:02.000 --> 00:43:08.000
minutes I will do it a third
time because it is possible to
00:43:06.000 --> 00:43:12.000
write this in still a more
condensed form.
00:43:09.000 --> 00:43:15.000
And the advantage of the more
condensed form is A,
00:43:13.000 --> 00:43:19.000
it takes only that much space
to write, and B,
00:43:16.000 --> 00:43:22.000
it applies to systems,
not just the two-by-two
00:43:20.000 --> 00:43:26.000
systems, but to end-by-end
systems.
00:43:23.000 --> 00:43:29.000
The method is exactly the same.
Let's write it out as it would
00:43:27.000 --> 00:43:33.000
apply to end-by-end systems.
The vector I started with is
00:43:34.000 --> 00:43:40.000
(x, y) and so on,
but I will simply abbreviate
00:43:39.000 --> 00:43:45.000
this, as is done in 18.02,
by x with an arrow over it.
00:43:45.000 --> 00:43:51.000
The matrix A I will abbreviate
with A, as I did before with
00:43:51.000 --> 00:43:57.000
capital A.
And then the system looks like
00:43:56.000 --> 00:44:02.000
x prime is equal to --
00:44:05.000 --> 00:44:11.000
x prime is what?
Ax.
00:44:06.000 --> 00:44:12.000
That is all there is to it.
00:44:10.000 --> 00:44:16.000
There is our green system.
Now notice in this form I did
00:44:15.000 --> 00:44:21.000
not even tell you whether this a
two-by-two matrix or an
00:44:20.000 --> 00:44:26.000
end-by-end.
And in this condensed form it
00:44:23.000 --> 00:44:29.000
will look the same no matter how
many equations you have.
00:44:30.000 --> 00:44:36.000
Your book deals from the
beginning with end-by-end
00:44:33.000 --> 00:44:39.000
systems.
That is, in my view,
00:44:36.000 --> 00:44:42.000
one of its weaknesses because I
don't think most students start
00:44:40.000 --> 00:44:46.000
with two-by-two.
Fortunately,
00:44:43.000 --> 00:44:49.000
the book double-talks.
The theory is end-by-end,
00:44:46.000 --> 00:44:52.000
but all the examples are
two-by-two.
00:44:49.000 --> 00:44:55.000
So just read the examples.
Read the notes instead,
00:44:53.000 --> 00:44:59.000
which just do two-by-two to
start out with.
00:44:58.000 --> 00:45:04.000
The trial solution is x equals
what?
00:45:01.000 --> 00:45:07.000
An unknown vector alpha times e
to the lambda t.
00:45:07.000 --> 00:45:13.000
Alpha is what we called a1 and
00:45:11.000 --> 00:45:17.000
a2 before.
Plug this into there and cancel
00:45:15.000 --> 00:45:21.000
the e to the lambda t's.
00:45:19.000 --> 00:45:25.000
What do you get?
Well, this is lambda alpha e to
00:45:24.000 --> 00:45:30.000
the lambda t equals A alpha e to
the lambda t.
00:45:36.000 --> 00:45:42.000
These two cancel.
And the system to be solved,
00:45:40.000 --> 00:45:46.000
A alpha equals lambda alpha.
00:45:46.000 --> 00:45:52.000
And now the question is how do
you solve that system?
00:45:51.000 --> 00:45:57.000
Well, you can tell if a book is
written by a scoundrel or not by
00:45:57.000 --> 00:46:03.000
how they go --
A book, which is in my opinion
00:46:02.000 --> 00:46:08.000
completely scoundrel,
simply says you subtract one
00:46:07.000 --> 00:46:13.000
from the other,
and without further ado writes
00:46:12.000 --> 00:46:18.000
A minus lambda,
and they tuck a little I in
00:46:16.000 --> 00:46:22.000
there and write alpha equals
zero.
00:46:19.000 --> 00:46:25.000
Why is the I put in there?
Well, this is what you would
00:46:25.000 --> 00:46:31.000
like to write.
What is wrong with this
00:46:28.000 --> 00:46:34.000
equation?
This is not a valid matrix
00:46:32.000 --> 00:46:38.000
equation because that is a
square end-by-end matrix,
00:46:36.000 --> 00:46:42.000
a square two-by-two matrix if
you like.
00:46:39.000 --> 00:46:45.000
This is a scalar.
You cannot subtract the scalar
00:46:42.000 --> 00:46:48.000
from a matrix.
It is not an operation.
00:46:45.000 --> 00:46:51.000
To subtract matrices they have
to be the same size,
00:46:49.000 --> 00:46:55.000
the same shape.
What is done is you make this a
00:46:52.000 --> 00:46:58.000
two-by-two matrix.
This is a two-by-two matrix
00:46:56.000 --> 00:47:02.000
with lambdas down the main
diagonal and I elsewhere.
00:47:01.000 --> 00:47:07.000
And the justification is that
lambda alpha is the same thing
00:47:05.000 --> 00:47:11.000
as the lambda I times alpha
because I is an identity matrix.
00:47:10.000 --> 00:47:16.000
Now, in fact,
jumping from here to here is
00:47:13.000 --> 00:47:19.000
not something that would occur
to anybody.
00:47:16.000 --> 00:47:22.000
The way it should occur to you
to do this is you do this,
00:47:21.000 --> 00:47:27.000
you write that,
you realize it doesn't work,
00:47:24.000 --> 00:47:30.000
and then you say to yourself I
don't understand what these
00:47:29.000 --> 00:47:35.000
matrices are all about.
I think I'd better write it all
00:47:33.000 --> 00:47:39.000
out.
And then you would write it all
00:47:36.000 --> 00:47:42.000
out and you would write that
equation on the left-hand board
00:47:39.000 --> 00:47:45.000
there.
Oh, now I see what it should
00:47:41.000 --> 00:47:47.000
look like.
I should subtract lambda from
00:47:44.000 --> 00:47:50.000
the main diagonal.
That is the way it will come
00:47:47.000 --> 00:47:53.000
out.
And then say,
00:47:48.000 --> 00:47:54.000
hey, the way to save lambda
from the main diagonal is put it
00:47:51.000 --> 00:47:57.000
in an identity matrix.
That will do it for me.
00:47:54.000 --> 00:48:00.000
In other words,
there is a little detour that
00:47:57.000 --> 00:48:03.000
goes from here to here.
And one of the ways I judge
00:48:01.000 --> 00:48:07.000
books is by how well they
explain the passage from this to
00:48:05.000 --> 00:48:11.000
that.
If they don't explain it at all
00:48:08.000 --> 00:48:14.000
and just write it down,
they have never talked to
00:48:11.000 --> 00:48:17.000
students.
They have just written books.
00:48:14.000 --> 00:48:20.000
Where did we get finally here?
The characteristic equation
00:48:17.000 --> 00:48:23.000
from that, I had forgotten what
color.
00:48:20.000 --> 00:48:26.000
That is in salmon.
The characteristic equation,
00:48:23.000 --> 00:48:29.000
then, is going to be the thing
which says that the determinant
00:48:27.000 --> 00:48:33.000
of that is zero.
That is the circumstances under
00:48:32.000 --> 00:48:38.000
which it is solvable.
In general, this is the way the
00:48:35.000 --> 00:48:41.000
characteristic equation looks.
And its roots,
00:48:38.000 --> 00:48:44.000
once again, are the
eigenvalues.
00:48:40.000 --> 00:48:46.000
And from then you calculate the
corresponding eigenvectors.
00:48:44.000 --> 00:48:50.000
Okay.
Go home and practice.
00:48:46.000 --> 00:48:52.000
In recitation you will practice
on both two-by-two and
00:48:50.000 --> 00:48:56.000
three-by-three cases,
and we will talk more next
00:48:53.000 --> 00:48:59.000
time.