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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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00:03:02,000 --> 00:03:08,000
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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00:06:02,000 --> 00:06:08,000
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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00:11:35,000 --> 00:11:41,000
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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00:12:09,000 --> 00:12:15,000
Well, times (x,
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00:12:12,000 --> 00:12:18,000
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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00:12:35,000 --> 00:12:41,000
An algebraic equation to be
solved for lambda a1 and a2.
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00:12:39,000 --> 00:12:45,000
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,
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of solving differential
equations, to solving algebraic
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00:12:58,000 --> 00:13:04,000
equations.
In some sense that is the only
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00:13:02,000 --> 00:13:08,000
method there is,
unless you do numerical stuff.
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You reduce the calculus to
algebra.
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00:13:08,000 --> 00:13:14,000
The Laplace transform is
exactly the same thing.
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All the work is algebra.
You turn the original
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00:13:14,000 --> 00:13:20,000
differential equation into an
algebraic equation for Y of s,
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you solve it,
and then you use more algebra
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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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00:13:34,000 --> 00:13:40,000
solving this system,
first of all,
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what is the system?
Let's write it out explicitly.
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Well, it is really two
equations, isn't it?
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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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00:13:52,000 --> 00:13:58,000
The other one says lambda a2 is
equal to 2 a1 minus 5 a2.
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00:14:06,000 --> 00:14:12,000
Now, purely,
if you want to classify that,
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00:14:09,000 --> 00:14:15,000
that is two equations and three
variables, three unknowns.
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The a1, a2, and lambda are all
unknown.
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00:14:16,000 --> 00:14:22,000
And, unfortunately,
if you want to classify them
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00:14:19,000 --> 00:14:25,000
correctly, they are nonlinear
equations because they are made
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00:14:24,000 --> 00:14:30,000
nonlinear by the fact that you
have multiplied two of the
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00:14:28,000 --> 00:14:34,000
variables.
Well, if you sit down and try
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00:14:32,000 --> 00:14:38,000
to hack away at solving those
without a plan,
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00:14:35,000 --> 00:14:41,000
you are not going to get
anywhere.
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00:14:37,000 --> 00:14:43,000
It is going to be a mess.
Also, two equations and three
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00:14:41,000 --> 00:14:47,000
unknowns is indeterminate.
You can solve three equations
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00:14:46,000 --> 00:14:52,000
and three unknowns and get a
definite answer,
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00:14:49,000 --> 00:14:55,000
but two equations and three
unknowns usually have an
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00:14:53,000 --> 00:14:59,000
infinity of solutions.
Well, at this point it is the
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00:14:56,000 --> 00:15:02,000
only idea that is required.
Well, this was a little idea,
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00:15:02,000 --> 00:15:08,000
but I assume one would think of
that.
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00:15:05,000 --> 00:15:11,000
And the idea that is required
here is, I think,
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00:15:09,000 --> 00:15:15,000
not so unnatural,
it is not to view these a1,
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00:15:13,000 --> 00:15:19,000
a2, and lambda as equal.
Not all variables are created
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00:15:17,000 --> 00:15:23,000
equal.
Some are more equal than
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00:15:20,000 --> 00:15:26,000
others.
a1 and a2 are definitely equal
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to each other,
and let's relegate lambda to
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00:15:27,000 --> 00:15:33,000
the background.
In other words,
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00:15:31,000 --> 00:15:37,000
I am going to think of lambda
as just a parameter.
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00:15:34,000 --> 00:15:40,000
I am going to demote it from
the status of variable to
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parameter.
If I demoted it further it
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00:15:41,000 --> 00:15:47,000
would just be an unknown
constant.
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That is as bad as you can be.
I am going to focus my
231
00:15:48,000 --> 00:15:54,000
attention on the a1,
a2 and sort of view the lambda
232
00:15:52,000 --> 00:15:58,000
as a nuisance.
Now, as soon as I do that,
233
00:15:55,000 --> 00:16:01,000
I see that these equations are
linear if I just look at them as
234
00:15:59,000 --> 00:16:05,000
equations in a1 and a2.
And moreover,
235
00:16:04,000 --> 00:16:10,000
they are not just linear,
they are homogenous.
236
00:16:07,000 --> 00:16:13,000
Because if I think of lambda
just as a parameter,
237
00:16:11,000 --> 00:16:17,000
I should rewrite the equations
this way.
238
00:16:14,000 --> 00:16:20,000
I am going to subtract this and
move the left-hand side to the
239
00:16:19,000 --> 00:16:25,000
right side, and it is going to
look like (minus 2 minus lambda)
240
00:16:23,000 --> 00:16:29,000
times a1 plus 2 a2 is equal to
zero.
241
00:16:28,000 --> 00:16:34,000
And the same way for the other
242
00:16:32,000 --> 00:16:38,000
one.
It is going to be 2a1 plus,
243
00:16:35,000 --> 00:16:41,000
what is the coefficient,
(minus 5 minus lambda) a2
244
00:16:39,000 --> 00:16:45,000
equals zero.
245
00:16:43,000 --> 00:16:49,000
That is a pair of simultaneous
linear equations for determining
246
00:16:48,000 --> 00:16:54,000
a1 and a2, and the coefficients
involved are parameter lambda.
247
00:16:53,000 --> 00:16:59,000
Now, what is the point of doing
that?
248
00:16:58,000 --> 00:17:04,000
Well, now the point is whatever
you learned about linear
249
00:17:02,000 --> 00:17:08,000
equations, you should have
learned the most fundamental
250
00:17:06,000 --> 00:17:12,000
theorem of linear equations.
The main theorem is that you
251
00:17:10,000 --> 00:17:16,000
have a square system of
homogeneous equations,
252
00:17:13,000 --> 00:17:19,000
this is a two-by-two system so
it is square,
253
00:17:16,000 --> 00:17:22,000
it always has the trivial
solution, of course,
254
00:17:20,000 --> 00:17:26,000
a1, a2 equals zero.
Now, we don't want that trivial
255
00:17:24,000 --> 00:17:30,000
solution because if a1 and a2
are zero, then so are x and y
256
00:17:28,000 --> 00:17:34,000
zero.
Now that is a solution.
257
00:17:32,000 --> 00:17:38,000
Unfortunately,
it is of no interest.
258
00:17:35,000 --> 00:17:41,000
If the solution were x,
y zero, it corresponds to the
259
00:17:39,000 --> 00:17:45,000
fact that this is an ice bath.
The yoke is at zero,
260
00:17:44,000 --> 00:17:50,000
the white is at zero and it
stays that way for all time
261
00:17:48,000 --> 00:17:54,000
until the ice melts.
So that is the solution we
262
00:17:52,000 --> 00:17:58,000
don't want.
We don't want the trivial
263
00:17:56,000 --> 00:18:02,000
solution.
Well, when does it have a
264
00:17:59,000 --> 00:18:05,000
nontrivial solution?
Nontrivial means non-zero,
265
00:18:03,000 --> 00:18:09,000
in other words.
If and only if this determinant
266
00:18:09,000 --> 00:18:15,000
is zero.
267
00:18:18,000 --> 00:18:24,000
In other words,
by using that theorem on linear
268
00:18:22,000 --> 00:18:28,000
equations, what we find is there
is a condition that lambda must
269
00:18:28,000 --> 00:18:34,000
satisfy, an equation in lambda
in order that we would be able
270
00:18:33,000 --> 00:18:39,000
to find non-zero values for a1
and a2.
271
00:18:38,000 --> 00:18:44,000
Let's write it out.
I will recopy it over here.
272
00:18:41,000 --> 00:18:47,000
What was it?
Negative 2 minus lambda,
273
00:18:44,000 --> 00:18:50,000
two, here it was 2 and minus 5
minus lambda.
274
00:18:49,000 --> 00:18:55,000
All right.
You have to expand the
275
00:18:51,000 --> 00:18:57,000
determinant.
In other words,
276
00:18:54,000 --> 00:19:00,000
we are trying to find out for
what values of lambda is this
277
00:18:58,000 --> 00:19:04,000
determinant zero.
Those will be the good values
278
00:19:03,000 --> 00:19:09,000
which lead to nontrivial
solutions for the a's.
279
00:19:06,000 --> 00:19:12,000
This is the equation lambda
plus 2.
280
00:19:09,000 --> 00:19:15,000
See, this is minus that and
minus that, the product of the
281
00:19:14,000 --> 00:19:20,000
two minus ones is plus one.
So it is lambda plus 2 times
282
00:19:18,000 --> 00:19:24,000
lambda plus 5,
283
00:19:21,000 --> 00:19:27,000
which is the product of the two
diagonal elements,
284
00:19:25,000 --> 00:19:31,000
minus the product of the two
anti-diagonal elements,
285
00:19:29,000 --> 00:19:35,000
which is 4, is equal to zero.
And if I write that out,
286
00:19:35,000 --> 00:19:41,000
what is that,
that is the equation lambda
287
00:19:39,000 --> 00:19:45,000
squared plus 7 lambda,
288
00:19:43,000 --> 00:19:49,000
5 lambda plus 2 lambda,
and then the constant term is
289
00:19:48,000 --> 00:19:54,000
10 minus 4 which is 6.
How many of you have long
290
00:19:52,000 --> 00:19:58,000
enough memories,
two-day memories that you
291
00:19:56,000 --> 00:20:02,000
remember that equation?
When I did the method of
292
00:20:02,000 --> 00:20:08,000
elimination, it led to exactly
the same equation except it had
293
00:20:10,000 --> 00:20:16,000
r's in it instead of lambda.
And this equation,
294
00:20:15,000 --> 00:20:21,000
therefore, is given the same
name and another color.
295
00:20:21,000 --> 00:20:27,000
Let's make it salmon.
296
00:20:30,000 --> 00:20:36,000
And it is called the
characteristic equation for this
297
00:20:34,000 --> 00:20:40,000
method.
All right.
298
00:20:36,000 --> 00:20:42,000
Now I am going to use now the
word from last time.
299
00:20:40,000 --> 00:20:46,000
You factor this.
From the factorization we get
300
00:20:44,000 --> 00:20:50,000
its root easily enough.
The roots are lambda equals
301
00:20:48,000 --> 00:20:54,000
negative 1 and
lambda equals negative 6
302
00:20:54,000 --> 00:21:00,000
by factoring the
equation.
303
00:20:57,000 --> 00:21:03,000
Now what I am supposed to do?
You have to keep the different
304
00:21:03,000 --> 00:21:09,000
parts of the method together.
Now I have found the only
305
00:21:08,000 --> 00:21:14,000
values of lambda for which I
will be able to find nonzero
306
00:21:12,000 --> 00:21:18,000
values for the a1 and a2.
For each of those values of
307
00:21:17,000 --> 00:21:23,000
lambda, I now have to find the
corresponding a1 and a2.
308
00:21:21,000 --> 00:21:27,000
Let's do them one at a time.
Let's take first lambda equals
309
00:21:26,000 --> 00:21:32,000
negative one.
My problem is now to find a1
310
00:21:32,000 --> 00:21:38,000
and a2.
Where am I going to find them
311
00:21:35,000 --> 00:21:41,000
from?
Well, from that system of
312
00:21:38,000 --> 00:21:44,000
equations over there.
I will recopy it over here.
313
00:21:42,000 --> 00:21:48,000
What is the system?
The hardest part of this is
314
00:21:47,000 --> 00:21:53,000
dealing with multiple minus
signs, but you had experience
315
00:21:52,000 --> 00:21:58,000
with that in determinants so you
know all about that.
316
00:21:58,000 --> 00:22:04,000
In other words,
there is the system of
317
00:22:01,000 --> 00:22:07,000
equations over there.
Let's recopy them here.
318
00:22:06,000 --> 00:22:12,000
Minus 2, minus minus 1 makes
minus 1.
319
00:22:09,000 --> 00:22:15,000
What's the other coefficient?
It is just plain old 2.
320
00:22:15,000 --> 00:22:21,000
Good.
There is my first equation.
321
00:22:18,000 --> 00:22:24,000
And when I substitute lambda
equals negative one
322
00:22:24,000 --> 00:22:30,000
for the second equation,
what do you get?
323
00:22:30,000 --> 00:22:36,000
2 a1 plus negative 5 minus
negative 1 makes negative 4.
324
00:22:37,000 --> 00:22:43,000
There is my system that will
find me a1 and a2.
325
00:22:43,000 --> 00:22:49,000
What is the first thing you
notice about it?
326
00:22:48,000 --> 00:22:54,000
You immediately notice that
this system is fake because this
327
00:22:56,000 --> 00:23:02,000
second equation is twice the
first one.
328
00:23:03,000 --> 00:23:09,000
Something is wrong.
No, something is right.
329
00:23:06,000 --> 00:23:12,000
If that did not happen,
if the second equation were not
330
00:23:12,000 --> 00:23:18,000
a constant multiple of the first
one then the only solution of
331
00:23:17,000 --> 00:23:23,000
the system would be a1 equals
zero, a2 equals zero because the
332
00:23:23,000 --> 00:23:29,000
determinant of the coefficients
would not be zero.
333
00:23:29,000 --> 00:23:35,000
The whole function of this
exercise was to find the value
334
00:23:33,000 --> 00:23:39,000
of lambda, negative 1,
for which the system would be
335
00:23:37,000 --> 00:23:43,000
redundant and,
therefore, would have a
336
00:23:40,000 --> 00:23:46,000
nontrivial solution.
Do you get that?
337
00:23:43,000 --> 00:23:49,000
In other words,
calculate the system out,
338
00:23:47,000 --> 00:23:53,000
just as I have done here,
you have an automatic check on
339
00:23:51,000 --> 00:23:57,000
the method.
If one equation is not a
340
00:23:54,000 --> 00:24:00,000
constant multiple of the other
you made a mistake.
341
00:24:00,000 --> 00:24:06,000
You don't have the right value
of lambda or you substituted
342
00:24:05,000 --> 00:24:11,000
into the system wrong,
which is frankly a more common
343
00:24:10,000 --> 00:24:16,000
error.
Go back, recheck first the
344
00:24:13,000 --> 00:24:19,000
substitution,
and if convinced that is right
345
00:24:17,000 --> 00:24:23,000
then recheck where you got
lambda from.
346
00:24:20,000 --> 00:24:26,000
But here everything is going
fine so we can now find out what
347
00:24:26,000 --> 00:24:32,000
the value of a1 and a2 are.
You don't have to go through a
348
00:24:32,000 --> 00:24:38,000
big song and dance for this
since most of the time you will
349
00:24:36,000 --> 00:24:42,000
have two-by-two equations and
now and then three-by-three.
350
00:24:40,000 --> 00:24:46,000
For two-by-two all you do is,
since we really have the same
351
00:24:44,000 --> 00:24:50,000
equation twice,
to get a solution I can assign
352
00:24:47,000 --> 00:24:53,000
one of the variables any value
and then simply solve for the
353
00:24:51,000 --> 00:24:57,000
other.
The natural thing to do is to
354
00:24:54,000 --> 00:25:00,000
make a2 equal one,
then I won't need fractions and
355
00:24:57,000 --> 00:25:03,000
then a1 will be a2.
So the solution is (2,
356
00:25:01,000 --> 00:25:07,000
1).
I am only trying to find one
357
00:25:04,000 --> 00:25:10,000
solution.
Any constant multiple of this
358
00:25:07,000 --> 00:25:13,000
would also be a solution,
as long as it wasn't zero,
359
00:25:12,000 --> 00:25:18,000
zero which is the trivial one.
And, therefore,
360
00:25:15,000 --> 00:25:21,000
this is a solution to this
system of algebraic equations.
361
00:25:20,000 --> 00:25:26,000
And the solution to the whole
system of differential equations
362
00:25:25,000 --> 00:25:31,000
is, this is only the (a1,
a2) part.
363
00:25:30,000 --> 00:25:36,000
I have to add to it,
as a factor,
364
00:25:32,000 --> 00:25:38,000
lambda is negative,
therefore, e to the minus t.
365
00:25:36,000 --> 00:25:42,000
There is our purple thing.
366
00:25:50,000 --> 00:25:56,000
See how I got it?
Starting with the trial
367
00:25:52,000 --> 00:25:58,000
solution, I first found out
through this procedure what the
368
00:25:57,000 --> 00:26:03,000
lambda's have to be.
Then I took the lambda and
369
00:26:00,000 --> 00:26:06,000
found what the corresponding a1
and a2 that went with it and
370
00:26:04,000 --> 00:26:10,000
then made up my solution out of
that.
371
00:26:06,000 --> 00:26:12,000
Now, quickly I will do the same
thing for lambda
372
00:26:11,000 --> 00:26:17,000
equals negative 6.
Each one of these must be
373
00:26:13,000 --> 00:26:19,000
treated separately.
They are separate problems and
374
00:26:17,000 --> 00:26:23,000
you are looking for separate
solutions.
375
00:26:19,000 --> 00:26:25,000
Lambda equals negative 6.
What do I do?
376
00:26:22,000 --> 00:26:28,000
How do my equations look now?
Well, the first one is minus 2
377
00:26:25,000 --> 00:26:31,000
minus negative 6 makes plus 4.
It is 4a1 plus 2a2 equals zero.
378
00:26:32,000 --> 00:26:38,000
Then I hold my breath while I
379
00:26:37,000 --> 00:26:43,000
calculate the second one to see
if it comes out to be a constant
380
00:26:44,000 --> 00:26:50,000
multiple.
I get 2a1 plus negative 5 minus
381
00:26:49,000 --> 00:26:55,000
negative 6, which makes plus 1.
And, indeed,
382
00:26:54,000 --> 00:27:00,000
one is a constant multiple of
the other.
383
00:27:00,000 --> 00:27:06,000
I really only have on equation
there.
384
00:27:04,000 --> 00:27:10,000
I will just write down
immediately now what the
385
00:27:09,000 --> 00:27:15,000
solution is to the system.
Well, the (a1,
386
00:27:13,000 --> 00:27:19,000
a2) will be what?
Now, it is more natural to make
387
00:27:19,000 --> 00:27:25,000
a1 equal 1 and then solve to get
an integer for a2.
388
00:27:25,000 --> 00:27:31,000
If a1 is 1, then a2 is negative
2.
389
00:27:30,000 --> 00:27:36,000
And I should multiply that by e
to the negative 6t
390
00:27:34,000 --> 00:27:40,000
because negative 6 is the
corresponding value.
391
00:27:38,000 --> 00:27:44,000
There is my other one.
And now there is a
392
00:27:41,000 --> 00:27:47,000
superposition principle,
which if I get a chance will
393
00:27:44,000 --> 00:27:50,000
prove for you at the end of the
hour.
394
00:27:47,000 --> 00:27:53,000
If not, you will have to do it
yourself for homework.
395
00:27:51,000 --> 00:27:57,000
Since this is a linear system
of equations,
396
00:27:54,000 --> 00:28:00,000
once you have two separate
solutions, neither a constant
397
00:27:58,000 --> 00:28:04,000
multiple of the other,
you can multiply each one of
398
00:28:02,000 --> 00:28:08,000
these by a constant and it will
still be a solution.
399
00:28:08,000 --> 00:28:14,000
You can add them together and
that will still be a solution,
400
00:28:11,000 --> 00:28:17,000
and that gives the general
solution.
401
00:28:13,000 --> 00:28:19,000
The general solution is the sum
of these two,
402
00:28:16,000 --> 00:28:22,000
an arbitrary constant.
I am going to change the name
403
00:28:19,000 --> 00:28:25,000
since I don't want to confuse it
with the c1 I used before,
404
00:28:22,000 --> 00:28:28,000
times the first solution which
is (2, 1) e to the negative t
405
00:28:26,000 --> 00:28:32,000
plus c2, another arbitrary
constant, times 1 negative 2 e
406
00:28:29,000 --> 00:28:35,000
to the minus 6t.
407
00:28:32,000 --> 00:28:38,000
Now you notice that is exactly
408
00:28:37,000 --> 00:28:43,000
the same solution I got before.
The only difference is that I
409
00:28:43,000 --> 00:28:49,000
have renamed the arbitrary
constants.
410
00:28:47,000 --> 00:28:53,000
The relationship between them,
c1 over 2,
411
00:28:53,000 --> 00:28:59,000
I am now calling c1 tilda,
and c2 I am calling c2 tilda.
412
00:29:00,000 --> 00:29:06,000
If you have an arbitrary
constant, it doesn't matter
413
00:29:04,000 --> 00:29:10,000
whether you divide it by two.
It is still just an arbitrary a
414
00:29:09,000 --> 00:29:15,000
constant.
It covers all values,
415
00:29:12,000 --> 00:29:18,000
in other words.
Well, I think you will agree
416
00:29:16,000 --> 00:29:22,000
that is a different procedure,
yet it has only one
417
00:29:20,000 --> 00:29:26,000
coincidence.
It is like elimination goes
418
00:29:24,000 --> 00:29:30,000
this way and comes to the
answer.
419
00:29:28,000 --> 00:29:34,000
And this method goes a
completely different route and
420
00:29:31,000 --> 00:29:37,000
comes to the answer,
except it is not quite like
421
00:29:34,000 --> 00:29:40,000
that.
They walk like this and then
422
00:29:36,000 --> 00:29:42,000
they come within viewing
distance of each other to check
423
00:29:40,000 --> 00:29:46,000
that both are using the same
characteristic equation,
424
00:29:44,000 --> 00:29:50,000
and then they again go their
separate ways and end up with
425
00:29:48,000 --> 00:29:54,000
the same answer.
426
00:29:58,000 --> 00:30:04,000
There is something special of
these values.
427
00:30:00,000 --> 00:30:06,000
You cannot get away from those
two values of lambda.
428
00:30:03,000 --> 00:30:09,000
Somehow they are really
intrinsically connected.
429
00:30:06,000 --> 00:30:12,000
Occurs the exponential
coefficient, and they are
430
00:30:09,000 --> 00:30:15,000
intrinsically connected with the
problem of the egg that we
431
00:30:12,000 --> 00:30:18,000
started with.
Now what I would like to do is
432
00:30:15,000 --> 00:30:21,000
very quickly sketch how this
method looks when I remove all
433
00:30:18,000 --> 00:30:24,000
the numbers from it.
In some sense,
434
00:30:20,000 --> 00:30:26,000
it becomes a little clearer
what is going on.
435
00:30:23,000 --> 00:30:29,000
And that will give me a chance
to introduce the terminology
436
00:30:26,000 --> 00:30:32,000
that you need when you talk
about it.
437
00:30:55,000 --> 00:31:01,000
Well, you have notes.
Let me try to write it down in
438
00:31:03,000 --> 00:31:09,000
general.
439
00:31:10,000 --> 00:31:16,000
I will first write it out
two-by-two.
440
00:31:13,000 --> 00:31:19,000
I am just going to sketch.
The system looks like (x,
441
00:31:18,000 --> 00:31:24,000
y) equals, I will still put it
up in colors.
442
00:31:30,000 --> 00:31:36,000
Except now, instead of using
twos and fives,
443
00:31:33,000 --> 00:31:39,000
I will use (a,
b; c, d).
444
00:31:40,000 --> 00:31:46,000
The trial solution will look
how?
445
00:31:44,000 --> 00:31:50,000
The trial is going to be (a1,
a2).
446
00:31:48,000 --> 00:31:54,000
That I don't have to change the
name of.
447
00:31:53,000 --> 00:31:59,000
I am going to substitute in,
and what the result of
448
00:31:59,000 --> 00:32:05,000
substitution is going to be
lambda (a1, a2).
449
00:32:06,000 --> 00:32:12,000
I am going to skip a step and
pretend that the e to the lambda
450
00:32:11,000 --> 00:32:17,000
t's have already
been canceled out.
451
00:32:16,000 --> 00:32:22,000
Is equal to (a,
b; c, d) times (a1,
452
00:32:19,000 --> 00:32:25,000
a2).
What does that correspond to?
453
00:32:22,000 --> 00:32:28,000
That corresponds to the system
as I wrote it here.
454
00:32:28,000 --> 00:32:34,000
And then we wrote it out in
terms of two equations.
455
00:32:31,000 --> 00:32:37,000
And what was the resulting
thing that we ended up with?
456
00:32:35,000 --> 00:32:41,000
Well, you write it out,
you move the lambda to the
457
00:32:38,000 --> 00:32:44,000
other side.
And then the homogeneous system
458
00:32:41,000 --> 00:32:47,000
is we will look in general how?
Well, we could write it out.
459
00:32:45,000 --> 00:32:51,000
It is going to look like a
minus lambda,
460
00:32:48,000 --> 00:32:54,000
b, c, d minus lambda.
461
00:32:50,000 --> 00:32:56,000
That is just how it looks there
462
00:32:53,000 --> 00:32:59,000
and the general calculation is
the same.
463
00:32:56,000 --> 00:33:02,000
Times (a1, a2) is equal to
zero.
464
00:33:05,000 --> 00:33:11,000
This is solvable nontrivially.
In other words,
465
00:33:13,000 --> 00:33:19,000
it has a nontrivial solution if
an only if the determinant of
466
00:33:25,000 --> 00:33:31,000
coefficients is zero.
467
00:33:35,000 --> 00:33:41,000
Let's now write that out,
calculate out once and for all
468
00:33:39,000 --> 00:33:45,000
what that determinant is.
I will write it out here.
469
00:33:43,000 --> 00:33:49,000
It is a minus lambda times d
minus lambda,
470
00:33:46,000 --> 00:33:52,000
the product of the diagonal
elements, minus the
471
00:33:50,000 --> 00:33:56,000
anti-diagonal minus bc is equal
to zero.
472
00:33:55,000 --> 00:34:01,000
And let's calculate that out.
473
00:34:00,000 --> 00:34:06,000
It is lambda squared minus a
lambda minus d lambda plus ad,
474
00:34:07,000 --> 00:34:13,000
the constant term from here,
negative bc from there,
475
00:34:14,000 --> 00:34:20,000
plus ad minus bc,
476
00:34:21,000 --> 00:34:27,000
where have I seen that before?
This equation is the general
477
00:34:29,000 --> 00:34:35,000
form using letters of what we
calculated using the specific
478
00:34:36,000 --> 00:34:42,000
numbers before.
Again, I will code it the same
479
00:34:45,000 --> 00:34:51,000
way with that color salmon.
Now, most of the calculations
480
00:34:54,000 --> 00:35:00,000
will be for two-by-two systems.
I advise you,
481
00:35:00,000 --> 00:35:06,000
in the strongest possible
terms, to remember this
482
00:35:03,000 --> 00:35:09,000
equation.
You could write down this
483
00:35:06,000 --> 00:35:12,000
equation immediately for the
matrix.
484
00:35:08,000 --> 00:35:14,000
You don't have to go through
all this stuff.
485
00:35:11,000 --> 00:35:17,000
For God's sakes,
don't say let the trial
486
00:35:13,000 --> 00:35:19,000
solution be blah,
blah, blah.
487
00:35:15,000 --> 00:35:21,000
You don't want to do that.
I don't want you to repeat the
488
00:35:19,000 --> 00:35:25,000
derivation of this every time
you go through a particular
489
00:35:23,000 --> 00:35:29,000
problem.
It is just like in solving
490
00:35:25,000 --> 00:35:31,000
second order equations.
You have a second order
491
00:35:29,000 --> 00:35:35,000
equation.
You immediately write down its
492
00:35:32,000 --> 00:35:38,000
characteristic equation,
then you factor it,
493
00:35:35,000 --> 00:35:41,000
you find its roots and you
construct the solution.
494
00:35:39,000 --> 00:35:45,000
It takes a minute.
The same thing,
495
00:35:42,000 --> 00:35:48,000
this takes a minute,
too.
496
00:35:43,000 --> 00:35:49,000
What is the constant term?
Ad minus bc,
497
00:35:47,000 --> 00:35:53,000
what is that?
Matrix is (a,
498
00:35:49,000 --> 00:35:55,000
b; c, d).
Ad minus bc is its determinant.
499
00:35:52,000 --> 00:35:58,000
This is the determinant of that
matrix.
500
00:35:55,000 --> 00:36:01,000
I didn't give the matrix a
name, did I?
501
00:36:00,000 --> 00:36:06,000
I will now give the matrix a
name A.
502
00:36:03,000 --> 00:36:09,000
What is this?
Well, you are not supposed to
503
00:36:07,000 --> 00:36:13,000
know that until now.
I will tell you.
504
00:36:10,000 --> 00:36:16,000
This is called the trace of A.
Put that down in your little
505
00:36:16,000 --> 00:36:22,000
books.
The abbreviation is trace A,
506
00:36:19,000 --> 00:36:25,000
and the word is trace.
The trace of a square matrix is
507
00:36:25,000 --> 00:36:31,000
the sum of the d elements down
its main diagonal.
508
00:36:31,000 --> 00:36:37,000
If it were a three-by-three
there would be three terms in
509
00:36:35,000 --> 00:36:41,000
whatever you are up to.
Here it is a plus b,
510
00:36:40,000 --> 00:36:46,000
the sum of the diagonal
elements.
511
00:36:43,000 --> 00:36:49,000
You can immediately write down
this characteristic equation.
512
00:36:48,000 --> 00:36:54,000
Let's give it a name.
This is a characteristic
513
00:36:52,000 --> 00:36:58,000
equation of what?
Of the matrix,
514
00:36:55,000 --> 00:37:01,000
now.
Not of the system,
515
00:36:57,000 --> 00:37:03,000
of the matrix.
You have a two-by-two matrix.
516
00:37:02,000 --> 00:37:08,000
You could immediately write
down its characteristic
517
00:37:06,000 --> 00:37:12,000
equation.
Watch out for this sign,
518
00:37:08,000 --> 00:37:14,000
minus.
That is a very common error to
519
00:37:11,000 --> 00:37:17,000
leave out the minus sign because
that is the way the formula
520
00:37:15,000 --> 00:37:21,000
comes out.
Its roots.
521
00:37:17,000 --> 00:37:23,000
If it is a quadratic equation
it will have roots;
522
00:37:20,000 --> 00:37:26,000
lambda1, lambda2 for the moment
let's assume are real and
523
00:37:25,000 --> 00:37:31,000
distinct.
524
00:37:37,000 --> 00:37:43,000
For the enrichment of your
vocabulary, those are called the
525
00:37:39,000 --> 00:37:45,000
eigenvalues.
526
00:37:50,000 --> 00:37:56,000
They are something which
belonged to the matrix A.
527
00:37:53,000 --> 00:37:59,000
They are two secret numbers.
You can calculate from the
528
00:37:56,000 --> 00:38:02,000
coefficients a,
b, and c, and d,
529
00:37:58,000 --> 00:38:04,000
but they are not in the
coefficients.
530
00:38:00,000 --> 00:38:06,000
You cannot look at a matrix and
see what its eigenvalues are.
531
00:38:05,000 --> 00:38:11,000
You have to calculate
something.
532
00:38:07,000 --> 00:38:13,000
But they are the most important
numbers in the matrix.
533
00:38:10,000 --> 00:38:16,000
They are hidden,
but they are the things that
534
00:38:13,000 --> 00:38:19,000
control how this system behaves.
Those are called the
535
00:38:17,000 --> 00:38:23,000
eigenvalues.
Now, there are various purists,
536
00:38:20,000 --> 00:38:26,000
there are a fair number of them
in the world who do not like
537
00:38:24,000 --> 00:38:30,000
this word because it begins
German and ends English.
538
00:38:29,000 --> 00:38:35,000
Eigenvalues were first
introduced by a German
539
00:38:32,000 --> 00:38:38,000
mathematician,
you know, around the time
540
00:38:35,000 --> 00:38:41,000
matrices came into being in
or so.
541
00:38:38,000 --> 00:38:44,000
A little while after
eigenvalues came into being,
542
00:38:41,000 --> 00:38:47,000
too.
And since all this happened in
543
00:38:44,000 --> 00:38:50,000
Germany they were named
eigenvalues in German,
544
00:38:47,000 --> 00:38:53,000
which begins eigen and ends
value.
545
00:38:50,000 --> 00:38:56,000
But people who do not like that
call them the characteristic
546
00:38:54,000 --> 00:39:00,000
values.
Unfortunately,
547
00:38:56,000 --> 00:39:02,000
it is two words and takes a lot
more space to write out.
548
00:39:02,000 --> 00:39:08,000
An older generation even calls
them something different,
549
00:39:06,000 --> 00:39:12,000
which you are not so likely to
see nowadays,
550
00:39:10,000 --> 00:39:16,000
but you will in slightly older
books.
551
00:39:13,000 --> 00:39:19,000
You can also call them the
proper values.
552
00:39:17,000 --> 00:39:23,000
Characteristic is not a
translation of eigen,
553
00:39:21,000 --> 00:39:27,000
but proper is,
but it means it in a funny
554
00:39:25,000 --> 00:39:31,000
sense which has almost
disappeared nowadays.
555
00:39:30,000 --> 00:39:36,000
It means proper in the sense of
belong to.
556
00:39:33,000 --> 00:39:39,000
The only example I can think of
is the word property.
557
00:39:37,000 --> 00:39:43,000
Property is something that
belongs to you.
558
00:39:40,000 --> 00:39:46,000
That is the use of the word
proper.
559
00:39:43,000 --> 00:39:49,000
It is something that belongs to
the matrix.
560
00:39:46,000 --> 00:39:52,000
The matrix has its proper
values.
561
00:39:49,000 --> 00:39:55,000
It does not mean proper in the
sense of fitting and proper or I
562
00:39:54,000 --> 00:40:00,000
hope you will behave properly
when we go to Aunt Agatha's or
563
00:39:59,000 --> 00:40:05,000
something like that.
But, as I say,
564
00:40:03,000 --> 00:40:09,000
by far the most popular thing,
slowly the word eigenvalue is
565
00:40:07,000 --> 00:40:13,000
pretty much taking over the
literature.
566
00:40:11,000 --> 00:40:17,000
Just because it's just one
word, that is a tremendous
567
00:40:15,000 --> 00:40:21,000
advantage.
Okay.
568
00:40:16,000 --> 00:40:22,000
What now is still to be done?
Well, there are those vectors
569
00:40:21,000 --> 00:40:27,000
to be found.
So the very last step would be
570
00:40:24,000 --> 00:40:30,000
to solve the system to find the
vectors a1 and a2.
571
00:40:35,000 --> 00:40:41,000
For each (lambda)i,
find the associated vector.
572
00:40:40,000 --> 00:40:46,000
The vector, we will call it
(alpha)i.
573
00:40:44,000 --> 00:40:50,000
That is the a1 and a2.
Of course it's going to be
574
00:40:50,000 --> 00:40:56,000
indexed.
You have to put another
575
00:40:53,000 --> 00:40:59,000
subscript on it because there
are two of them.
576
00:41:00,000 --> 00:41:06,000
And a1 and a2 is stretched a
little too far.
577
00:41:04,000 --> 00:41:10,000
By solving the system,
and the system will be the
578
00:41:09,000 --> 00:41:15,000
system which I will write this
way, (a minus lambda,
579
00:41:15,000 --> 00:41:21,000
b, c, d minus lambda).
580
00:41:19,000 --> 00:41:25,000
It is just that system that was
581
00:41:24,000 --> 00:41:30,000
over there, but I will recopy
it, (a1, a2) equals zero,
582
00:41:30,000 --> 00:41:36,000
zero.
And these are called the
583
00:41:34,000 --> 00:41:40,000
eigenvectors.
Each of these is called the
584
00:41:39,000 --> 00:41:45,000
eigenvector associated with or
belonging to,
585
00:41:44,000 --> 00:41:50,000
again, in that sense of
property.
586
00:41:48,000 --> 00:41:54,000
Eigenvector,
let's say belonging to,
587
00:41:52,000 --> 00:41:58,000
I see that a little more
frequently, belonging to lambda
588
00:41:58,000 --> 00:42:04,000
i.
So we have the eigenvalues,
589
00:42:01,000 --> 00:42:07,000
the eigenvectors and,
of course, the people who call
590
00:42:04,000 --> 00:42:10,000
them characteristic values also
call these guys characteristic
591
00:42:08,000 --> 00:42:14,000
vectors.
I don't think I have ever seen
592
00:42:11,000 --> 00:42:17,000
proper vectors,
but that is because I am not
593
00:42:13,000 --> 00:42:19,000
old enough.
I think that is what they used
594
00:42:16,000 --> 00:42:22,000
to be called a long time ago,
but not anymore.
595
00:42:20,000 --> 00:42:26,000
And then, finally,
the general solution will be,
596
00:42:24,000 --> 00:42:30,000
by the superposition principle,
(x, y) equals the arbitrary
597
00:42:30,000 --> 00:42:36,000
constant times the first
eigenvector times the eigenvalue
598
00:42:35,000 --> 00:42:41,000
times the e to the corresponding
eigenvalue.
599
00:42:40,000 --> 00:42:46,000
And then the same thing for the
second one, (a1,
600
00:42:44,000 --> 00:42:50,000
a2), but now the second index
will be 2 to indicate that it
601
00:42:50,000 --> 00:42:56,000
goes with the eigenvalue e to
the lambda 2t.
602
00:42:56,000 --> 00:43:02,000
I have done that twice.
And now in the remaining five
603
00:43:02,000 --> 00:43:08,000
minutes I will do it a third
time because it is possible to
604
00:43:06,000 --> 00:43:12,000
write this in still a more
condensed form.
605
00:43:09,000 --> 00:43:15,000
And the advantage of the more
condensed form is A,
606
00:43:13,000 --> 00:43:19,000
it takes only that much space
to write, and B,
607
00:43:16,000 --> 00:43:22,000
it applies to systems,
not just the two-by-two
608
00:43:20,000 --> 00:43:26,000
systems, but to end-by-end
systems.
609
00:43:23,000 --> 00:43:29,000
The method is exactly the same.
Let's write it out as it would
610
00:43:27,000 --> 00:43:33,000
apply to end-by-end systems.
The vector I started with is
611
00:43:34,000 --> 00:43:40,000
(x, y) and so on,
but I will simply abbreviate
612
00:43:39,000 --> 00:43:45,000
this, as is done in 18.02,
by x with an arrow over it.
613
00:43:45,000 --> 00:43:51,000
The matrix A I will abbreviate
with A, as I did before with
614
00:43:51,000 --> 00:43:57,000
capital A.
And then the system looks like
615
00:43:56,000 --> 00:44:02,000
x prime is equal to --
616
00:44:05,000 --> 00:44:11,000
x prime is what?
Ax.
617
00:44:06,000 --> 00:44:12,000
That is all there is to it.
618
00:44:10,000 --> 00:44:16,000
There is our green system.
Now notice in this form I did
619
00:44:15,000 --> 00:44:21,000
not even tell you whether this a
two-by-two matrix or an
620
00:44:20,000 --> 00:44:26,000
end-by-end.
And in this condensed form it
621
00:44:23,000 --> 00:44:29,000
will look the same no matter how
many equations you have.
622
00:44:30,000 --> 00:44:36,000
Your book deals from the
beginning with end-by-end
623
00:44:33,000 --> 00:44:39,000
systems.
That is, in my view,
624
00:44:36,000 --> 00:44:42,000
one of its weaknesses because I
don't think most students start
625
00:44:40,000 --> 00:44:46,000
with two-by-two.
Fortunately,
626
00:44:43,000 --> 00:44:49,000
the book double-talks.
The theory is end-by-end,
627
00:44:46,000 --> 00:44:52,000
but all the examples are
two-by-two.
628
00:44:49,000 --> 00:44:55,000
So just read the examples.
Read the notes instead,
629
00:44:53,000 --> 00:44:59,000
which just do two-by-two to
start out with.
630
00:44:58,000 --> 00:45:04,000
The trial solution is x equals
what?
631
00:45:01,000 --> 00:45:07,000
An unknown vector alpha times e
to the lambda t.
632
00:45:07,000 --> 00:45:13,000
Alpha is what we called a1 and
633
00:45:11,000 --> 00:45:17,000
a2 before.
Plug this into there and cancel
634
00:45:15,000 --> 00:45:21,000
the e to the lambda t's.
635
00:45:19,000 --> 00:45:25,000
What do you get?
Well, this is lambda alpha e to
636
00:45:24,000 --> 00:45:30,000
the lambda t equals A alpha e to
the lambda t.
637
00:45:36,000 --> 00:45:42,000
These two cancel.
And the system to be solved,
638
00:45:40,000 --> 00:45:46,000
A alpha equals lambda alpha.
639
00:45:46,000 --> 00:45:52,000
And now the question is how do
you solve that system?
640
00:45:51,000 --> 00:45:57,000
Well, you can tell if a book is
written by a scoundrel or not by
641
00:45:57,000 --> 00:46:03,000
how they go --
A book, which is in my opinion
642
00:46:02,000 --> 00:46:08,000
completely scoundrel,
simply says you subtract one
643
00:46:07,000 --> 00:46:13,000
from the other,
and without further ado writes
644
00:46:12,000 --> 00:46:18,000
A minus lambda,
and they tuck a little I in
645
00:46:16,000 --> 00:46:22,000
there and write alpha equals
zero.
646
00:46:19,000 --> 00:46:25,000
Why is the I put in there?
Well, this is what you would
647
00:46:25,000 --> 00:46:31,000
like to write.
What is wrong with this
648
00:46:28,000 --> 00:46:34,000
equation?
This is not a valid matrix
649
00:46:32,000 --> 00:46:38,000
equation because that is a
square end-by-end matrix,
650
00:46:36,000 --> 00:46:42,000
a square two-by-two matrix if
you like.
651
00:46:39,000 --> 00:46:45,000
This is a scalar.
You cannot subtract the scalar
652
00:46:42,000 --> 00:46:48,000
from a matrix.
It is not an operation.
653
00:46:45,000 --> 00:46:51,000
To subtract matrices they have
to be the same size,
654
00:46:49,000 --> 00:46:55,000
the same shape.
What is done is you make this a
655
00:46:52,000 --> 00:46:58,000
two-by-two matrix.
This is a two-by-two matrix
656
00:46:56,000 --> 00:47:02,000
with lambdas down the main
diagonal and I elsewhere.
657
00:47:01,000 --> 00:47:07,000
And the justification is that
lambda alpha is the same thing
658
00:47:05,000 --> 00:47:11,000
as the lambda I times alpha
because I is an identity matrix.
659
00:47:10,000 --> 00:47:16,000
Now, in fact,
jumping from here to here is
660
00:47:13,000 --> 00:47:19,000
not something that would occur
to anybody.
661
00:47:16,000 --> 00:47:22,000
The way it should occur to you
to do this is you do this,
662
00:47:21,000 --> 00:47:27,000
you write that,
you realize it doesn't work,
663
00:47:24,000 --> 00:47:30,000
and then you say to yourself I
don't understand what these
664
00:47:29,000 --> 00:47:35,000
matrices are all about.
I think I'd better write it all
665
00:47:33,000 --> 00:47:39,000
out.
And then you would write it all
666
00:47:36,000 --> 00:47:42,000
out and you would write that
equation on the left-hand board
667
00:47:39,000 --> 00:47:45,000
there.
Oh, now I see what it should
668
00:47:41,000 --> 00:47:47,000
look like.
I should subtract lambda from
669
00:47:44,000 --> 00:47:50,000
the main diagonal.
That is the way it will come
670
00:47:47,000 --> 00:47:53,000
out.
And then say,
671
00:47:48,000 --> 00:47:54,000
hey, the way to save lambda
from the main diagonal is put it
672
00:47:51,000 --> 00:47:57,000
in an identity matrix.
That will do it for me.
673
00:47:54,000 --> 00:48:00,000
In other words,
there is a little detour that
674
00:47:57,000 --> 00:48:03,000
goes from here to here.
And one of the ways I judge
675
00:48:01,000 --> 00:48:07,000
books is by how well they
explain the passage from this to
676
00:48:05,000 --> 00:48:11,000
that.
If they don't explain it at all
677
00:48:08,000 --> 00:48:14,000
and just write it down,
they have never talked to
678
00:48:11,000 --> 00:48:17,000
students.
They have just written books.
679
00:48:14,000 --> 00:48:20,000
Where did we get finally here?
The characteristic equation
680
00:48:17,000 --> 00:48:23,000
from that, I had forgotten what
color.
681
00:48:20,000 --> 00:48:26,000
That is in salmon.
The characteristic equation,
682
00:48:23,000 --> 00:48:29,000
then, is going to be the thing
which says that the determinant
683
00:48:27,000 --> 00:48:33,000
of that is zero.
That is the circumstances under
684
00:48:32,000 --> 00:48:38,000
which it is solvable.
In general, this is the way the
685
00:48:35,000 --> 00:48:41,000
characteristic equation looks.
And its roots,
686
00:48:38,000 --> 00:48:44,000
once again, are the
eigenvalues.
687
00:48:40,000 --> 00:48:46,000
And from then you calculate the
corresponding eigenvectors.
688
00:48:44,000 --> 00:48:50,000
Okay.
Go home and practice.
689
00:48:46,000 --> 00:48:52,000
In recitation you will practice
on both two-by-two and
690
00:48:50,000 --> 00:48:56,000
three-by-three cases,
and we will talk more next
691
00:48:53,000 --> 00:48:59,000
time.