WEBVTT
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LINAN CHEN: Hi.
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Welcome back to recitation.
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In the lecture, you've learned
eigenvalues and eigenvectors
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of a matrix.
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One of the many important
applications of them
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is solving a higher-order
linear differential equation
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with constant coefficients.
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A typical example is like what
I've written on the board here.
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y is a function of t,
and y and its derivatives
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satisfy this equation.
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As you can see, it involves
y, y prime, and all the way
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to its third derivative.
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So our first goal is to solve
this differential equation
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for its general solution
using the method of matrix.
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So the very first
thing that we should do
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is to find out which matrix
we should be working with.
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So after that, we also
want to say something about
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the explanation of
this matrix A*t.
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We want to find out the
first column of this matrix
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exponential.
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Why don't you hit the pause
now, and try to write down
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this matrix A by yourself.
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But before you continue,
make sure you come back
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to this video and
check with me you've
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got the correct A. I'll
see you in a while.
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OK, let's work together
to transform this problem
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into linear algebra.
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The idea is to put
y double prime,
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y prime, and y
together as a vector.
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And let me call this vector u.
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So of course, vector u
is also a function in t.
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So this is vector u.
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If this is u, what's
going to be u prime?
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OK, u prime is going to
be-- so I take derivative
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of every coordinate here that's
going to be y triple prime,
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y double prime, and y prime.
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So this is our u prime t.
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And my goal is to write
u prime as a matrix,
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call it A, times
vector u itself.
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So I want to put a matrix here.
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And I want to create this
matrix by incorporating
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this differential equation.
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If you move everything
except y triple prime
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to the right-hand
side of the equation,
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you can read y triple prime
is equal to negative 2 times
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y double prime-- so y
triple prime is negative
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2 times y double prime--
plus y prime-- that's
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plus 1 times y prime-- plus 2y.
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That's 2y, right?
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That gives you the first row.
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Then look at the
second coordinate,
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this y double prime.
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y double prime is simply itself.
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So you read y double prime is
equal to 1 y double prime, then
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0, 0.
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That's the second row.
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Well, same thing
happens to the last row.
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y prime is again itself.
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So that's 0, 1, and 0.
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That is our matrix A. Did
you get the right answer?
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So we have transformed
this equation,
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this third-order ordinary
differential equation of y
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into a first-order
differential equation of u(t).
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Although u(t) is a
vector, but if we
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can solve this
equation for u, we
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have all the information
we need for y.
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So let's plan on
solving this equation.
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In order to solve
this equation, we
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will need the eigenvalues and
eigenvectors of this matrix A.
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Again, this is a good
practice for you.
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Why don't you pause
the video again,
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and try to complete this
problem on your own.
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When you're ready, I'm going
to come back and show you
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how I did it.
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Let's finish up
everything together.
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So as we said, we need the
eigenvalues and eigenvectors
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of matrix A, and that involves
computing the determinant
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of the following matrix.
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So I want to compute the
determinant of A minus lambda
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times the identity matrix I.
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Let's write it out.
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That's the determinant
of -2 minus lambda, 1 2;
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1, negative lambda, 0;
and 0, 1, negative lambda.
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So we need the determinant of
this three by three matrix.
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Do it in your favorite way.
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You can either use the
big summation formula,
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or you can do by cofactor
along any row or any column.
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The correct answer
should be this
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is equal to 1 minus
lambda times 1 plus lambda
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times 2 plus lambda.
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And this polynomial has
three roots: 1, -1, and -2.
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These are the eigenvalues
we're looking for.
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So let me write it here.
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Lambda_1 is equal to 1.
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Lambda_2 is equal to -1.
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And lambda_3 is equal to -2.
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So now what we need is the
eigenvector corresponding
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to each eigenvalue.
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Let's take lambda_1 for example.
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The eigenvector of A
corresponding to lambda_1 is
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in the null space of the
matrix A minus lambda_1*I,
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so in this case it's A minus
I. So it's in the null space
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of this matrix.
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In other words, we are
looking for a vector,
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let's call it [a, b, c], a
column vector a, b, and c, such
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that this matrix multiplying
[a, b, c] gives me 0.
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So if you write it
out, that's going
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to be A minus I is
[-3, 1, 2; 1, -1, 0; 0, 1, -1]
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times [a; b; c] is equal to 0.
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OK.
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Could we choose
constants a, b, c
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such that this is always true?
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Well if you read the last
row, so the last dot product,
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it says that b has
to be equal to c.
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And if you read the
second row it says
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that a has to be equal to b.
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Which means a is equal
to b is equal to c.
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And if this relation is
true, the first product
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is always going to be 0.
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So that simply
means we can choose
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the first eigenvector,
the eigenvector
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corresponding to
lambda_1, to be x_1 is
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equal to [1, 1, 1] transpose.
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So we choose the
first eigenvector
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to be the column vector with
all the coordinates being 1.
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And you can do the same thing
to lambda_2 and lambda_3.
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But please allow me to
skip the computation here.
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I'm going to write out
the answer for you.
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So x_2 the eigenvector
corresponding
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to the second
eigenvalue, is going
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to be equal to 1, -1, and 1.
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And x_3 is going
to be 4, -2, and 1.
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Now we've got everything
we need in order
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to create the general
solutions for u(t)
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So we have eigenvalues, we have
the corresponding eigenvectors.
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What should be u(t)?
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The general solution for u(t)
is equal to some constant C_1
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times e to the power
lambda_1*t-- so in this case,
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e to the power t.
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Then times the first
eigenvector, x_1.
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Plus some other constant
C_2 times e to the power
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lambda_2*t--
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so e to the power
negative t-- times x_2.
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That's the second eigenvector.
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Then plus some other constant,
C_3 times e to the power
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lambda_3t-- so negative
2t-- times x_3.
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That gives you the
general solution for u.
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As we just said, if
you know what u is,
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you have all the
information you need for y.
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Just in case you're
curious about what y is,
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you can just read
the last coordinate
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of x_1, x_2, and x_3.
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And you can see that
all of them are 1.
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So y(t) is simply equal to C_1
e to the power lambda t plus C_2
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e to the power negative t plus
C_3 e to the power negative 2t.
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And the choice of C_1, C_2, and
C_3 is completely arbitrary.
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So that completes the first
part of this question.
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In the second part, we
want to say something about
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the exponential of A*t.
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So let me first give you
the recipe to cook up
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the exponential of A*t.
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The exponential of A*t is
equal to the product of three
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matrices.
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So you usually we
denote them by S times e
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to the power capital
lambda t times S inverse.
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And you may ask what S is,
and what this matrix is.
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So S is the matrix that has x_1,
x_2, and x_3 being its column
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vectors.
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So S is x_1, x_2, and x_3.
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Let me copy it down here.
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So that's 1, 1, 1;
1 -1, 1; 4, -2, 1.
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And the matrix in the middle,
e to the power lambda*t is
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a diagonal matrix.
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So e to the power lambda*t,
it's a diagonal matrix,
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and its diagonal entries
are given by e to the power
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lambda_1*t-- so that's
e to the power t--
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then e to the power
lambda_2*t-- negative t--
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and e to the power
lambda_3*t-- negative 2t.
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0 everywhere else.
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So that's e to the
power lambda*t.
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Then the exponential
of this At is
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given by the product of
these three matrices.
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It looks a bit complicated
because it involves
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the inverse of S.
But luckily, we only
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want the first
column of the result.
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So if we consider this
product, we can see:
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the product of the
first two matrices
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is relatively easy, because
this is a diagonal matrix,
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and we know that S is
given by these columns.
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So the result of the
product of these two
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is simply multiplying
the columns
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of S by these
coefficients respectively.
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So you expect to get e
to the power lambda_t
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x_1 times e to
the power-- sorry.
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The second column should be e
to the power negative t, x_2.
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The third column should be e
to the power negative 2t, x_3.
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And here, what we
should put is S inverse.
00:14:07.570 --> 00:14:10.680
But we don't need
everything from S inverse,
00:14:10.680 --> 00:14:13.770
because as we just
said, we only need
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the first column of this
result. And the first column
00:14:17.660 --> 00:14:20.050
of this product is
going to be given
00:14:20.050 --> 00:14:23.940
by linear combinations
of these columns,
00:14:23.940 --> 00:14:27.790
and the coefficients are going
to be given by the first column
00:14:27.790 --> 00:14:29.430
S inverse.
00:14:29.430 --> 00:14:32.720
So our goal should
be just to get
00:14:32.720 --> 00:14:37.320
the first column of S inverse.
00:14:37.320 --> 00:14:40.290
Then what is the first
column of S inverse?
00:14:40.290 --> 00:14:44.040
Well, the formula
for S inverse is
00:14:44.040 --> 00:14:47.510
S inverse is going to be the
reciprocal of the determinant
00:14:47.510 --> 00:14:50.770
of S, so 1 over
determinant of S,
00:14:50.770 --> 00:14:57.130
times the transpose of a
matrix C. This matrix C,
00:14:57.130 --> 00:15:02.680
the entries of this matrix C are
given by cofactors of matrix S.
00:15:02.680 --> 00:15:05.660
And then you take transpose,
you divide everything
00:15:05.660 --> 00:15:11.270
by the determinant of S. The
result will be S inverse.
00:15:11.270 --> 00:15:15.700
And we only need the first
column of this matrix.
00:15:15.700 --> 00:15:18.220
Let's try to write
the first column out.
00:15:18.220 --> 00:15:20.930
Well again, do it
in your favorite way
00:15:20.930 --> 00:15:25.830
to compute the determinant of S.
The result should be 1 over 6.
00:15:25.830 --> 00:15:28.650
So the determinant of S is 6.
00:15:28.650 --> 00:15:32.080
Then what is the first
column of C transpose?
00:15:32.080 --> 00:15:37.360
Well we can read it from here.
00:15:37.360 --> 00:15:40.600
This spot, the (1,
1) spot, should be
00:15:40.600 --> 00:15:44.560
the cofactor of this spot here.
00:15:44.560 --> 00:15:49.800
That negative 1 minus
negative 2, which is 1,
00:15:49.800 --> 00:15:51.390
so we put 1 here.
00:15:55.230 --> 00:16:03.660
Now this spot will be the
cofactor of this entry here.
00:16:03.660 --> 00:16:08.390
so that's 1 minus
negative 2, that's 3.
00:16:08.390 --> 00:16:11.940
But this is (1, 2) entry, so
you should put a negative sign
00:16:11.940 --> 00:16:12.700
in the front.
00:16:16.850 --> 00:16:22.590
Then the last spot should be
the cofactor of this entry here,
00:16:22.590 --> 00:16:26.520
which is 1 minus
negative 1, that's 2.
00:16:29.220 --> 00:16:30.360
Something else here.
00:16:34.240 --> 00:16:35.710
Two warnings.
00:16:35.710 --> 00:16:39.700
First, don't forget
this transpose sign.
00:16:39.700 --> 00:16:44.430
Second, don't forget
this negative sign.
00:16:44.430 --> 00:16:47.360
We've got the first
column of S inverse,
00:16:47.360 --> 00:16:49.320
and that's all we need.
00:16:49.320 --> 00:16:50.720
So we put it here.
00:16:50.720 --> 00:16:57.780
That's 1 over 6, -1/2, and 1/3.
00:17:02.750 --> 00:17:04.210
That's good enough for me.
00:17:04.210 --> 00:17:08.700
Now I can read out the first
column of exponential of A*t.
00:17:08.700 --> 00:17:13.050
So the first column of
the exponential of A*t,
00:17:13.050 --> 00:17:16.060
I'm going to write it here.
00:17:16.060 --> 00:17:20.690
That's going to be equal
to the linear combination
00:17:20.690 --> 00:17:21.990
of these columns.
00:17:21.990 --> 00:17:25.154
So that's 1/6 of
the first column,
00:17:25.154 --> 00:17:30.580
that's e to the power
t over 6 times x_1.
00:17:30.580 --> 00:17:37.700
Plus this times this, so that's
going to be minus 1/2 of e
00:17:37.700 --> 00:17:42.040
to the power
negative t times x_2.
00:17:42.040 --> 00:17:50.660
Then plus 1/3 of e to the
power negative 2t times x_3.
00:17:50.660 --> 00:17:53.940
That's the first column
of the exponential A*t.
00:17:53.940 --> 00:17:55.980
And then with the
other two columns.
00:18:00.750 --> 00:18:02.520
That's the answer.
00:18:02.520 --> 00:18:07.740
If you want more practice, you
can certainly complete this S
00:18:07.740 --> 00:18:11.720
inverse, and then you can
also complete the exponential
00:18:11.720 --> 00:18:13.280
of A*t.
00:18:13.280 --> 00:18:16.360
But I will leave
the rest to you.
00:18:16.360 --> 00:18:19.120
OK, I hope this
example shows you
00:18:19.120 --> 00:18:23.160
that linear algebra can be
a powerful tool in solving
00:18:23.160 --> 00:18:25.760
higher-order ordinary
differential equations
00:18:25.760 --> 00:18:27.890
with constant coefficients.
00:18:27.890 --> 00:18:32.110
And we have demonstrated the
standard procedure to do it,
00:18:32.110 --> 00:18:34.720
and we also practiced
how to calculate
00:18:34.720 --> 00:18:37.140
the exponential of a matrix.
00:18:37.140 --> 00:18:40.140
Thanks for watching,
and see you next time.