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PROFESSOR: Welcome back.

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So in this session, we're
going to use the matrix

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method to solve this linear
system of differential

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equations.

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These are x dot equals 6x plus
5y, and y dot equals x plus 2y.

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So why don't you
take a few minutes

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to write down the
system in matrix form

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and go through the matrix
method to solve it.

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And I'll be right back.

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Welcome back.

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So let's write down this
system in matrix form.

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You would have a vector with
entries x and y prime equals

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a matrix with entries 6, 5; 1,
2 multiplying the column vector

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[x, y].

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So now, we did big
part of the work.

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The matrix method
tells us that we

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need to find the
eigenvalues of this matrix

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to be able to basically
diagonalize it and seek

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eigenvectors so that then we
can just read off the solutions

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and write the
solution of the system

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as a linear combination of the
eigenvectors that we found.

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So let's look for the
eigenvalues first.

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The eigenvalues
would be computed

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by seeking the determinant
of this matrix in this form:

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6 minus lambda, 5;
1, 2 minus lambda.

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We're going to have an equation
on lambda, solve for lambda,

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and the solutions will
be our eigenvalues.

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So the determinant would be
6 minus lambda multiplying 2

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minus lambda minus 5,
1 dot 5, equals to 0.

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So here, the lambda that lambda
gives us a lambda squared.

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We have minus 6*lambda
minus 2*lambda,

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which would be minus 8*lambda.

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And then, we would have 2
dot 6, which is 12, minus 5,

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which gives us 7.

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So quadratic equation
in lambda, and you

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can factorize it and find the
solutions, which is lambda_1

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equals to 1,
lambda_2 equals to 7.

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So we're done with
the first part.

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These are our eigenvalues.

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They're not repeated.

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They're just completely
different and real valued.

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So now, we're going to look
at the eigenvectors associated

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to each eigenvalue.

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So first eigenvector would be
associated with lambda_1 equals

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to 1.

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So we would be
solving this system.

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We would be solving this system
with a new matrix, 6 minus 1.

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I'm going to spell out this
one so that-- 2 minus 1.

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So this is just our lambda,
multiplying an unknown vector

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with components a_1 and
a_2, equals to zero vector.

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And basically here, the
unknowns are a_1 and a_2.

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So this is simply 5, 5; 1, and
1, [a1, a2] equals to 0, 0.

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So as you saw before,
here, basically, we

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can read off the
equation as being

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5a_1 plus 5a_2 equals to
0 and another one which

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is a_1 plus a_2 equals to 0.

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They're the same equations.

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So really, we just have
a_1 plus a_2 equals to 0.

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And so our vector
v_1 could be picked

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to just have component 1,
for example, a_1 equals to 1.

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And its second component
would just be minus 1.

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That would be one
pick for our v_1.

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We could normalize this
vector if you wanted to.

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I'm just going to keep
it like this for now.

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So if we look now for the
second eigenvector corresponding

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to the second
eigenvalue of 7, I would

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be looking for the
components of these vectors

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by doing a similar solving
for the same thing.

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And I'm going to spell it out
again so that you see where

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the terms are coming from.

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It's just 6 minus the
value of my lambda...

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[0, 0].

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So here, we have 6
minus 7, which is 1, 5.

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And then, we have 1 and 2
minus 7, which is minus 5.

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So really, what do we have
is an equation minus 1

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plus 5a_2 in both cases.

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So we can pick a
value for a_1 or a_2

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and write down a vector v_2,
in for example the form of a_1

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equals to-- let's
pick a_2 equals to 1.

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And we would have a_1
equals to 5, for example.

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Again, if you wanted an
orthonormal basis formed

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by your v_1, v_2, you would just
normalize these two vectors.

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So here, basically, we can
then rewrite the solution

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to the original system as
being linear combinations

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of-- so I'm just going to
write it in vector form.

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The first vector 1--
I'll keep it in v_1, v_2,

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that way you see it.

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And then, I'll go
into the component.

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We'd have v_1 exponential
of the value of lambda

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we found that
corresponds to v_1.

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So it would be 1 dot t.

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And then, v_2
exponential of the lambda

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value that corresponds to v_2.

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And then, basically,
we just have

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constants of integration here.

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And so the solution
to this problem

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would be linear
combination of the vectors

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by the basis of our
eigenvectors and multiplied

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by the exponentials assigned
a value of the eigenvalues

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that we found when we
looked for the eigenvalues

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of the matrix of the system.

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So here, just know that
like for the 1D problem

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that we saw before,
we're building

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a solution based on linear
combination of lucky guesses

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that we used.

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And in the one equation case,
we used a guess of e to lambda*t

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in 1D.

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Here, in this case, we
had a guess of a vector v

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and the form of
lambda*t that we used.

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And then, basically, we
just solved for the lambdas,

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and solved for the v's, and
did a linear combination

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of all the solutions.

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Like we did before in the 1D
case, we solved the lambda.

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We had different
values of lambda.

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We did a linear combinations
of the exponentials.

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So that ends this problem.

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And here, the key is just
to go through the method

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of diagonalizing your matrix.

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Basically, it's finding
the eigenvalues,

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and then computing the
eigenvectors associated

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with that, and
writing your solutions

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in terms of a linear combination
of the solution that you found.