We want to compute the matrix exponential
e to the power of tA,
where matrix A is the two by two matrix
with entries 1, -2, -2, 1.
And then we're asked to determine the general solution
to x prime equals A times x.
We begin by determining the eigenvalues
and corresponding eigenvectors.
If we have two linearly independent eigenvectors,
we follow the general procedure outlined below.
So again, the first step is to determine the eigenvalues
by setting up the equation of the determinant
of the difference of A and lambda I equals zero,
and then solve for lambda.
So here we have our setup.
Next we simplify.
Next we evaluate the two by two determinant,
which is equal to 1 minus lambda times 1 minus lambda
minus -2 times 2.
Simplifying and factoring,
we have a factor of lambda minus 3
and a factor of lambda plus 1,
giving us two eigenvalues of lambda sub 1 equals 3
and lambda sub 2 equals -1.
And now we determine the corresponding eigenvectors.
Recall we do this by setting up the equation,
the difference of A and lambda I times vector v
equals a zero vector,
and then solve for our vector v.
For lambda sub 1 equals 3,
here we have the setup.
Next, we simplify.
Notice in this case,
both equations are the same.
We have -2 v1 minus 2 v2 equals zero,
indicating v1 equals -v2.
So if we let v2 equal -1, notice v1 is positive 1,
and we have the eigenvector v1 equals the vector 1, -1.
And now we do the same for lambda sub 2 equals -1.
Again, here's the setup.
Next we simplify.
Again, we have two equivalent equations
where the first equation now is 2 v1 minus 2 v2 equals zero,
which indicates v1 equals v2.
For that v2 equal 1, we have the second eigenvector v2
is the vector 1, 1.
Because we have two linearly independent eigenvectors,
to find the matrix exponential e to the power of tA,
we follow the general procedure outlined here on the right,
where in step one we need to write the matrix A
in the form of E times D times E-inverse,
where matrix E is the matrix formed using the eigenvectors
as the columns and matrix D is the diagonal matrix,
where along the main diagonal we have the eigenvalues.
So for A equal E D E-inverse, here's the setup.
Again, we have matrix A on the left equals matrix E,
where the first column is the eigenvector v1
and the second column is eigenvector v2.
Then we have times matrix D, where we have the eigenvalues
along the main diagonal, all the entries are zero.
Then we have times E-inverse.
Now that we've completed step one,
let's continue on the next slide.
Once we have the equation A equals E D E-inverse,
if we take a look at our notes,
we exponentiate both sides with the base of e,
and then in our lesson, we prove
that e to the power of E D E-inverse
is equal to E times e to the D times E-inverse.
This gives us e to the A equals E times e to the D
times E-inverse.
If we include a factor of t,
we have our final formula for the matrix exponential,
e to the power of tA.
e to the power of tA
is equal to E times the matrix exponential,
e to the power of tD, times E-inverse.
So going back to our problem,
e to the power of tA is equal to matrix E
times e to the power of tD is the diagonal matrix,
where now, the entries along the main diagonal
are e to the 3t and e to the -t,
using the eigenvalues.
Then we have times E-inverse,
where E-inverse is equal to 1/2 times the two by two matrix
with entries 1, -1, 1, 1.
Now we multiply.
For the next step, I move the 1/2 to the front
and then multiply the three matrices.
The result is 1/2 times the two by two matrix,
where the entries are e to the 3t plus e to the -t,
e to the 3t plus e to the -t,
and e to the 3t plus e to the -t.
And then performing the scalar multiplication,
we have the matrix exponential
e to the power of tA here on the right.
And now we're asked to find the general solution
to x prime equals A times x.
Let's do this on the next slide.
Since the exponential is a fundamental matrix solution,
the general solution
is x equals e to the power of tA times c,
where c is a constant vector,
which indicates the general solution is x equals, again,
the matrix exponential e to the power of tA
times the constant vector, c1, c2.
While this is the general solution,
we could eliminate the denominators of 2
by letting c3 equal 2c1 and c4 equal to 2c2,
and write the general solution
in the form shown below instead.
Either form would be correct for the general solution.
I hope you found this helpful.