We have to compute the eigenvalues
and eigenvectors of the given two by two matrix, matrix A,
and then solve the system represented
by the vector equation x prime equals A times x.
Let's begin by determining the eigenvalues.
To do this, we set up the equation,
the determinant of the difference
of A and lambda I equals 0
and then we solve for lambda.
Setting this up, again we have the determinant
of the difference of matrix A
and lambda times the two by two identity matrix equals 0.
Simplifying inside the parenthesis,
we get the determinant of the two by two matrix
with entries 1 minus lambda, 1,
negative 1, and negative lambda.
And now we evaluate the determinant
which is equal to lambda sub 1, minus negative lambda,
minus 1 times negative 1.
Simplifying and combining like terms,
we have lambda squared minus lambda plus 1 equals 0.
This is not factorable and therefore to solve for lambda
we apply the quadratic formula or complete the square.
This gives us lambda equals one half
plus or minus i square root 3 divided by 2.
So now we have our eigenvalues.
We have lambda sub one equals
one half plus i square root 3 divided by 2.
And lambda sub two equals
one half minus i square root 3 divided by 2.
The next step is determening the corresponding eigenvectors
for each eigenvalue.
Beginning with lambda sub one,
we set up the equation the difference of A and lambda i
times vector v equals a zero vector,
and then determining vector v
which will give us a corresponding eigenvector.
So again, using lambda sub one equals
one half plus i square root 3 divided by 2.
We set up our equation,
simplifying inside the parenthesis here,
the result is the two by two matrix with entries
one half minus i square root 3 divided by 2,
1 in the first row,
over the second row negative 1,
and negative one half minus i square root 3 divided by 2.
You may want to pause the video
and just verify this two by two matrix.
So again, now we have this two by two matrix
times vector v equals a zero vector.
Let's continue on the next slide.
Recall this represents a system of equations
that has an infinite number of solutions
or a dependent system
and therefore we can determine an eigenvector
using just one of the equations.
The first equation is one half minus i square root 3
times v1 plus one times v2 equals zero.
Solving this equation for v2,
we have v2 equals negative one half
plus i square root 3 divided by 2 times v1.
To find a corresponding eigenvector,
let's let v1 equal negative 2
so that v2 doesn't have any fractions.
Notice if v1 is equal to negative 2,
then v2 is equal to positive 1 minus i square root 3.
This is a corresponding eigenvector
for the eigenvalue lambda sub one.
And now I move along to lambda sub two.
Remember, lambda sub two is equal
to the complex conjugate of lambda sub one
or lambda sub one again is
one half plus i square root 3 divided by 2
and therefore lambda sub two which we already know
is one half minus i square root 3 divided by 2.
But we can also determine a corresponding eigenvector,
the vector, v2 by determining the complex conjugate
of the eigenvector v sub one.
We do this by taking the complex conjugate
of each entry of vector v1.
The complex conjugate of negative 2 is just negative 2
and the complex conjugate of 1 minus i square root 3
is 1 plus i square root 3.
Now we have a corresponding eigenvector,
vector v2, for lambda sub two.
And now for part b,
we want to solve the system x prime equals A times x.
Because we have two complex eigenvalues,
this does affect how we're going to solve the system.
Looking in our notes below,
if we have a first order linear
homogeneous constant coefficient system of ODEs
in the form of x prime equals P times x
and P is a two by two matrix with complex eigenvalues.
Step one, for each pair of complex eigenvalues
we can find the general solution using
just lambda sub one and vector v sub one.
Step two, after we determine lambda sub one and v sub one,
we write the vector equation X one of t
equals vector v1 times e to the power of
a plus ib times T.
Step three, we apply Euler's formula.
Step four, we write the result in the form of
x one of t equals x three of t plus i times x four of t.
And it works out that x three of t
and x four of t are two linearly independent solutions
to the system of equations.
And therefore the general solution is
x of t equals c one times x three of t
plus c two times x four of t.
Where x three of t is the real part of vector v1
times e to the power of lambda sub one t
and x four of t is the imaginary part of vector v1
times e to the power of lambda sub one t
So now to determine the general solution,
we only need to consider x one of t,
where again x one of t is equal to
v1 of t times e to the power of lambda sub one t.
This gives us x one of t equals the eigenvector v1
times e to the power of one half
plus i square root 3 divided by 2 times t.
Before we apply Euler's formula,
we write e to the power of one half
plus i square root 3 divided by 2 times t
as e to the power of t divided by 2
or e to the power of one half t times
e to the power of i square root 3 divided by 2 times t.
And now we apply Euler's formula on
e raised to the power of i square root 3
divided by 2 times t.
So using the first formula,
notice that e to the power of i square root 3 divided by 2t
is equal to cosine of square root 3 divided by 2t
plus i sine square root 3 divided by 2t.
And now we perform the substitution,
and now we're going to multiply
negative 2 times cosine of square root 3 divided by 2t
plus i sine square root 3 divided by 2t,
gives us the first row.
And then the complex number
one minus i square root 3 times cosine square root 3
divided by 2t plus i sine square root 3 divided by 2t
gives us the second row.
And we still have times e to the power of t divided by 2.
And now we're going to break this
two by one matrix up into two matrices
where in the first matrix we have the real parts,
and the second matrix we have the imaginary parts.
Which gives us x one of t equals this sum.
Again, we have the real part in the first matrix
and the imaginary part in the second matrix.
And now we can identify x three of t and x four of t
which we can use to form the general solution.
Going to the next slide,
we can now state the general solution
where the general solution is
x of t equals c1 times x three of t
plus c2 times x four of t.
And here we have our general solution.
I hope you found this helpful.