We're asked to find the general solution
to the system given by the equation
in the form of x-prime equals A times x,
or matrix A, is a three-by-three matrix.
Where in the first row, the entries are 2,-5,0.
For the second row, the entries are 0,2,0.
And for the third row, the entries are -1,4,1.
We're recall to use the eigenvalue method
to solve the system.
The first step is determine the eigenvalues of matrix A.
We do this by setting up the equation,
the determinant of the difference of A and lambda-I
equal 0, and then we solve for lambda.
So below we have the setup.
To evaluate the determinant,
it's best to use row two
because row two has two 0 entries.
The result is the square of the quantity
2 minus lambda times the quantity 1 minus lambda
equals 0, giving us eigenvalues of lambda sub-1 equals 1
and lambda sub-2 equals 2.
Notice lambda sub-2 equals 2 has multiplicity 2.
It may be helpful to write the eigenvalues
as lambda sub-1 equals 1,
and lambda sub-2 and lambda sub-3 equal 2.
The next step is to determine corresponding eigenvectors
for each eigenvalue.
Recall, we do this by setting up the equation,
the difference of A and lambda-I times vector v
equals a zero vector, then determine a vector v.
Recall because lambda sub-2 equals 2 has multiplicity 2,
if we're not able to determine two linearly independent
eigenvectors using this equation,
we'll have to follow the procedure outlined below.
Let's begin with lambda sub-1 equals 1.
We begin by setting up the equation
for lambda sub-1 equals 1 and simplifying.
Now we need to solve this system.
I'm going to go ahead and use an augmented matrix.
Here we have the augmented matrix for the system,
and now we write it in reduced row echelon form,
which I've already done here on the right.
Notice v3 is a free variable.
The first row indicates that v1 equals 0,
the second row indicates v2 equals 0,
and, again, v3 is a free variable.
So for example, if we let v3 equal 1,
we have the corresponding eigenvector,
the vector v1, as the vector 0,0,1.
This also indicates one independent solution
is x1 equals the eigenvector 0,0,1
times e to the power of 1t or just e to the t.
And now let's move along to lambda sub-2 equals 2.
Again, recall lambda sub-2 equals 2 has multiplicity 2.
So again, the first step is to set up the equation
and simplify.
And now we solve the system using an augmented matrix
as before.
So here we have the augmented matrix.
And here, we have the augmented matrix
and reduced row echelon form.
Again, notice v3 is a free variable.
Row one indicates that v1 equals negative v3.
The second row indicates that v2 equals 0.
So if we let v3 equal -1,
we have the eigenvector v2 equals the vector 1,0,-1.
And therefore, another independent solution
is x2 equals the eigenvector 1,0,-1
times e to the power of 2t.
At this point, if we try to choose a different v3,
we just get a multiple of the eigenvector v2.
Therefore, we are not going to be able
to determine two linearly independent eigenvectors
for the eigenvalue lambda sub-2
which has multiplicity 2.
And therefore, the eigenvalue lambda equals 2
has a defect of one.
To find a vector v3 and another independent solution,
we need to set up and solve the equation,
the difference of A and lambda-I times the vector v3
equals the eigenvector v2.
You may want to pause the video and review the notes below.
Let's go ahead and set up this equation on the next slide.
Again, the equation is a difference of A and lambda-I
times the vector v3 equals the eigenvector v2.
So here we have the setup.
Again, notice on the right we don't have the 0 vector,
we have the vector v2.
Simplifying, we now need to solve the system.
Again, we use an augmented matrix, which we have here.
And now we write the augmented matrix
in reduced row echelon form.
Again, notice v3 is a free variable.
From row one, we have that v1 equals -v3 plus 1/5.
From row two, we have v2 equals -1/5
And again, v3 is a free variable.
So we let v3 equal 0.
We have v1 equals 1/5 and v2 equals -1/5
giving us the vector v3
is equal to the vector 1/5,-1/5,0.
And therefore x3,
which will be a third independent solution,
is equal to the sum of the vector v3 and the eigenvector v2
times t all times e to the power of 2t.
And now we have all the information we need
to write the general solution.
We have the general solution is x of t equals c1
times the eigenvector v1,
which is the vector 0,0,1 times e of the t,
plus c2 times the eigenvector v2,
which is the vector 1,0,-1 times e to the 2t,
plus c3 times, using the eigenvector v2 and the vector v3,
we have the vector v3,
which is the vector 1/5,-1/5,0 plus eigenvector v2,
which is the vector 1,0,-1 times t,
all times e to the power 2t.
This is the general solution to the given system.
I hope you found this helpful.