Your answers are close. I suspect there are differences with the signs of your values, but you are still missing some parts of the general solution because of mistakes with the signs of values.
When you find your Eigenvalues, you should produce a set of three values, but please pay close attention to what they are:
The first eigenvalue will be 1.
The second and third eigenvalues should be -2. This implies that we are dealing with a repeated eigenvalue. In that case, we will have to create two linearly independent solutions if we only produce one eigenvector that is supposed to be associated with our repeated eigenvalue.
Why are the eigenvalues these? Let's look at the factored determinant of the matrix we set up to find our eigenvalues:
In the rawest form, we have
I suggest to check the simplifications and arithmetic involved with the determinant of the matrix created from the difference of the numerical and identity matrix to get your determinant in this form before continuing.
When simplifying, you can use the rational root theorem to figure out the factored expression of the determinant, which should be the following:
Let's label these eigenvalues separately to make things easier on our part:
Let λ3 = 3
We'll start by taking the first eigenvalue and plugging it into our determinant matrix (A-λI) to find our first eigenvector. I'm going to work in variables of k for each eigenvector we use for reference. For the first modified determinant matrix, it should look like the following:
| 1 4 3 | * |k1| = 0
| -4 -7 -3| * |k2| = 0
| 3 3 0| * |k3| = 0
When putting into reduced row echelon form, the matrix should look like the following (I included the zeros as part of this matrix):
| 1 0 -1 0 |
| 0 1 1 0 |
| 0 0 0 0 |
Let's take this matrix and put it into system of equations form, noting that from left to right, the first column is recognized as k1, and the third recognized as k3.
k1 - k3 = 0
k2 + k3 = 0
When simplifying the equations, we will let k1 = 1 to make the other numbers of the eigenvector whole. This should give us an eigenvector with the following values:
k1 = 1
k2 = -1
k3 = 1
Let's move on to the repeated eigenvalue now:
When plugging in our eigenvalue of -2 into the determinant matrix, we are left with a modified matrix with the following values:
| 4 4 3| |k1|
| -4 -4 -3| * |k2| = 0
| 3 3 3| |k3|
When putting this matrix into reduced row echelon form, we get the following (noting that the zeroes have also been included in this one too):
|1 1 0 0|
|0 0 1 0|
|0 0 0 0|
Putting this into a system of equations, we get the following:
k1 + k2 = 0
k3 = 0
We will assume k1 is equal to one for this eigenvector as well. Therefore, the other values of the eigenvector should be the following:
k1 = 1
k2 = -1
k3 = 0
NOTE: This eigenvector is the first eigenvector we have for our repeated eigenvalue. Since we cannot produce another linearly independent eigenvector from this system of equations since we do not have only one equation to work with, we will have to make another solution which will give us another eigenvector as part of that new second solution too.
In order to make our other linearly independent eigenvector, we will have to make a new matrix using our modified matrix using the eigenvector value of -2. Instead of incorporating zeroes into this matrix, we will include the values of our first eigenvector we previously created:
| 4 4 3 1|
|-4 -4 -3 -1|
| 3 3 3 0|
Putting this matrix into reduced row echelon form, we get:
|1 1 0 1|
|0 0 1 -1|
|0 0 0 0|
Putting this into a system of equations:
k1 + k2 = 1
k3 = -1
We'll assume k1 is 1 here too:
In that case:
k2 = 0
k3 = -1
We're at the final stages. Now we'll put our whole solution into the precise form we need.
The general composition of each part of the solution is we'll need a constant multiplied by the respective eigenvector, and a function in terms of e to the power of the respective eigenvalue multiplied by t:
Xn = Cn * [eigenvector] * eeigenvalue*t
where Xn is the nth part of the general solution, Cn is the nth constant of the particular part of the solution, and e is Euler's number.
For the repeated eigenvalue part of the solution with the secondary eigenvector, we need that part of the solution in the following form:
Xn = Cn * ( [first eigenvector of the repeated eigenvalue]*t* eeigenvalue*t + [second eigenvalue]* eeigenvalue*t )
Thus, the final solution should look like the following:
X = C1*[eigenvector for λ = 1]*et + C2*[first eigenvector for λ=-2]*e-2t + C3*([first eigenvector for λ=-2]*t*e-2t + [second eigenvector for λ=-2]*e-2t)
Problems like these can be drawn out and take a lot of time and patience to be careful with getting the final result. Always be sure to check each part of your solution before moving forwards. Hope this helps!