Solving a nonhomogenous system of eqns with one eigenvalue

I have the system:

$\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]= \left[\begin{array}{@{}c@{}} 3&2 \\ -2 & -1 \end{array} \right]\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]+\left[\begin{array}{@{}c@{}} 2e^{-t} \\ e^{-t} \end{array} \right]$

Which I should solve using the fundamental matrix.

So I start with obtaining the homogenous solution:

I find the eigenvalues; \begin{pmatrix} 3-\lambda&2 \\ -2 & -1-\lambda \end{pmatrix}

which gives the determinant: $\lambda^2-2\lambda+1=0$. Thus $\lambda_1=1$. Plugging that in the matrix in the original equation, I get that x=y. So a solution to the homogenous system would be: $y_h=e^{t}\left[\begin{array}{@{}c@{}} 1 \\ 1 \end{array} \right]$

Since there is no second solution to the determinant, I would ideally form the fundamental matrix:

\begin{pmatrix} e^{t} & e^0 \\ e^{t} & e^0 \end{pmatrix}

but this is to no avail. So how do I find the solution of this nonhomogenous system using the fundamental matrix with one eigenvalue?

Thanks

UPDATE:

I set up the generalized eigenvector formula \begin{equation} v_2(A-\lambda I)=v_2 \begin{pmatrix} 3-\lambda&2 \\ -2 & -1-\lambda \end{pmatrix}=v_1 \end{equation}

\begin{equation} v_2(A-\lambda I)=v_1= \begin{vmatrix} 3-\lambda&2 & | 1 \\ -2 & -1-\lambda & |-1 \end{vmatrix} \end{equation}

I now get as given by Moo, with Gaussian elimination, the matrix:

\begin{equation} \begin{vmatrix} 1 &1 & | 1/2 \\ 0 & 0 & |0 \end{vmatrix} \end{equation}

and have the second eigenvector: $e_2=e^{t}\left[\begin{array}{@{}c@{}} \frac{1}{2} \\ 0 \end{array} \right]$ .

So the homogeneous solution is:

\begin{equation} y_h=e^{\lambda_1 t}e_1+e^{\lambda_2t}e_2=e^{t}\left[\begin{array}{@{}c@{}} 1 \\ -1 \end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}} \frac{1}{2} \\ 0 \end{array} \right] \end{equation}

At this stage, it remains to find the particular solution. We know that it must be in the form of:

\begin{equation} y_p=Ce^{-t} \end{equation}

and thus the general solution is:

\begin{equation} y_p=y_h+Ce^{-t}=e^{t}\left[\begin{array}{@{}c@{}} 1 \\ -1 \end{array} \right]+e^{t}\left[\begin{array}{@{}c@{}} \frac{1}{2} \\ 0 \end{array} \right]+Ce^{-t} \end{equation}

But can this be said?

You have what is called a deficient matrix, so you need to find a generalized eigenvector.

We have the system

$$\left[\begin{array}{@{}c@{}} x' \\ y' \end{array} \right]= \left[\begin{array}{@{}c@{}} 3&2 \\ -2 & -1 \end{array} \right]\left[\begin{array}{@{}c@{}} x \\ y \end{array} \right]+\left[\begin{array}{@{}c@{}} 2e^{-t} \\ e^{-t} \end{array} \right]$$

We find a repeated eigenvalue of $\lambda_{1,2} = 1$ and we can find a single eigenvector of

$$v_1 = \begin{bmatrix} -1 \\ 1 \end{bmatrix}$$

Finding generalized eigenvectors is not a simple topic and requires work to learn the ins and outs, but in this case, we will use this example.

Solve for $v_2$ using the row-reduced-echelon-form (RREF) of $[A-\lambda I]v_2 = [A -I]v_2 = v_1 $

We get the augmented matrix

$$ \left[\begin{array}{rr|r} 1 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] $$

We can choose

$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$

Update For the eigenvalues, we find

$$|A - \lambda I| = \begin{vmatrix} -\lambda +3 & 2 \\ -2 & -\lambda -1 \\ \end{vmatrix} = (-\lambda+3)(-\lambda - 1) -2(-2) = \lambda ^2-2 \lambda +1 = 0$$

This results in

$$\lambda_{1, 2} = 1$$

To find the generalized eigenvector, we solve (you are actually using the eigenvalue $\lambda = 1$ below)

$$[A - \lambda I]v_2 = [A - 1 I]v_2 = \begin{bmatrix} 2 & 2 \\ -2 & -2 \\ \end{bmatrix}v_2 = v_1 = \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}$$

That is

$$\begin{bmatrix} 2 & 2 \\ -2 & -2 \\ \end{bmatrix}v_2 = \begin{bmatrix} -1 \\ 1 \\ \end{bmatrix}$$

As an augmented matrix, this is

$$ \left[\begin{array}{rr|r} 2 & 2 & -1 \\ -2 & -2 & 1 \end{array}\right] $$

The RREF (Gaussian Elimination) is

$$ \left[\begin{array}{rr|r} 1 & 1 & \dfrac{1}{2} \\ 0 & 0 & 0 \end{array}\right] $$

From this, we can choose

$$v_2 = \begin{bmatrix} -\dfrac{1}{2} \\ 0 \end{bmatrix}$$