# Further explanation needed for this first order system of linear equations which is as follows:

I was trying the following problem which was as follows:

Consider the first order system of linear equations:
$\frac{dX}{dt}=AX; \space A=\begin{pmatrix} 3 &2 \\ -2&-1 \end{pmatrix}; X=\space \begin{pmatrix} x_1(t)\\ x_2(t) \end{pmatrix}$.
Then which of the following options are correct?

1. The coeffecient matrix $A$ has a repeated eigenvalue $\lambda =1.$
2. There is only one linearly independent eigenvector $X_1=\space \begin{pmatrix} 1\\ -1 \end{pmatrix}$
3. The general solution of the ODE is $(aX_1+bX_2)e^t,$ where $a,b$ are arbitrary constants and $X_1=\space \begin{pmatrix} 1\\ -1 \end{pmatrix},X_2=\space \begin{pmatrix} t\\ \frac{1}{2}-t \end{pmatrix}$

4.The vector in the option (3) given above are linearly independent.

My Attempt: Option (1) is true as the characteristic equation of $A$ is given by $(\lambda-1)^2=0$.
Option (2) also appears to be correct as $(A-I)X=0$ gives $X=c(1,-1)^t, c$ being a scalar.

Option (4) is also correct as $\begin{vmatrix} 1 & t\\ -1 & \frac{1}{2}-t \end{vmatrix} \neq 0$ and so the vectors $X_1,X_2$ are L.I.
But I am stuck on option (3) .

Can someone point me in the right direction? Thanks in advance for your time.

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I get different eigenvalues for $A$. –  copper.hat Apr 19 '13 at 14:37

In your notation you have $$aX_1e^t+bX_2e^t.$$ To prove that this expression is the general solution you need to show two facts:
• First, you need to show that both $X_1e^t$ and $X_2e^t$ solve the system