General solutions of the system $X'=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $

First note that solutions of this linear system will approach zero if

(i) both eigenvalues are real and negative, or
(ii) the systems has two complex eigenvalues $r=\lambda\pm i \mu$, and the real part is negative ($\lambda<0$).

Lets computer the eigenvalues now:

\[(1)     det \begin{pmatrix} a-r & b \\ c & d-r \end{pmatrix} = (a-r)(d-r)-bc=r^2-(a+d)r+(ad-bc)=0.\]
Now note that if $x_1,x_2$ ? are two solutions of the quadratic equation $x^2+\alpha x+\beta$. Then $x_1+x_2=-\alpha$ and $x_1 x_2=\beta.$  Hence from (1) we see that

\[(2)   r_1+r_2=a+d <0.    \text{and}    r_1 r_2 =ad-bc>0,\]

where $r_1, r_2$? are two eigenvalues of the coefficient matrix. Now note that

(i) if $r_1$ ? and $r_2$ ? are real numbers, then by (2) the sum of $r_1$ ? and $r_2$ ? is negative and their product is positive. Therefore $r_1,r_2$ ? must be both negative.
(ii) If $r_1,r_2$? are complex valued. Then $r_1,r_2=\lambda \pm i\mu$, and hence
\[2\lambda=r_1+r_2=a+d<0 \Rightarrow \lambda<0.\]
Hence the real part of the complex eigenavlues are negative and therefore the solutions will again approach the origin as $t \rightarrow \infty$.