Systems of Differential Equations · Easy problem

For $A=\begin{pmatrix}2&3\\4&1\end{pmatrix}$, the vector $\mathbf v=\begin{pmatrix}3\\-4\end{pmatrix}$ is an eigenvector, so $\mathbf x(t)=e^{\lambda t}\mathbf v$ is a straight-line solution of $\mathbf x'=A\mathbf x$. Find $\lambda$.