Here are two ways to solve the problem:
First, if you make $P$ the matrix of eigenvectors and $D$ the diagonal matrix with the eigenvalues, we have $ A = PDP^{-1} $
You have eigenvalues, but you can pick whatever eigenvectors you want. So, let the eigenvectors be $[1,1]$ and $[3,4]$ so that $P$ is $\begin{pmatrix} 1 & 3 \\ 1 & 4 \end{pmatrix}$. Now see if $A=PDP^{-1}$ has any zero entries. If it does, try different eigenvectors.
A dirtier way can be had by realizing that the determinant is the product of the eigenvalues and the trace is the sum of the eigenvalues. Thus, you want your matrix to have determinant -12 and trace -1.
If we write $A = \begin{pmatrix} a & b \\ c & d \end{pmatrix}$, we want $ad-bc=-12$ and $a+d=-1$. So, let's take $a = 1$. Then $d=-2$ from the trace formula. Plugging these into the determinant formula yields $-2-bc=-12$, so $bc=10$. Take $b=2$ and $c=5$ for one possible solution.