matrices - Find matrix from Eigenvectors and Eigenvalues

A matrix $A$ has eigenvectors
$v_1 = \left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)$
$v_2 = \left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)$

with corresponding eigenvalues $\lambda_1$= 2 and $\lambda_2$= -3, respectively.

Determine Ab for the vector b = $\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)$

I know how to find eigenvalues and eigenvectors from a given matrix A, but not this one,
the vector A is a 2x1 matrix, determinant does not exist here, so how to find the matrix A as stated in the question?

Answer

By definition of eigenvalue and eigenvector, we have
$$\tag{1}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)=2\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)\mbox{ and }A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=-3\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right).$$
Now, since
$$\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right),$$
we have
$$A\left( \begin{array}{c} 1 \\ 1 \\ \end{array} \right)=\frac{2}{3}A\left( \begin{array}{c} 2 \\ 1 \\ \end{array} \right)-\frac{1}{3}A\left( \begin{array}{c} 1 \\ -1 \\ \end{array} \right)=....\mbox{(using $(1)$)}$$