Is there an elegant way to determine $Av$ given $Au_1, Au_2$, and $Au_3$ for a $3\times3$ matrix $A$? Let A be a 3x3 matrix such that ${A} \begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 7 \\ -13 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} = \begin{pmatrix} -6 \\ 0 \\ 4 \end{pmatrix}, \quad \ {A} \begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 3 \\ -11 \end{pmatrix}.$
Find $\ {A} \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$
My original approach to this problem was to let $\ {A} $ be some 3x3 matrix with variable elements, and solve for each one, But i was just wondering if there was a more elegant way to solve it.
Thanks in advance!

Note that the given conditions can be generalized:
$$\ {A} \begin{pmatrix} 3&4&5 \\ 4&5&-9 \\ 5&6&1 \end{pmatrix}=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -13&4&-11 \end{pmatrix}.$$
Find $A$:
$$A=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -13&4&-11 \end{pmatrix}\cdot \begin{pmatrix} 3&4&5 \\ 4&5&-9 \\ 5&6&1 \end{pmatrix}^{-1}=\begin{pmatrix} 2&-6&3 \\ 7&0&3 \\ -\frac{97}{4}&-\frac{19}{2}&\frac{99}{4} \end{pmatrix}.$$
At last:
$$A\cdot \begin{pmatrix}3\\-11\\-1\end{pmatrix}=\begin{pmatrix}69\\18\\7\end{pmatrix}.$$
