<s>[INST] Find $A \left(\begin{smallmatrix} 3 \\ -11 \\ -1 \end{smallmatrix}\right).$ Let $A$ be a $3 \times 3$ 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
$$A \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}.$$

I'm sure there's a way to arrange the equations so that we get the result, but I've tried adding and subtracting but none leads to $\begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$.
 [/INST] You're thinking the right thing.  There is a way to do it systematically.  Essentially you want to find $x,y,z$ such that
$$x\begin{pmatrix} 3 \\ 4 \\ 5 \end{pmatrix}+ y\begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix} + z\begin{pmatrix} 5 \\ -9 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ -11 \\ -1 \end{pmatrix}$$
This amounts to solving this system of equations for $x,y,z$:
\begin{align*}
  3x + 4y + 5z &= 3\\
  4x + 5y - 9z &= -11\\
  5x + 6y + z &= -1
\end{align*}
Once you have $x,y,z$ you'll be able to express $\begin{pmatrix} 3 \\ -11 \\ -1\end{pmatrix}$ as a linear combination of those other $3$ vectors above.  Then use linearity of matrix-vector multiplication to get the desired result.  Specifically, you'll want to use the fact that if $A$ is a matrix, $a,b,c$ are scalars, and $\vec u, \vec v, \vec w$ are vectors, then $$A \left[a\vec u + b\vec v + c\vec w\right] = a\cdot A\vec u + b\cdot A\vec v + c\cdot A\vec w.$$
Let me know if you require further assistance.
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