Multiply inverse of matricies Let ${A}$ be a $2 \times 2$ 
matrix such that ${A}$ * $\begin{pmatrix} 3 \\ -8 \end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \text{and} \quad {A} * \begin{pmatrix} 5 \\ 7 \end{pmatrix} = \begin{pmatrix} 0 \\ 1 \end{pmatrix}.$
Find
${A}^{-1} *\begin{pmatrix} -2 \\ -1 \end{pmatrix}.$
What is the easiest way to start this problem?
Thanks

First, notice that $A$ is invertible since it transforms a basis of $\Bbb R^2$ to a basis of $\Bbb R^2$. Now write
$$A^{-1 }\cdot(-2,-1)^T=-2A^{-1}\cdot(1,0)^T-A^{-1}\cdot(0,1)^T$$
and the result follows.
