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
 A: Given

$$
\mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
= \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad
\mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \begin{pmatrix} 0 \\ 1 \end{pmatrix}
$$

IF $\mathbf{A}^{-1}$ exists, we can write

$$
\mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
= \mathbf{A}^{-1} \begin{pmatrix} 1 \\ 0 \end{pmatrix} \quad \textrm{and} \quad
\mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \mathbf{A}^{-1} \begin{pmatrix} 0 \\ 1 \end{pmatrix}
$$

Note that

$$
\mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= -2 \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 3 \\ -8 \end{pmatrix}
- \mathbf{A}^{-1} \mathbf{A} \begin{pmatrix} 5 \\ 7 \end{pmatrix}
$$

So

$$
\mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= -2 \begin{pmatrix} 3 \\ -8 \end{pmatrix}
- \begin{pmatrix} 5 \\ 7 \end{pmatrix}
= \begin{pmatrix} -11 \\ 9 \end{pmatrix}
$$

Thus

$$
\bbox[16px,border:2px solid #800000] { \mathbf{A}^{-1} \begin{pmatrix} -2 \\ -1 \end{pmatrix}
= \begin{pmatrix} -11 \\ 9 \end{pmatrix} }
$$

A: 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.
