# Solution

Suppose $M$ is a linear transformation that both operates on, and produces 2-dimensional vectors. If the following are true $$M\begin{pmatrix}2\\0\end{pmatrix} = \begin{pmatrix}1\\3\end{pmatrix} \quad \quad \textrm{and} \quad \quad M\begin{pmatrix}4\\6\end{pmatrix} = \begin{pmatrix}1\\-3\end{pmatrix}$$ Find $\displaystyle{M\begin{pmatrix}2\\5\end{pmatrix}}$

One way to proceed would be to realize that $M$ will have some matrix form $$M = \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right]$$ where

$$\left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left(\begin{array}{c} 2 \\ 0 \end{array} \right) = \left(\begin{array}{c} 1 \\ 3 \end{array} \right) \quad \textrm{ and } \quad \left[ \begin{array}{cc} a & b \\ c & d \end{array} \right] \left(\begin{array}{c} 4 \\ 6 \end{array} \right) = \left(\begin{array}{c} 1 \\ -3 \end{array} \right)$$

This leads to the system of equations given by

$$\begin{array}{rcl} 2a &=& 1\\ 2c &=& 3\\ 4a + 6b &=& 1\\ 4c + 6d &=& -3 \end{array}$$

Using the first two equations, we quickly find $a=1/2$ and $c = 3/2$.

Substituting these values into the second pair of equations, we find $b=-1/6$ and $d=-3/2$.

Thus, $$M = \left[ \begin{array}{cc} 1/2 & -1/6 \\ 3/2 & -3/2 \end{array} \right]$$ and $$M\begin{pmatrix} 2 \\ 5 \end{pmatrix} = \left[ \begin{array}{cc} 1/2 & -1/6 \\ 3/2 & -3/2 \end{array} \right] \begin{pmatrix} 2 \\ 5 \end{pmatrix} = \begin{pmatrix}1/6 \\ -9/2 \end{pmatrix}$$