What is the rotation matrice which transform the point (3,-2) in (2,3)?
well...
You can go through the lengthy process of finding the angle of (3,-2) and the angle of (2,3) and subtracting but if you just look at a quick sketch
it sure looks like there's $\pi/2$ radians between those two points.
That gets you a rotation matrix of
$R=\begin{pmatrix}0 &-1 \\1 &\phantom{-}0 \end{pmatrix}$
and checking
$R \begin {pmatrix}\phantom{-}3 \\-2\end{pmatrix} = \begin{pmatrix}2 \\ 3\end{pmatrix}$
as desired.
It's not THAT lengthly a process...
$\displaystyle \begin{align*} \mathbf{a} = (3, -2) \end{align*}$ and $\displaystyle \begin{align*} \mathbf{b} = (2, 3) \end{align*}$, so $\displaystyle \begin{align*} \mathbf{a} \cdot \mathbf{b} = 3\cdot 2 + (-2) \cdot 3 = 6 - 6 = 0 \end{align*}$.
Since the dot product of the two vectors is 0, they are perpendicular.
Even if they weren't perpendicular, it's then just a case of using $\displaystyle \begin{align*} \mathbf{a} \cdot \mathbf{b} = \left| \mathbf{a} \right| \left| \mathbf{b} \right| \cos{ \left( \theta \right) } \end{align*}$, also not a massive amount of work...
Even so, I think a much easier way to approach this problem would just be to set up $\displaystyle \begin{align*} \left[ \begin{matrix} a & b \\ c & d \end{matrix} \right] \left[ \begin{matrix} \phantom{-}3 \\ -2 \end{matrix} \right] = \left[ \begin{matrix} 2 \\ 3 \end{matrix} \right] \end{align*}$, and either filling in the numbers in the transformation matrix by observation, or if you have to, multiplying it out and seeing what numbers are needed...
Finally - there is no such thing as a "matrice". The singular is "matrix", the plural is "matrices".