A squirrel buries some acorns in several holes. Lamenting that she forgot where half the acorns were last winter, she decides to use a rule to remember the locations: if one hole has the coordinates $(x,y)$, then the next hole will have the coordinates \[(y-2x,kx + 3y).\] She chose these numbers carefully so that the rule would cycle through all the holes, i.e. if $(x,y)$ are the coordinates of the final hole, then the rule gives the coordinates of the first hole, and this works no matter where she chose to dig the first hole. So if the squirrel can remember the location of one of the holes, she can find all of them. But once winter comes, she has forgotten the value of $k$. What might it be? Show hint
We can represent the rule using a matrix: \[\begin{bmatrix}y-2x\\kx+3y\end{bmatrix} = \begin{bmatrix}-2&1\\k&3\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}\] If $(x_1,y_1)$ are the coordinates of the first hole, then the second hole will be found at \[\begin{bmatrix}x_2\\y_2\end{bmatrix} = \begin{bmatrix}-2&1\\k&3\end{bmatrix}\begin{bmatrix}x_1\\y_1\end{bmatrix}\] The third hole will be found at \begin{align*} \begin{bmatrix}x_3\\y_3\end{bmatrix} &= \begin{bmatrix}-2&1\\k&3\end{bmatrix}\begin{bmatrix}x_2\\y_2\end{bmatrix} \\ &= \begin{bmatrix}-2&1\\k&3\end{bmatrix}^2\begin{bmatrix}x_1\\y_1\end{bmatrix} \end{align*} And so on, with the $n$th hole at \[\begin{bmatrix}x_n\\y_n\end{bmatrix} =\begin{bmatrix}-2&1\\k&3\end{bmatrix}^{n-1}\begin{bmatrix}x_1\\y_1\end{bmatrix}\] Now using the fact that the last hole maps back to the first hole, if there are $n$ holes then \begin{align*} \begin{bmatrix}-2&1\\k&3\end{bmatrix} \begin{bmatrix}x_n\\y_n\end{bmatrix} &= \begin{bmatrix}x_1\\y_1\end{bmatrix}\\ \begin{bmatrix}-2&1\\k&3\end{bmatrix} \begin{bmatrix}-2&1\\k&3\end{bmatrix}^{n-1}\begin{bmatrix}x_1\\y_1\end{bmatrix} &= \begin{bmatrix}x_1\\y_1\end{bmatrix}\\ \begin{bmatrix}-2&1\\k&3\end{bmatrix}^n \begin{bmatrix}x_1\\y_1\end{bmatrix} &= \begin{bmatrix}x_1\\y_1\end{bmatrix} \end{align*} If this equation were only true for certain values of $x_1$ and $y_1$, then we’d be dealing with eigenvectors (see Section 12.4.1: Eigenvalues), but because it’s true no matter where the first hole is, we can conclude that: \begin{align*} \begin{bmatrix}-2&1\\k&3\end{bmatrix}^n &= \boldsymbol{I}\\ \det\left(\begin{bmatrix}-2&1\\k&3\end{bmatrix}^n\right) &= 1\\ \det\left(\begin{bmatrix}-2&1\\k&3\end{bmatrix}\right)^n &= 1\\ \det\left(\begin{bmatrix}-2&1\\k&3\end{bmatrix}\right) &= \pm1\\ -6-k &= \pm 1\\ k &= -7,-5 \end{align*} We actually don’t know at this stage if these values of $k$ work; just because $\det(\boldsymbol{M}^n) = 1$ doesn’t mean that $\boldsymbol{M}^n = \boldsymbol{I}$. If you test them out, you would find that $k=-7$ works with six holes, but $k=-5$ won’t work with any number of holes.