Find the number of 2×2 matrices A with every entry in \-1,0,1\ such that A^3=I, where I is the identity matrix.

By Cayley-Hamilton, $A^2=tA-dI$ with $t=\operatorname{tr}A$ and $d=\det A$. From $A^3=I$ we get $(\det A)^3=1$, so $d=1$ for integer entries. Then the eigenvalues $\lambda$ satisfy $\lambda^3=1$ and $\lambda_1\lambda_2=1$, so either both equal $1$, or they are $\omega,\bar\omega$ with $t=-1$. The case $t=2$ with $A$ diagonalisable gives only $A=I$ (a non-identity matrix with both eigenvalues $1$ is not of finite order). For $t=-1,\ d=1$ we need trace $-1$ and determinant $1$; enumerating entries from $\{-1,0,1\}$ gives exactly four matrices, e.g. $\begin{pmatrix}0&-1\\1&-1\end{pmatrix}$ and $\begin{pmatrix}-1&1\\-1&0\end{pmatrix}$. Together with $I$ the count is $5$, which a direct enumeration of all $81$ matrices confirms.