 Solution

Reduce the matrix modulo two. It is $A=1\oplus \left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, so $A\ne I$ but $(A-I{)}^2=0$. Thus the minimal polynomial is $\boxed{m=(X+1{)}^2}$. Its monic divisors are $1,X+1,(X+1{)}^2$, giving exactly $\boxed{3\text{ ideals}}$ in the quotient.

This ring has the four elements $0,I,A,I+A$. Coprimality with $X+1$ is equivalent to the representing polynomial taking value one at $1$. Thus precisely $I$ and $A$ are units: $\boxed{2\text{ invertible matrices}}$. Indeed $A^2=I$, whereas $(I+A{)}^2=0$.