= Solution
Reduce the <matrix> modulo two. It is $A=1\oplus\begin{pmatrix}0&1\\1&0\end{pmatrix}$, 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 $(I+A)^2=0$.
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