Let M=beginpmatrix2&-1\\1&0endpmatrix and let n be a positive integer.

The characteristic polynomial of $M$ is $\lambda^2-2\lambda+1=(\lambda-1)^2$, so $(M-I)^2=0$. Writing $M=I+N$ with $N^2=0$, the binomial theorem gives $M^n=I+nN=I+n(M-I)=nM-(n-1)I$: (A) is true. Hence $M^{10}=\begin{pmatrix}11&-10\\10&-9\end{pmatrix}$ with trace $2$ (indeed $\operatorname{tr}M^n=2$ for all $n$), so (B) is false. Since $\det M=1$, $M^{-1}=\begin{pmatrix}0&1\\-1&2\end{pmatrix}$, which differs from the matrix in (C) in the signs of the off-diagonal entries: false. Finally the off-diagonal entries of $M^n$ are $-n$ and $n$, which are unequal for every $n\ge1$, so $M^n$ is never symmetric: (D) false.