Prove this matrix is neither unipotent nor nilpotent. The question asks me to prove that the matrix,
$$A=\begin{bmatrix}1 & 1\\0 & 1\end{bmatrix}$$
is neither unipotent nor nilpotent. However, can't I simply row reduce this to the identity matrix:
$$A=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}$$
which shows that it clearly is unipotent since $A^k=I$ for all $k \in \Bbb Z^+$?
Is there something wrong with the question or should I treat A as the first matrix rather than reducing it (I don't see how it would then not be unipotent considering the two matrices here are equivalent)?

HINT: As Calvin Lin pointed out in the comments, $(A-I)^2=0$, so $A$ is unipotent. To show that $A$ is not nilpotent, show by induction on $n$ that
$$A^n=\begin{bmatrix}1 & n\\0 & 1\end{bmatrix}\;.$$
