No you're not. What if your matrix $A$ was a $3\times 3$ zero matrix (all elements are $0$). Then your equation would be valid but the characteristic polynomial is $\lambda^3 = 0$.
However you know that if $\lambda$ is an eigenvalue of $A$ then $\lambda=0$ or $\lambda =1$, which are the only two roots of $P$, where
$$P(\lambda) = \lambda^3 - 2\lambda^2 + \lambda = \lambda (\lambda-1)^2\,.$$
So the characteristic polynomial can have at most $0$ and $1$ as roots. So all you know is that the characteristic polynomial $Q$ must be in the form
$$Q(\lambda) = \lambda^n (\lambda-1)^m$$
with $n\in\mathbb{N}$, $m\in\mathbb{N}$ such that $m+n=N$, where $N$ is the number of rows (and so columns) of $A$. In your case $N=3$.