Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 &0& 1 \end{pmatrix}^t.$ Let $T: \mathbb{R}^3 \to \mathbb{R}^3$ be defined by $T(v)=\begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix}v.$ Determine the minimal polynomial of $T$ with respect to $\begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$
$T(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 1 \\ -5 \\ -4 \end{pmatrix}$
$T^2(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 1 \\ -5 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -3 \\ -2 \end{pmatrix}$
$T^3(v)= \begin{pmatrix} 2 & -1 & 1 \\ -3 & 4 & -5 \\ -3 & 3 & -4 \end{pmatrix} \begin{pmatrix} 3 \\ -3 \\ -2 \end{pmatrix} = \begin{pmatrix} 7 \\ -11 \\ -10 \end{pmatrix}$
The corresponding matrix is
$A = (v,T(v),T^2(v),T^3(v))= \begin{pmatrix} 0 & 1 & 3 & 7 \\ 0 & -5 & -3 & - 11 \\ 1 & -4 & -2 & -10 \end{pmatrix}$
After row reducing this matrix, we obtain $ \begin{pmatrix} 1 & 0 & 0 & -2 \\ 0 & 1 & 0 & 1\\ 0 & 0 & 1 & 2 \end{pmatrix}$, and so $\text{null}(A) = \text{span} \left\{\begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \begin{pmatrix} 2 \\ -1 \\ 0 \\ 1 \end{pmatrix} \right\}.$ From this we obtain polynomials $f(x) = x^2$ and $g(x) = x^3-x-2$ and since $\deg f(x) < \deg g(x),$ $x^2$ is the minimal polynomial. 
The back of my book says the minimal polynomial is $x^2-3x+2.$ What am I doing wrong? I assume it is an arithmetical error, but I redid everything and got the same answer...

You got the nullspace of $A$ wrong. It is one-dimensional, since the rank of $A$ is $3$. The row-reduced form that you have indicates that the first three columns are linearly independent, and the fourth one is a certain linear combination of them. This yields
$$
\operatorname{null}(A) = \operatorname{span}( (2,-1,-2,1)^t)
$$
and the minimal polynomial of $v$ is 
$$p(x) = x^3-2x^2-x+2$$



*

*Being of the same degree as the order of the matrix, this is also its minimal polynomial, and characteristic polynomial.

*I don't know where  the book's answer comes from: it's easy to check that $T^2v-3Tv+2v\ne 0$.
