Find the minimal polynomial for the given matrix. The first part is trivial. How to go about the second part?

 A: You can check that
$$A^2=\begin{pmatrix}
0 &c &ac\\
0&b&ab+c\\
1&a&a^2+b
\end{pmatrix}
$$
Suppose that the minimal polynomial is $m(x) = x^2+\alpha x+\beta$. Then $m(A) = 0$. Notice that 
$$(m(A))_{11} = 0+0+\beta=0\Rightarrow \beta=0$$
$$(m(A))_{21} = 0+\alpha+0=0\Rightarrow \alpha=0$$
Thus, $A^2=0$ which is clearly false.
By a simpler argument you can conclude that $m(x)$ is not a linear polynomial either. Hence, $m(x)$ must be at least of degree 3. But the characteristic polynomial is of degree 3 and we are done.
A: If $[e_1,e_2,e_3]$ is the standard basis, then it is immediate that $A\cdot e_1=e_2$ and $A^2\cdot e_1=A\cdot e_2=e_3$. Therefore $(p_0I+p_1A+p_2A^2)\cdot e_1=p_0e_1+p_1e_2+p_2e_3=(p_0,p_1,p_2)$, and no nonzero polynomial$~P$ of degree${}<3$ annihilates $A$, since $P[A]\cdot e_1\neq(0,0,0)$. One also has $A^3\cdot e_1=(c,b,a)$, and it follows that for the polynomial $M=X^3-aX^2-bX-c$ one has $M[A]\cdot e_1=(0,0,0)$. To see that $M$ is in fact the minimal polynomial (i.e., $M[A]=0$) you can choose one of the following


*

*Simply compute $M[A]$, and find it to be the zero matrix;

*Use the Cayley-Hamilton theorem to get that some monic degree$~3$ polynomial annihilates $A$, and it cannot be anything but $M$ (this also gives you the first point of the question with little computation);

*Observe that $M[A]$ commutes with $A$, so $M[A]\cdot e_{1+i}=M[A]\cdot A^i \cdot e_1 = A^i\cdot M[A]\cdot e_1=A^i\cdot 0$ is zero for $i=1,2$, and $M[A]$ kills all three standard basis vectors.


All in all it is easier to see that $M$ is the minimal polynomial than that it is the characteristic polynomial.
