Is there any easy method to find the minimal polynomial of this matrix? 
Consider $$A = \begin{bmatrix} 0 &4&1&-2\\-1&4&0&-1\\0&0&1&0 \\-1&3&0&0 \end{bmatrix}$$
Find the minimal polynomial of $A$ .

Is there any  easy/tricky  method  to find minimal polynomial of this  matrix so that I can save my time in examination hall?
Any Hints/solution
Thank you!
 A: I think there is no easy method to find the minimal polynomial of this matrix, but rather the following idea might help:
To find the characteristic polynomial for this matrix is easy, since the third row has three zeros. so expanding the determinant of $A-xI$ in the third row to see, $$\rho_A(x)=(1-x).\begin{vmatrix} -x &4&-2\\-1&4-x&-1\\-1&3&-x\end{vmatrix}=-(1-x)(x-1)^2(x-2)=(x-1)^3(x-2)$$ so $\rho_A(x)=0$ if $(x-1)^3=0$ or $(x-2)=0$ and hence $$\sigma(A)=\{1,2\}$$
But minimal polynomial $m_A(x)$ and characteristic polynomial $\rho_A(x)$ have same irreducible factors, so $$m_A(x)\in \Big\{(x-1)(x-2),(x-1)^2(x-2),(x-1)^3(x-2)\Big\}$$

Note that $m_A(x)=\rho_A(x) \iff \dim(\text{each eigenspace})=1$. This is not the case in this matrix , so  $m_A(x) \neq (x-1)^3(x-2)=\rho_A(x)$ 

Now if $m_A(x)=(x-1)(x-2)$, then $(A-I)(A-2I)=0$ which means $$\text{Im}(A-2I) \subset \text{ker}(A-I)$$ so $\text{rank}(A-2I)=3 \leq \text{null}(A-I)=2$, which is false 

Hence $$m_A(x)=(x-1)^2(x-2)\;[\text{check!}]$$
