Find $A^5-4A^4+7A^3+11A^2-A-10I$ 
If$$A=  \begin{bmatrix}
    1 & 4   \\
    2 & 3
  \end{bmatrix},$$then find $A^5-4A^4+7A^3+11A^2-A-10I$ where $I$ is Identity matrix of $2^{nd}$ order. 
  Answer should come in terms of $A$ and $I$.

My approach: I thought it would might end up in a pattern so I found 
$$A^2=\begin{bmatrix}    9 & 16   \\    8 & 17  \end{bmatrix}$$
and similarly $A^3$ but got no such pattern. I don't want to evaluate it to up to $A^5$. Any better approach or solution is much appreciated.

The characteristic polynomial of that matrix is $\lambda^2-4\lambda-5$. So, by the Hamilton-Cayley theorem, $A^2-4A-5\operatorname{Id}=0$. In other words, $A^2=4A+5\operatorname{Id}$. Can you take it from here?
